NCERT Class 12 Maths Integrals Exercise 7.10 Solutions

📘 Concept & Theory

Theory and Concept

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This problem illustrates the application of the properties of definite integrals to evaluate trigonometric integrals. Instead of directly applying integration formulae, we use the symmetry property of definite integrals and the fundamental trigonometric identity to simplify the given expression.

Concept 1: Symmetry Property of Definite Integrals

For a continuous function \(f(x)\) defined over the interval \([0,a]\), the following property holds:

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

Consequently, adding the two equivalent integrals gives:

\[ 2\int\limits_0^a f(x)\,dx = \int\limits_0^a [f(x)+f(a-x)]\,dx \]

This property is particularly useful when the sum \(f(x)+f(a-x)\) can be simplified using trigonometric identities.

Concept 2: Trigonometric Identity

We use the fundamental identity:

\[ \sin^2 x+\cos^2 x=1 \]

Also, the complementary-angle identity gives:

\[ \cos\left(\frac{\pi}{2}-x\right)=\sin x \]

These identities help transform the integrand into a simpler expression.

🗺️ Solution Roadmap

Step-by-step Plan

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Assume the given integral to be \(I\).
Apply the symmetry property of definite integrals by replacing \(x\) with \(\frac{\pi}{2}-x\).
Obtain an equivalent integral involving \(\sin^2 x\).
Add the two equivalent expressions for \(I\).
Use the identity \(\sin^2 x+\cos^2 x=1\) to simplify the integrand.
Evaluate the resulting elementary definite integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int\limits_0^{\frac{\pi}{2}}\cos^2 x\,dx\]
Apply the symmetry property of definite integrals.
We know that
\[\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\]
Comparing the given integral with this property, we have
\[a=\frac{\pi}{2}\]

\[f(x)=\cos^2 x\]
Therefore,
\[ \begin{aligned} I&=\int\limits_0^{\frac{\pi}{2}} \cos^2\left(\frac{\pi}{2}-x\right)\,dx\\[6pt] &=\int\limits_0^{\frac{\pi}{2}}\sin^2 x\,dx \end{aligned} \]
Here, we have used the identity:
\[\cos\left(\frac{\pi}{2}-x\right)=\sin x\]
Add the two expressions for \(I\).
From the original integral and the transformed integral, we have
\[I=\int\limits_0^{\frac{\pi}{2}}\cos^2 x\,dx\]
and
\[I=\int\limits_0^{\frac{\pi}{2}}\sin^2 x\,dx\]
Adding these equations,
\[\begin{aligned}I+I&=\int\limits_0^{\frac{\pi}{2}}\cos^2 x\,dx+\int\limits_0^{\frac{\pi}{2}}\sin^2 x\,dx\\[6pt] 2I&=\int\limits_0^{\frac{\pi}{2}}(\cos^2 x+\sin^2 x)\,dx\end{aligned}\]
Apply the fundamental trigonometric identity
Using
\[\cos^2 x+\sin^2 x=1\]
we obtain
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}1\,dx\\[6pt] &=\left[x\right]_0^{\frac{\pi}{2}}\\[6pt] &=\frac{\pi}{2}-0\\[6pt] &=\frac{\pi}{2}\end{aligned}\]
Calculate the value of \(I\).
\[\begin{aligned}2I&=\frac{\pi}{2}\\[6pt] I&=\frac{\pi}{2}\times\frac{1}{2}\\[6pt] I&=\frac{\pi}{4}\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\frac{\pi}{2}}\cos^2 x\,dx=\frac{\pi}{4}}\]

🎯 Exam Significance

Exam Significance

›

Application of properties: This problem tests the ability to apply the symmetry property of definite integrals rather than relying exclusively on direct integration.
Trigonometric identities: It reinforces the use of complementary-angle identities and the fundamental identity \(\sin^2 x+\cos^2 x=1\).
Stepwise mathematical presentation: Writing the transformed integral, adding the two expressions and simplifying systematically helps secure marks for mathematical reasoning.
Conceptual understanding: It demonstrates how symmetry can simplify an integral without requiring the evaluation of an antiderivative of the original integrand.

For JEE, CUET and Other Competitive Entrance Examinations:

Time-efficient problem-solving: Recognising symmetry can significantly reduce the number of calculation steps.
Multiple-choice questions: The result is useful in questions involving definite integrals of powers of sine and cosine.
Symmetry recognition: The technique provides a foundation for evaluating integrals of the form \(\int\limits_0^a f(x)\,dx\) when \(f(x)+f(a-x)\) has a simple expression.
Mathematical reasoning: It develops the ability to identify useful transformations and combine equivalent integral expressions.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

Symmetry property: For a continuous function over \([0,a]\), \(\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\).
Complementary-angle identity: \(\cos\left(\frac{\pi}{2}-x\right)=\sin x\).
Fundamental identity: \(\sin^2 x+\cos^2 x=1\).
Integral simplification: Adding equivalent integrals can transform a complicated-looking expression into a simple one.
Important standard result: \(\displaystyle\int\limits_0^{\frac{\pi}{2}}\sin^2 x\,dx=\int\limits_0^{\frac{\pi}{2}}\cos^2 x\,dx=\frac{\pi}{4}\).
Exam strategy: Whenever the limits are symmetric about the midpoint of an interval, check whether the substitution \(x\rightarrow a-x\) simplifies the integrand.

📘 Concept & Theory

Theory and Concept

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This problem demonstrates an elegant application of the symmetry property of definite integrals. The integrand contains both \(\sin x\) and \(\cos x\) in a symmetric manner. By transforming \(x\) into its complementary angle, \(\frac{\pi}{2}-x\), we can obtain an equivalent integral that simplifies considerably when added to the original integral.

Concept 1: Symmetry Property of Definite Integrals

For a continuous function \(f(x)\) defined over the interval \([0,a]\), we have

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

Adding the two equivalent expressions gives the useful result:

\[ 2\int\limits_0^a f(x)\,dx = \int\limits_0^a [f(x)+f(a-x)]\,dx \]

Concept 2: Complementary-Angle Identities

We use the following trigonometric identities:

\[ \sin\left(\frac{\pi}{2}-x\right)=\cos x \]

\[ \cos\left(\frac{\pi}{2}-x\right)=\sin x \]

Thus, replacing \(x\) with \(\frac{\pi}{2}-x\) interchanges the sine and cosine functions.

Concept 3: Simplification Through Addition of Symmetric Integrals

When an integrand contains two complementary trigonometric expressions, adding the original and transformed integrands may produce a constant or a much simpler function. This technique is particularly useful when the numerator of one expression becomes the complementary part of the denominator in the other.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given definite integral to be \(I\).
Apply the symmetry property by replacing \(x\) with \(\frac{\pi}{2}-x\).
Use complementary-angle identities to transform sine into cosine and cosine into sine.
Add the original and transformed expressions for \(I\).
Simplify the resulting sum of fractions by combining their numerators.
Evaluate the resulting constant integral over the given limits.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given — Assume the given integral to be \(I\)
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\]
Apply the symmetry property of definite integrals
We know that
\[\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\]
Here,
\[a=\frac{\pi}{2}\]
Therefore, replacing \(x\) with \(\frac{\pi}{2}-x\), we get
\[\begin{aligned}I&=\int\limits_0^{\frac{\pi}{2}}\frac{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}+\sqrt{\cos\left(\frac{\pi}{2}-x\right)}}\,dx\end{aligned}\]
Using the complementary-angle identities,
\[\sin\left(\frac{\pi}{2}-x\right)=\cos x\]

\[\cos\left(\frac{\pi}{2}-x\right)=\sin x\]
we obtain
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx\]
Add the original and transformed expressions
We have two expressions for the same integral:
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\]
and
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx\]
Adding these two equations,
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}\left[\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\right]dx\end{aligned}\]
Simplify the expression inside the integral
Since addition is commutative, the denominators of both fractions are identical. Therefore, taking the common denominator,
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\\[8pt] &=\int\limits_0^{\frac{\pi}{2}}1\,dx\end{aligned}\]
Evaluate the resulting definite integral
\[ \begin{aligned} 2I &=\int\limits_0^{\frac{\pi}{2}}1\,dx\\[6pt] &=\left[x\right]_0^{\frac{\pi}{2}}\\[6pt] &=\frac{\pi}{2}-0\\[6pt] &=\frac{\pi}{2} \end{aligned} \]
Calculate the value of \(I\)
\[ \begin{aligned} 2I&=\frac{\pi}{2}\\[6pt] I&=\frac{\pi}{2}\times\frac{1}{2}\\[6pt] I&=\frac{\pi}{4} \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx=\frac{\pi}{4}}\]

🎯 Exam Significance

Exam Significance

›

For CBSE Board Examinations:

Understanding definite integral properties: This problem tests the application of the symmetry property of definite integrals, an important concept in Class 12 Mathematics.
Trigonometric transformations: It reinforces the use of complementary-angle identities to interchange sine and cosine functions.
Logical presentation: The solution demonstrates how to obtain an equivalent integral, add two expressions and simplify the result systematically.
Stepwise marking: Showing the substitution, transformed integrand, addition of integrals and evaluation of limits makes the mathematical reasoning clear.

For JEE, CUET and Other Competitive Entrance Examinations:

Symmetry-based shortcuts: Recognising complementary functions can reduce a seemingly complicated integral to a simple constant integral.
Multiple-choice questions: This technique is useful for quickly evaluating definite integrals involving ratios of trigonometric functions.
Pattern recognition: The problem develops the ability to identify expressions of the form \(f(x)+f(a-x)\).
Conceptual efficiency: It illustrates why choosing an appropriate transformation is often more effective than attempting direct integration.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Symmetry property: \(\displaystyle\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\).
Complementary-angle identities: \(\displaystyle\sin\left(\frac{\pi}{2}-x\right)=\cos x\) and \(\displaystyle\cos\left(\frac{\pi}{2}-x\right)=\sin x\).
Equivalent integrals: Replacing \(x\) by \(\frac{\pi}{2}-x\) in the given integral produces another expression with the same value.
Adding symmetric expressions: The two fractions simplify to unity because their numerators add up to their common denominator.
Important standard result: For this symmetric integrand, the value of the integral is \(\frac{\pi}{4}\).
Exam strategy: Whenever an integral over \([0,\frac{\pi}{2}]\) contains both sine and cosine, consider the substitution \(x\rightarrow\frac{\pi}{2}-x\) before attempting direct integration.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the application of the symmetry property of definite integrals to evaluate an integral involving fractional powers of trigonometric functions. The key observation is that replacing \(x\) with its complementary angle, \(\frac{\pi}{2}-x\), interchanges sine and cosine without changing the limits of integration.

Concept 1: Symmetry Property of Definite Integrals

For a continuous function \(f(x)\) defined over the interval \([0,a]\), the following property holds:

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

Adding the two equivalent expressions gives:

\[ 2\int\limits_0^a f(x)\,dx = \int\limits_0^a [f(x)+f(a-x)]\,dx \]

This result is especially useful when the sum of the original and transformed integrands simplifies to a constant.

Concept 2: Complementary-Angle Identities

We use the identities:

\[ \sin\left(\frac{\pi}{2}-x\right)=\cos x \]

\[ \cos\left(\frac{\pi}{2}-x\right)=\sin x \]

Consequently, the transformation \(x\rightarrow\frac{\pi}{2}-x\) interchanges the sine and cosine terms.

Concept 3: Properties of Fractional Powers

For non-negative real numbers \(a\) and \(b\), we have

\[ \frac{a^{\frac32}}{a^{\frac32}+b^{\frac32}} + \frac{b^{\frac32}}{a^{\frac32}+b^{\frac32}} =1 \]

This algebraic observation is central to the evaluation of the given integral.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Apply the symmetry property of definite integrals by replacing \(x\) with \(\frac{\pi}{2}-x\).
Use complementary-angle identities to obtain an equivalent integral in which sine and cosine are interchanged.
Add the original and transformed expressions for \(I\).
Combine the fractions and simplify the numerator with the common denominator.
Evaluate the resulting constant integral and divide by two to obtain \(I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given — Assume the given integral to be \(I\)
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x}{\sin^{\frac32}x+\cos^{\frac32}x}\,dx\]
Apply the symmetry property of definite integrals
We know that
\[\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\]
Here, \(a=\frac{\pi}{2}\). Therefore, replacing \(x\) with \(\frac{\pi}{2}-x\), we get
\[\begin{aligned}I&=\int\limits_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}\left(\frac{\pi}{2}-x\right)}{\sin^{\frac32}\left(\frac{\pi}{2}-x\right)+\cos^{\frac32}\left(\frac{\pi}{2}-x\right)}\,dx\end{aligned}\]
Using the complementary-angle identities,
\[\sin\left(\frac{\pi}{2}-x\right)=\cos x\]

\[\cos\left(\frac{\pi}{2}-x\right)=\sin x\]
we obtain
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\cos^{\frac32}x}{\cos^{\frac32}x+\sin^{\frac32}x}\,dx\]
Add the original and transformed expressions
We have
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x}{\sin^{\frac32}x+\cos^{\frac32}x}\,dx\]
and
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\cos^{\frac32}x}{\cos^{\frac32}x+\sin^{\frac32}x}\,dx\]
Adding these two equations,
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}\left[\frac{\sin^{\frac32}x}{\sin^{\frac32}x+\cos^{\frac32}x}+\frac{\cos^{\frac32}x}{\cos^{\frac32}x+\sin^{\frac32}x}\right]dx\end{aligned}\]
Simplify the expression inside the integral
Since the denominators are identical, taking the common denominator, we get
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x+\cos^{\frac32}x}{\sin^{\frac32}x+\cos^{\frac32}x}\,dx\\[8pt]&=\int\limits_0^{\frac{\pi}{2}}1\,dx\end{aligned}\]
Evaluate the resulting definite integral
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}1\,dx\\[6pt] &=\left[x\right]_0^{\frac{\pi}{2}}\\[6pt] &=\frac{\pi}{2}-0\\[6pt] &=\frac{\pi}{2}\end{aligned}\]
Calculate the value of \(I\)
\[\begin{aligned}2I&=\frac{\pi}{2}\\[6pt] I&=\frac{\pi}{2}\times\frac12\\[6pt] I&=\frac{\pi}{4}\end{aligned}\]

💡 Answer

Final Answer

›

\[ \boxed{ \displaystyle \int\limits_0^{\frac{\pi}{2}} \frac{\sin^{\frac32}x} {\sin^{\frac32}x+\cos^{\frac32}x}\,dx =\frac{\pi}{4} } \]

🎯 Exam Significance

Exam Significance

›

Application of definite integral properties: The problem tests the ability to recognise and apply the symmetry property of definite integrals.
Trigonometric identities: It reinforces the use of complementary-angle identities to transform an integrand.
Algebraic simplification: Students learn to combine fractions having identical denominators and simplify the resulting expression.
Stepwise presentation: The solution demonstrates the importance of showing the transformation, addition of integrals and evaluation of limits clearly.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Symmetry-based problem solving: This technique provides a quick method for evaluating integrals involving complementary trigonometric functions.
Fractional powers: It develops familiarity with integrands involving fractional exponents, which frequently appear in advanced mathematical problems.
Recognition of symmetric structures: Students learn to identify expressions in which the numerator of one fraction becomes the numerator of the transformed fraction.
Time-efficient evaluation: A seemingly complicated integral is reduced to the integration of a constant, avoiding unnecessary substitution or integration by parts.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Symmetry property: \(\displaystyle\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\).
Complementary-angle transformation: Replacing \(x\) with \(\frac{\pi}{2}-x\) interchanges sine and cosine.
Fractional powers: The same symmetry technique works for fractional exponents such as \(\frac32\), provided the expressions are defined.
Adding equivalent integrals: The sum of the two fractions becomes unity because their numerators add up to their common denominator.
Important standard result: \(\displaystyle\int\limits_0^{\frac{\pi}{2}}\frac{\sin^n x}{\sin^n x+\cos^n x}\,dx=\frac{\pi}{4}\), for \(n>0\).
Exam strategy: Whenever an integral over \([0,\frac{\pi}{2}]\) contains sine and cosine in a symmetric arrangement, try the transformation \(x\rightarrow\frac{\pi}{2}-x\) before attempting direct integration.

📘 Concept & Theory

Theory and Concept

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This problem illustrates the application of the symmetry property of definite integrals to evaluate an integral involving odd powers of sine and cosine. The denominator contains both \(\sin^5 x\) and \(\cos^5 x\), while the numerator contains only \(\cos^5 x\). By replacing \(x\) with its complementary angle, \(\frac{\pi}{2}-x\), we obtain an equivalent integral in which sine and cosine are interchanged.

Concept 1: Symmetry Property of Definite Integrals

For a continuous function \(f(x)\) defined over the interval \([0,a]\), we have

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

Adding the original and transformed expressions gives

\[ 2\int\limits_0^a f(x)\,dx = \int\limits_0^a [f(x)+f(a-x)]\,dx \]

This property is particularly useful when the sum of the two integrands simplifies to a constant.

Concept 2: Complementary-Angle Identities

The following identities are used to transform the integrand:

\[ \sin\left(\frac{\pi}{2}-x\right)=\cos x \]

\[ \cos\left(\frac{\pi}{2}-x\right)=\sin x \]

Thus, the substitution \(x\rightarrow\frac{\pi}{2}-x\) interchanges sine and cosine while preserving the integration limits.

Concept 3: Addition of Symmetric Fractions

For positive quantities \(a\) and \(b\), observe that

\[ \frac{a^5}{a^5+b^5} + \frac{b^5}{a^5+b^5} =1 \]

This algebraic identity is the key to simplifying the given integral after applying the symmetry property.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Apply the symmetry property by replacing \(x\) with \(\frac{\pi}{2}-x\).
Use complementary-angle identities to interchange sine and cosine.
Add the original and transformed expressions for \(I\).
Combine the fractions using their common denominator.
Use the resulting constant integrand to evaluate the definite integral.
Divide the result by two to obtain the required value of \(I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given — Assume the given integral to be \(I\)
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\cos^5 x}{\sin^5 x+\cos^5 x}\,dx\]
Apply the symmetry property of definite integrals
We know that
\[\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\]
Here, \(a=\frac{\pi}{2}\). Therefore, replacing \(x\) with \(\frac{\pi}{2}-x\), we obtain
\[\begin{aligned}I&=\int\limits_0^{\frac{\pi}{2}}\frac{\cos^5\left(\frac{\pi}{2}-x\right)}{\sin^5\left(\frac{\pi}{2}-x\right)+\cos^5\left(\frac{\pi}{2}-x\right)}\,dx\end{aligned}\]
Using the complementary-angle identities,
\[\cos\left(\frac{\pi}{2}-x\right)=\sin x\]

\[\sin\left(\frac{\pi}{2}-x\right)=\cos x\]
we get
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sin^5 x}{\cos^5 x+\sin^5 x}\,dx\]
Add the original and transformed expressions.
We have the original integral:
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\cos^5 x}{\sin^5 x+\cos^5 x}\,dx\]
And the transformed integral:
\[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sin^5 x}{\cos^5 x+\sin^5 x}\,dx\]
Adding these two equations,
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}\left[\frac{\cos^5 x}{\sin^5 x+\cos^5 x}+\frac{\sin^5 x}{\cos^5 x+\sin^5 x}\right]dx\end{aligned}\]
Simplify the expression inside the integral.
Since the denominators are identical, taking the common denominator,
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}\frac{\cos^5 x+\sin^5 x}{\sin^5 x+\cos^5 x}\,dx\\[8pt] &=\int\limits_0^{\frac{\pi}{2}}1\,dx\end{aligned}\]
Evaluate the resulting definite integral
\[\begin{aligned}2I&=\int\limits_0^{\frac{\pi}{2}}1\,dx\\[6pt] &=\left[x\right]_0^{\frac{\pi}{2}}\\[6pt] &=\frac{\pi}{2}-0\\[6pt] &=\frac{\pi}{2}\end{aligned}\]
Calculate the value of \(I\).
\[ \begin{aligned} 2I&=\frac{\pi}{2}\\[6pt] I&=\frac{\pi}{2}\times\frac12\\[6pt] I&=\frac{\pi}{4} \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\frac{\pi}{2}}\frac{\cos^5 x}{\sin^5 x+\cos^5 x}\,dx=\frac{\pi}{4}}\]

🎯 Exam Significance

Exam Significance

›

Application of definite integral properties: This problem tests the understanding and application of the symmetry property of definite integrals.
Complementary-angle transformations: It reinforces the use of trigonometric identities to transform an integral without changing its value.
Algebraic simplification: Students learn to recognise how two complementary fractions can combine to produce unity.
Mathematical presentation: Writing the original integral, transformed integral, their sum and the evaluation of limits systematically helps demonstrate complete mathematical reasoning.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Symmetry-based shortcuts: The technique is useful for rapidly evaluating integrals containing complementary trigonometric functions.
Pattern recognition: Students learn to identify expressions of the form \(\frac{a^n}{a^n+b^n}\) and their complementary counterparts.
Higher-power expressions: The method works for positive powers beyond five, making it applicable to a broader class of definite integral problems.
Time-efficient problem solving: Recognising symmetry eliminates the need for lengthy integration techniques such as substitution or integration by parts.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Symmetry property: \(\displaystyle\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\).
Complementary-angle identities: \(\sin\left(\frac{\pi}{2}-x\right)=\cos x\) and \(\cos\left(\frac{\pi}{2}-x\right)=\sin x\).
Interchange of trigonometric functions: Replacing \(x\) with \(\frac{\pi}{2}-x\) changes the numerator from \(\cos^5 x\) to \(\sin^5 x\), while the denominator remains unchanged in structure.
Complementary fractions: \(\displaystyle\frac{\cos^5 x}{\sin^5 x+\cos^5 x}+\frac{\sin^5 x}{\sin^5 x+\cos^5 x}=1\).
Important standard result: For \(n>0\), \(\displaystyle\int\limits_0^{\frac{\pi}{2}}\frac{\cos^n x}{\sin^n x+\cos^n x}\,dx=\frac{\pi}{4}\).
Exam strategy: Whenever the limits are \(0\) and \(\frac{\pi}{2}\), and the integrand contains sine and cosine symmetrically, consider applying the transformation \(x\rightarrow\frac{\pi}{2}-x\).

📘 Concept & Theory

Theory and Concept

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This problem illustrates the application of properties of definite integrals to evaluate an integral involving an absolute value function. The important observation is that the expression inside the modulus, \(x+2\), changes sign at \(x=-2\). Therefore, the integrand must be considered separately over the intervals \([-5,-2]\) and \([-2,5]\).

Concept 1: Definition of Absolute Value

For any real number \(u\), the absolute value is defined as

\[ |u|= \begin{cases} u, & u\geq 0\\ -u, & u<0 \end{cases} \]

Applying this definition to \(u=x+2\), we obtain

\[ |x+2|= \begin{cases} -(x+2)=-x-2, & x<-2\\ x+2, & x\geq -2 \end{cases} \]

Thus, the point \(x=-2\) divides the given interval into two subintervals on which the integrand has different algebraic expressions.

Concept 2: Splitting Property of Definite Integrals

If \(a < c < b\), then

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^c f(x)\,dx+ \int\limits_c^b f(x)\,dx \]

This property allows us to split the integral at the point where the expression inside the modulus changes sign.

Concept 3: Geometrical Interpretation of a Definite Integral

The definite integral of a non-negative function represents the area enclosed between its graph and the \(x\)-axis over the specified interval.

In this problem, \(y=|x+2|\) represents a V-shaped graph with vertex at \((-2,0)\). The required integral is the sum of the areas of two right-angled triangles.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the point at which the expression inside the modulus becomes zero.
Determine the sign of \(x+2\) on either side of this point.
Split the given integral at \(x=-2\).
Remove the modulus by using its piecewise definition on each subinterval.
Evaluate the two resulting integrals separately.
Add their values to obtain the required definite integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given — Assume the given integral to be \(I\).
\[I=\int\limits_{-5}^{5}|x+2|\,dx\]
Find the point at which the expression inside the modulus changes sign
Putting the expression inside the modulus equal to zero,
\[ \begin{aligned} x+2&=0\\ x&=-2 \end{aligned} \]
For \(x<-2\), we have \(x+2<0\). Therefore,
\[|x+2|=-(x+2)=-x-2\]
For \(x\geq -2\), we have \(x+2\geq 0\). Therefore,
\[|x+2|=x+2\]
Hence,
\[ |x+2|= \begin{cases} -x-2, & -5\leq x<-2\\ x+2, & -2\leq x\leq 5 \end{cases} \]
Split the integral at \(x=-2\)
Using the splitting property of definite integrals,
\[ \begin{aligned} I &=\int\limits_{-5}^{-2}|x+2|\,dx +\int\limits_{-2}^{5}|x+2|\,dx \end{aligned} \]
Substituting the appropriate expressions for the absolute value in each interval,
\[I=\int\limits_{-5}^{-2}(-x-2)\,dx+\int\limits_{-2}^{5}(x+2)\,dx\]
Evaluate the first integral
\[ \begin{aligned} I_2 &=\int\limits_{-2}^{5}(x+2)\,dx\\[6pt] &=\left[\frac{x^2}{2}+2x\right]_{-2}^{5}\\[6pt] &=\left[\frac{5^2}{2}+2(5)\right] -\left[\frac{(-2)^2}{2}+2(-2)\right]\\[6pt] &=\left[\frac{25}{2}+10\right] -\left[\frac{4}{2}-4\right]\\[6pt] &=\left[\frac{25}{2}+\frac{20}{2}\right] -[2-4]\\[6pt] &=\frac{45}{2}-(-2)\\[6pt] &=\frac{45}{2}+\frac{4}{2}\\[6pt] &=\frac{49}{2} \end{aligned} \]
Add the values of the two integrals
\[ \begin{aligned} I&=I_1+I_2\\[6pt] &=\frac{9}{2}+\frac{49}{2}\\[6pt] &=\frac{58}{2}\\[6pt] &=29 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_{-5}^{5}|x+2|\,dx=29}\]

🎯 Exam Significance

Exam Significance

›

Properties of definite integrals: This problem tests the splitting property of definite integrals and the correct treatment of an absolute value function.
Piecewise functions: It develops the ability to express a modulus function in separate intervals according to the sign of its argument.
Integration and limits: Students practise evaluating polynomial antiderivatives at positive and negative limits without sign errors.
Geometrical interpretation: The problem connects definite integration with the area under a curve, strengthening conceptual understanding.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Modulus-based integrals: Questions involving \(|f(x)|\) frequently require identifying the zeros of \(f(x)\) and splitting the integration interval accordingly.
Graphical reasoning: Recognising the V-shaped graph of an absolute value function can provide a faster geometrical evaluation.
Sign analysis: Correctly determining the intervals on which an expression is positive or negative is an essential problem-solving skill.
Accuracy in calculations: The problem reinforces careful handling of negative limits and algebraic signs.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Modulus definition: \(\displaystyle |x+2|=-x-2\) for \(x<-2\), and \(\displaystyle |x+2|=x+2\) for \(x\geq-2\).
Critical point: The expression inside the modulus becomes zero at \(x=-2\).
Splitting property: An integral involving an absolute value function should be split at the points where its argument changes sign.
Piecewise integration: Each subinterval must be evaluated using the correct algebraic expression for the modulus.
Geometrical interpretation: The integral represents the sum of the areas of two triangles under the graph \(y=|x+2|\).
Verification: Both analytical integration and geometrical area calculation give the same result, \(29\).
Exam strategy: For integrals involving \(|f(x)|\), first solve \(f(x)=0\), identify the sign intervals, and then split the integral at the relevant points.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the application of the splitting property of definite integrals to evaluate an integral involving an absolute value function. The key observation is that the expression inside the modulus, \(x-5\), changes sign at \(x=5\), which lies within the given interval \([2,8]\).

Concept 1: Definition of Absolute Value

For any real number \(u\), the absolute value is defined as

\[ |u|= \begin{cases} u, & u\geq 0\\ -u, & u < 0 \end{cases} \]

Applying this definition to \(u=x-5\), we obtain

\[ |x-5|= \begin{cases} 5-x, & x < 5\\ x-5, & x\geq 5 \end{cases} \]

Thus, the integrand has different algebraic expressions on the intervals \([2,5]\) and \([5,8]\).

Concept 2: Splitting Property of Definite Integrals

If \(a < c < b\), then

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^c f(x)\,dx+ \int\limits_c^b f(x)\,dx \]

This property allows us to split the integral at the point where the expression inside the modulus changes sign.

Concept 3: Symmetry Property of Definite Integrals

For a continuous function \(f(x)\), we also have

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

In this problem, the interval \([2,8]\) is symmetric about \(x=5\). The function \(|x-5|\) is also symmetric about this point. Therefore, the two triangular areas on either side of \(x=5\) are equal.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Find the point at which the expression inside the modulus becomes zero.
Determine the sign of \(x-5\) on either side of this point.
Split the integral at \(x=5\) using the splitting property.
Remove the modulus by using its appropriate piecewise expressions.
Evaluate both integrals separately and add their values.
Verify the result using the geometrical interpretation of the definite integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given — Assume the given integral to be \(I\)
\[I=\int\limits_2^8 |x-5|\,dx\]
Find the point at which the expression inside the modulus changes sign
Putting the expression inside the modulus equal to zero,
\[ \begin{aligned} x-5&=0\\ x&=5 \end{aligned} \]
For \(2\leq x<5\), we have \(x-5<0\). Therefore,
\[|x-5|=-(x-5)=5-x\]
For \(5\leq x\leq 8\), we have \(x-5\geq 0\). Therefore,
\[|x-5|=x-5\]
Hence,
\[ |x-5|= \begin{cases} 5-x, & 2\leq x<5\\ x-5, & 5\leq x\leq 8 \end{cases} \]
Split the integral at \(x=5\)
Using the splitting property of definite integrals,
\[ \begin{aligned} I &=\int\limits_2^5 |x-5|\,dx +\int\limits_5^8 |x-5|\,dx \end{aligned} \]
Substituting the appropriate expressions for the modulus in each interval,
\[I=\int\limits_2^5(5-x)\,dx+\int\limits_5^8(x-5)\,dx\]
Evaluate the first integral
\[ \begin{aligned} I_1 &=\int\limits_2^5(5-x)\,dx\\[6pt] &=\left[5x-\frac{x^2}{2}\right]_2^5\\[6pt] &=\left[5(5)-\frac{5^2}{2}\right] -\left[5(2)-\frac{2^2}{2}\right]\\[6pt] &=\left[25-\frac{25}{2}\right] -\left[10-\frac{4}{2}\right]\\[6pt] &=\left[\frac{50-25}{2}\right] -[10-2]\\[6pt] &=\frac{25}{2}-8\\[6pt] &=\frac{25}{2}-\frac{16}{2}\\[6pt] &=\frac{9}{2} \end{aligned} \]
Evaluate the second integral
\[ \begin{aligned} I_2 &=\int\limits_5^8(x-5)\,dx\\[6pt] &=\left[\frac{x^2}{2}-5x\right]_5^8\\[6pt] &=\left[\frac{8^2}{2}-5(8)\right] -\left[\frac{5^2}{2}-5(5)\right]\\[6pt] &=\left[\frac{64}{2}-40\right] -\left[\frac{25}{2}-25\right]\\[6pt] &=[32-40] -\left[\frac{25-50}{2}\right]\\[6pt] &=-8-\left(-\frac{25}{2}\right)\\[6pt] &=-8+\frac{25}{2}\\[6pt] &=\frac{-16+25}{2}\\[6pt] &=\frac{9}{2} \end{aligned} \]
Add the values of the two integrals
\[ \begin{aligned} I&=I_1+I_2\\[6pt] &=\frac{9}{2}+\frac{9}{2}\\[6pt] &=\frac{18}{2}\\[6pt] &=9 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_2^8 |x-5|\,dx=9}\]

🎯 Exam Significance

Exam Significance

›

Properties of definite integrals: This problem tests the splitting property and the evaluation of integrals involving absolute value functions.
Piecewise representation: Students learn to express a modulus function using different algebraic expressions over appropriate intervals.
Integration techniques: It reinforces the evaluation of elementary polynomial integrals with positive and negative limits.
Geometrical interpretation: It connects definite integration with the calculation of areas under a curve.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Modulus-based integrals: Such questions require identifying the zeros of the expression inside the modulus and splitting the integration interval appropriately.
Symmetry recognition: The interval \([2,8]\) is symmetric about \(x=5\), and the function \(|x-5|\) is symmetric about the same point.
Graphical problem solving: Recognising congruent triangular regions provides a quick alternative to direct integration.
Accuracy and sign analysis: The problem develops precision in handling piecewise functions and negative algebraic expressions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Critical point: The expression inside the modulus becomes zero at \(x=5\).
Piecewise definition: \(\displaystyle |x-5|=5-x\) for \(x<5\), and \(\displaystyle |x-5|=x-5\) for \(x\geq5\).
Splitting property: The integral must be divided at the point where the expression inside the modulus changes sign.
Symmetry: The interval \([2,8]\) is symmetric about \(x=5\), and the two triangular areas are equal.
Geometrical interpretation: The integral represents the sum of the areas of two congruent right-angled triangles.
Verification: Both analytical integration and geometrical calculation give the same result, \(9\).
Exam strategy: For integrals involving \(|x-a|\) over an interval symmetric about \(a\), exploit symmetry after identifying the point where the modulus changes sign.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the application of the symmetry property of definite integrals to evaluate an integral involving an algebraic expression raised to an arbitrary power \(n\). The key idea is to replace \(x\) with \(1-x\), which transforms the given integral into another equivalent integral. Adding the two expressions allows us to simplify the problem and obtain a standard result.

Concept 1: Symmetry Property of Definite Integrals

For a continuous function \(f(x)\) defined over the interval \([0,a]\), we have

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

For the particular interval \([0,1]\), this property becomes

\[ \int\limits_0^1 f(x)\,dx = \int\limits_0^1 f(1-x)\,dx \]

This property is useful when the transformation \(x\rightarrow 1-x\) produces a simpler or complementary expression.

Concept 2: Addition of Equivalent Integrals

If two integrals have the same value, they can be added to obtain a simpler expression:

\[ 2I=\int\limits_0^1 [f(x)+f(1-x)]\,dx \]

In the given problem, this technique transforms the original integral into one involving the factor \(x(1-x)^n+(1-x)x^n\).

Concept 3: Standard Integral of a Power Function

For \(n>-1\), we know that

\[ \int\limits_0^1 x^n\,dx = \left[\frac{x^{n+1}}{n+1}\right]_0^1 = \frac{1}{n+1} \]

This standard result will be used after simplifying the given integral.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Apply the symmetry property by replacing \(x\) with \(1-x\).
Obtain a second expression for \(I\).
Add the original and transformed integrals.
Factorise the resulting algebraic expression to obtain \(x(1-x)[x^{n-1}+(1-x)^{n-1}]\), for \(n\geq1\).
Use the substitution-free symmetry property to simplify the expression and evaluate the remaining elementary integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given — Assume the given integral to be \(I\)
\[I=\int\limits_0^1 x(1-x)^n\,dx\]
Apply the symmetry property of definite integrals
We know that
\[\int\limits_0^a f(x)\,dx=\int\limits_0^a f(a-x)\,dx\]
Here, \(a=1\). Therefore, replacing \(x\) with \(1-x\), we obtain
\[ \begin{aligned} I &=\int\limits_0^1 (1-x)\left[1-(1-x)\right]^n\,dx\\[6pt] &=\int\limits_0^1 (1-x)x^n\,dx \end{aligned} \]
Add the original and transformed expressions
We have
\[I=\int\limits_0^1 x(1-x)^n\,dx\]
and
\[I=\int\limits_0^1 (1-x)x^n\,dx\]
Adding these two equations,
\[\begin{aligned}2I&=\int\limits_0^1\left[x(1-x)^n+(1-x)x^n\right]dx\end{aligned}\]
Factorise the expression inside the integral
Taking \(x(1-x)\) as the common factor,
\[\begin{aligned}2I&=\int\limits_0^1 x(1-x)\left[(1-x)^{n-1}+x^{n-1}\right]dx\end{aligned}\]
Expanding the expression inside the integral,
\[\begin{aligned}2I&=\int\limits_0^1\left[x(1-x)^n+x^n(1-x)\right]dx\end{aligned}\]
Now observe that the integrand can also be written as the derivative of a suitable expression. In particular,
\[\frac{d}{dx}\left(\frac{(1-x)^{n+1}}{n+1}\right)=-(1-x)^n\]
However, a simpler route is to evaluate the original integral directly by putting \(t=1-x\).
Evaluate the integral using substitution
Let
\[t=1-x\]
Then
\[dt=-dx,\qquad x=1-t\]
Changing the limits:
- When \(x=0\), \(t=1\)
- When \(x=1\), \(t=0\)
Therefore,
\[ \begin{aligned} I &=\int\limits_1^0 (1-t)t^n(-dt)\\[6pt] &=\int\limits_0^1 (1-t)t^n\,dt\\[6pt] &=\int\limits_0^1(t^n-t^{n+1})\,dt\\[6pt] &=\left[ \frac{t^{n+1}}{n+1} -\frac{t^{n+2}}{n+2} \right]_0^1\\[6pt] &=\left(\frac{1}{n+1}-\frac{1}{n+2}\right)-0\\[6pt] &=\frac{n+2-(n+1)}{(n+1)(n+2)}\\[6pt] &=\frac{1}{(n+1)(n+2)} \end{aligned} \]
Note: The substitution above is included as a direct verification. The required evaluation can also be obtained entirely through the symmetry property and the fundamental theorem of calculus.
Obtain the final value
\[\boxed{I=\frac{1}{(n+1)(n+2)}}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^1 x(1-x)^n\,dx=\frac{1}{(n+1)(n+2)}}\]

Condition: The result is valid for \(n>-1\), so that the improper behaviour at \(x=1\), if any, remains integrable. For the usual NCERT interpretation where \(n\) is a positive integer, the condition is automatically satisfied.

🎯 Exam Significance

Exam Significance

›

Application of symmetry: The problem tests the ability to use the property \(x\rightarrow1-x\) to transform a definite integral.
Algebraic manipulation: It develops proficiency in factorisation and simplifying expressions involving arbitrary powers.
Integration of power functions: It reinforces the application of the power rule for definite integration.
Parameter-based integration: It introduces the evaluation of definite integrals containing a parameter \(n\), an important extension of elementary integration.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Parameter-dependent integrals: The result is useful in questions involving a variable exponent and definite integration.
Symmetry techniques: Students learn to recognise when the transformation \(x\rightarrow1-x\) can simplify an integral.
Beta function connection: This integral is a special case of the Beta function, which is useful in advanced integral calculus.
Efficient evaluation: The result can be applied directly to similar integrals without repeating lengthy calculations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Symmetry property: \(\displaystyle\int\limits_0^1 f(x)\,dx=\int\limits_0^1 f(1-x)\,dx\).
Useful transformation: Replacing \(x\) with \(1-x\) interchanges the factors \(x\) and \(1-x\).
Power rule: \(\displaystyle\int\limits_0^1 x^n\,dx=\frac{1}{n+1}\), for \(n>-1\).
Important standard result: \(\displaystyle\int\limits_0^1 x^m(1-x)^n\,dx=\frac{m!\,n!}{(m+n+1)!}\), for non-negative integers \(m,n\).
Special case: Setting \(m=1\) gives \(\displaystyle\int\limits_0^1 x(1-x)^n\,dx=\frac{1}{(n+1)(n+2)}\).
Exam strategy: For integrals over \([0,1]\) containing both \(x\) and \(1-x\), consider applying the symmetry transformation \(x\rightarrow1-x\).

📘 Concept & Theory

Theory and Concept

›

This problem demonstrates the application of the symmetry property of definite integrals to evaluate a logarithmic-trigonometric integral. The key idea is to transform the integrand using the substitution \(x\rightarrow \frac{\pi}{4}-x\), simplify the resulting trigonometric expression and then add the two equivalent integrals.

Concept 1: Symmetry Property of Definite Integrals

For a continuous function \(f(x)\) defined over the interval \([a,b]\), the following property holds:

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx \]

For the interval \(\left[0,\frac{\pi}{4}\right]\), this becomes

\[ \int\limits_0^{\pi/4}f(x)\,dx = \int\limits_0^{\pi/4} f\left(\frac{\pi}{4}-x\right)dx \]

Concept 2: Trigonometric Identity

We use the identity

\[ \tan\left(\frac{\pi}{4}-x\right) = \frac{1-\tan x}{1+\tan x} \]

This identity helps simplify the transformed logarithmic expression.

Concept 3: Logarithmic Properties

The following properties of logarithms are useful:

\[ \log a+\log b=\log(ab) \]

\[ \log\left(\frac{a}{b}\right) =\log a-\log b \]

These properties allow us to combine the two logarithmic integrands into a simpler expression.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Apply the symmetry property by replacing \(x\) with \(\frac{\pi}{4}-x\).
Use the tangent subtraction identity to simplify the transformed integrand.
Add the original and transformed integrals.
Apply logarithmic identities to simplify the expression.
Evaluate the resulting elementary definite integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given — Assume the given integral to be \(I\)
\[I=\int\limits_0^{\pi/4}\log(1+\tan x)\,dx\]
Apply the symmetry property of definite integrals
Using
\[\int\limits_a^b f(x)\,dx=\int\limits_a^b f(a+b-x)\,dx\]
we replace \(x\) with \(\frac{\pi}{4}-x\). Thus,
\[\begin{aligned}I&=\int\limits_0^{\pi/4}\log\left[1+\tan\left(\frac{\pi}{4}-x\right)\right]dx\end{aligned}\]
Simplify the transformed integrand
Using the tangent subtraction identity,
\[\tan\left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x}\]
Therefore,
\[\begin{aligned}I&=\int\limits_0^{\pi/4}\log\left(1+\frac{1-\tan x}{1+\tan x}\right)dx\\[8pt] &=\int\limits_0^{\pi/4}\log\left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right)dx\\[8pt] &=\int\limits_0^{\pi/4}\log\left(\frac{2}{1+\tan x}\right)dx\end{aligned}\]
Add the original and transformed integrals.
We have
\[I=\int\limits_0^{\pi/4}\log(1+\tan x)\,dx\]
and
\[I=\int\limits_0^{\pi/4}\log\left(\frac{2}{1+\tan x}\right)dx\]
Adding these two equations,
\[\begin{aligned}2I&=\int\limits_0^{\pi/4}\left[\log(1+\tan x)+\log\left(\frac{2}{1+\tan x}\right)\right]dx\end{aligned}\]
Apply the logarithmic property
Using \(\log a+\log b=\log(ab)\),
\[\begin{aligned}2I&=\int\limits_0^{\pi/4}\log\left[(1+\tan x)\cdot\frac{2}{1+\tan x}\right]dx\\[8pt] &=\int\limits_0^{\pi/4}\log 2\,dx\end{aligned}\]
Since \(\log 2\) is a constant,
\[ \begin{aligned} 2I &=\log 2\int\limits_0^{\pi/4}dx\\[6pt] &=\log 2\left[x\right]_0^{\pi/4}\\[6pt] &=\log 2\left(\frac{\pi}{4}-0\right)\\[6pt] &=\frac{\pi}{4}\log 2 \end{aligned} \]
Calculate the value of \(I\).
Dividing both sides by \(2\),
\[ \begin{aligned} I&=\frac{1}{2}\cdot\frac{\pi}{4}\log 2\\[6pt] &=\frac{\pi}{8}\log 2 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\pi/4}\log(1+\tan x)\,dx=\frac{\pi}{8}\log 2}\]

🎯 Exam Significance

Exam Significance

›

Application of symmetry: This question tests the use of the property \(x\rightarrow a+b-x\) in definite integrals.
Trigonometric identities: It reinforces the application of tangent subtraction and complementary-angle identities.
Logarithmic simplification: Students learn to combine logarithmic expressions effectively before integration.
Conceptual integration: The problem demonstrates that a definite integral can often be evaluated without finding an antiderivative.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Symmetry-based evaluation: Recognising the appropriate transformation can substantially reduce calculation.
Logarithmic-trigonometric integrals: The question combines two important areas of calculus.
Algebraic manipulation: It tests the ability to simplify a transformed integrand before proceeding.
Time-efficient problem solving: The symmetry method provides a direct alternative to integration by substitution or integration by parts.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Symmetry property: \(\displaystyle\int\limits_a^b f(x)\,dx=\int\limits_a^b f(a+b-x)\,dx\).
Useful transformation: For the interval \(\left[0,\frac{\pi}{4}\right]\), replace \(x\) with \(\frac{\pi}{4}-x\).
Tangent identity: \(\displaystyle\tan\left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x}\).
Logarithmic identity: \(\log a+\log b=\log(ab)\).
Important result: \(\displaystyle\int\limits_0^{\pi/4}\log(1+\tan x)\,dx=\frac{\pi}{8}\log 2\).
Exam strategy: Whenever an integral contains trigonometric functions over an interval with complementary angles, check whether the symmetry transformation can simplify the integrand.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the use of the symmetry property of definite integrals to transform an integral into an equivalent form that is easier to evaluate. The transformation \(x\rightarrow 2-x\) interchanges the expressions \(x\) and \(2-x\), allowing us to rewrite the given integral in a convenient form.

Concept 1: Symmetry Property of Definite Integrals

For a continuous function \(f(x)\) defined over the interval \([a,b]\),

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx \]

For the interval \([0,2]\), this becomes

\[ \int\limits_0^2 f(x)\,dx = \int\limits_0^2 f(2-x)\,dx \]

Concept 2: Integration of Fractional Powers

We use the power rule for integration:

\[ \int x^m\,dx=\frac{x^{m+1}}{m+1}+C, \qquad m\neq -1 \]

In particular,

\[ \int x^{1/2}\,dx=\frac{2}{3}x^{3/2}+C \]

\[ \int x^{3/2}\,dx=\frac{2}{5}x^{5/2}+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Apply the symmetry property using \(x\rightarrow 2-x\).
Rewrite the transformed integral in terms of fractional powers of \(x\).
Expand the expression and integrate each term separately.
Substitute the limits and simplify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given — Assume the given integral to be \(I\).
\[I=\int\limits_0^2 x\sqrt{2-x}\,dx\]
Apply the symmetry property of definite integrals
Using the property
\[\int\limits_a^b f(x)\,dx=\int\limits_a^b f(a+b-x)\,dx\]
we replace \(x\) with \(2-x\). Thus,
\[ \begin{aligned} I &=\int\limits_0^2 (2-x)\sqrt{2-(2-x)}\,dx\\[6pt] &=\int\limits_0^2 (2-x)\sqrt{x}\,dx \end{aligned} \]
Express the integrand in terms of powers of \(x\)
We have
\[ \begin{aligned} I &=\int\limits_0^2 (2-x)\sqrt{x}\,dx\\[6pt] &=\int\limits_0^2 \left(2\sqrt{x}-x\sqrt{x}\right)dx\\[6pt] &=\int\limits_0^2 \left(2x^{1/2}-x^{3/2}\right)dx \end{aligned} \]
Integrate each term separately
Applying the power rule,
\[ \begin{aligned} I &=2\int\limits_0^2 x^{1/2}\,dx -\int\limits_0^2 x^{3/2}\,dx\\[8pt] &=2\left[\frac{x^{3/2}}{3/2}\right]_0^2 -\left[\frac{x^{5/2}}{5/2}\right]_0^2\\[8pt] &=2\left[\frac{2}{3}x^{3/2}\right]_0^2 -\left[\frac{2}{5}x^{5/2}\right]_0^2\\[8pt] &=\left[ \frac{4}{3}x^{3/2} -\frac{2}{5}x^{5/2} \right]_0^2 \end{aligned} \]
Apply the limits of integration
\[\begin{aligned}I&=\left[\frac{4}{3}x^{3/2}-\frac{2}{5}x^{5/2}\right]_0^2\\[8pt] &=\left(\frac{4}{3}\cdot 2^{3/2}-\frac{2}{5}\cdot 2^{5/2}\right)-\left(\frac{4}{3}\cdot 0^{3/2}-\frac{2}{5}\cdot 0^{5/2}\right)\\[8pt] &=\frac{4}{3}\cdot 2\sqrt{2}-\frac{2}{5}\cdot 4\sqrt{2}\\[8pt] &=\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}\\[8pt] &=8\sqrt{2}\left(\frac{1}{3}-\frac{1}{5}\right)\\[8pt] &=8\sqrt{2}\left(\frac{5-3}{15}\right)\\[8pt] &=\frac{16\sqrt{2}}{15}\end{aligned}\]
\[\boxed{\displaystyle\int\limits_0^2 x\sqrt{2-x}\,dx=\frac{16\sqrt{2}}{15}}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^2 x\sqrt{2-x}\,dx=\frac{16\sqrt{2}}{15}}\]

🎯 Exam Significance

Exam Significance

›

Symmetry property: The problem tests the application of \(x\rightarrow a+b-x\) to transform a definite integral.
Fractional powers: It reinforces the integration of expressions involving square roots and fractional indices.
Algebraic simplification: Students practise converting radical expressions into powers before integration.
Definite integration: It develops accuracy in applying upper and lower limits after finding an antiderivative.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Transformation techniques: Recognising the symmetry of the integration interval can simplify otherwise unfamiliar integrals.
Radical integrals: The problem strengthens the ability to handle algebraic functions involving square roots.
Efficient evaluation: The symmetry property converts the original integrand into a more convenient algebraic form.
Conceptual understanding: It connects definite-integral properties with elementary integration techniques.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Symmetry property: \(\displaystyle\int\limits_a^b f(x)\,dx=\int\limits_a^b f(a+b-x)\,dx\).
For the interval \([0,2]\): Replace \(x\) with \(2-x\).
Power rule: \(\displaystyle\int x^m\,dx=\frac{x^{m+1}}{m+1}+C\), for \(m\neq-1\).
Radical conversion: \(\sqrt{x}=x^{1/2}\) and \(x\sqrt{x}=x^{3/2}\).
Important result: \(\displaystyle\int\limits_0^2 x\sqrt{2-x}\,dx=\frac{16\sqrt{2}}{15}\).
Exam strategy: When an integral contains both \(x\) and \(a-x\), consider applying the symmetry property before attempting direct integration.

📘 Concept & Theory

Theory and Concept

›

This problem combines the symmetry property of definite integrals with the double-angle identity of trigonometry and the properties of logarithms. The key observation is that the expression inside the logarithms can be simplified using the identity \(\sin 2x=2\sin x\cos x\).

Concept 1: Symmetry Property of Definite Integrals

For a suitable integrable function \(f(x)\) over the interval \([a,b]\),

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx \]

For the interval \(\left[0,\frac{\pi}{2}\right]\), this becomes

\[ \int\limits_0^{\pi/2}f(x)\,dx = \int\limits_0^{\pi/2} f\left(\frac{\pi}{2}-x\right)dx \]

Concept 2: Double-Angle Identity

\[ \sin 2x=2\sin x\cos x \]

Taking logarithms on both sides,

\[ \log\sin 2x = \log 2+\log\sin x+\log\cos x \]

Concept 3: Logarithmic Properties

\[ \log a-\log b=\log\left(\frac{a}{b}\right) \]

\[ \log(ab)=\log a+\log b \]

These identities help reduce the original integrand to a simpler logarithmic expression.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Use the double-angle identity to express \(\log\sin 2x\) in terms of \(\log\sin x\) and \(\log\cos x\).
Simplify the integrand using logarithmic properties.
Apply the symmetry property by replacing \(x\) with \(\frac{\pi}{2}-x\).
Add the two equivalent integral expressions.
Evaluate the resulting integral using the interval length.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Assume the given integral to be \(I\).
\[ I=\int\limits_0^{\pi/2} \left(2\log\sin x-\log\sin 2x\right)dx \]
Apply the double-angle identity
We know that
\[\sin 2x=2\sin x\cos x\]
Taking logarithms,
\[\log\sin 2x=\log 2+\log\sin x+\log\cos x\]
Substituting this expression into the given integral,
\[\begin{aligned}I&=\int\limits_0^{\pi/2}\left[2\log\sin x-\log 2-\log\sin x-\log\cos x\right]dx\\[6pt] &=\int\limits_0^{\pi/2}\left[\log\sin x-\log\cos x-\log 2\right]dx\end{aligned}\]
Using the logarithmic quotient property,
\[I=\int\limits_0^{\pi/2}\log\left(\frac{\sin x}{\cos x}\right)dx-\int\limits_0^{\pi/2}\log 2\,dx\]
Therefore,
\[I=\int\limits_0^{\pi/2}\log\tan x\,dx-\int\limits_0^{\pi/2}\log 2\,dx\]
Apply the symmetry property
Let
\[J=\int\limits_0^{\pi/2}\log\tan x\,dx\]
Using the symmetry property with \(x\rightarrow\frac{\pi}{2}-x\),
\[\begin{aligned}J&=\int\limits_0^{\pi/2}\log\tan\left(\frac{\pi}{2}-x\right)dx\\[6pt] &=\int\limits_0^{\pi/2}\log\cot x\,dx\end{aligned}\]
Since
\[\cot x=\frac{1}{\tan x}\]
we get
\[ \begin{aligned} J &=\int\limits_0^{\pi/2}\log\left(\frac{1}{\tan x}\right)dx\\[6pt] &=-\int\limits_0^{\pi/2}\log\tan x\,dx\\[6pt] &=-J \end{aligned} \]
Thus,
\[J=-J\]
or
\[\boxed{J=0}\]
Evaluate the remaining constant integral
Returning to the original expression,
\[ \begin{aligned} I &=J-\int\limits_0^{\pi/2}\log 2\,dx\\[6pt] &=0-\log 2\int\limits_0^{\pi/2}dx\\[6pt] &=-\log 2\left[x\right]_0^{\pi/2}\\[6pt] &=-\log 2\left(\frac{\pi}{2}-0\right)\\[6pt] &=-\frac{\pi}{2}\log 2 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\pi/2}\left(2\log\sin x-\log\sin 2x\right)dx=-\frac{\pi}{2}\log 2}\]

🎯 Exam Significance

Exam Significance

›

Symmetry property: The question tests the use of the transformation \(x\rightarrow\frac{\pi}{2}-x\) in definite integrals.
Trigonometric identities: It reinforces the double-angle identity and the complementary-angle relationship between tangent and cotangent.
Logarithmic properties: Students practise simplifying logarithmic expressions before integration.
Improper integrals: The logarithmic terms have singularities at the endpoints, but these are integrable. The integral is understood in the improper sense.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Symmetry-based evaluation: Recognising that \(\log\tan x\) changes sign under complementary-angle substitution is the central insight.
Logarithmic-trigonometric integrals: The problem combines logarithmic identities with trigonometric transformations.
Efficient problem-solving: The integral is evaluated without requiring a complicated antiderivative.
Improper-integral awareness: Students learn to handle endpoint singularities correctly while applying definite-integral properties.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Symmetry property: \(\displaystyle\int\limits_a^b f(x)\,dx=\int\limits_a^b f(a+b-x)\,dx\).
Double-angle identity: \(\sin 2x=2\sin x\cos x\).
Complementary-angle identity: \(\displaystyle\tan\left(\frac{\pi}{2}-x\right)=\cot x\).
Important result: \(\displaystyle\int\limits_0^{\pi/2}\log\tan x\,dx=0\).
Final evaluation: \(\displaystyle\int\limits_0^{\pi/2}\left(2\log\sin x-\log\sin 2x\right)dx=-\frac{\pi}{2}\log 2\).
Exam strategy: For logarithmic trigonometric integrals over \(\left[0,\frac{\pi}{2}\right]\), check whether the complementary-angle transformation simplifies the integrand or changes its sign.

📘 Concept & Theory

Theory and Concept

›

This problem demonstrates the use of the symmetry property of definite integrals and the standard trigonometric identity for the square of sine. The key observation is that \(\sin^2 x\) is an even function. Consequently, its integral over a symmetric interval can be evaluated by considering only the positive half of the interval.

Concept 1: Even Functions

A function \(f(x)\) is called an even function if

\[ f(-x)=f(x) \]

Since

\[ \sin(-x)=-\sin x \]

we have

\[ \sin^2(-x)=(-\sin x)^2=\sin^2 x \]

Therefore, \(\sin^2 x\) is an even function.

Concept 2: Integral of an Even Function

If \(f(x)\) is an even function, then

\[ \int\limits_{-a}^{a}f(x)\,dx = 2\int\limits_0^a f(x)\,dx \]

Thus, for the given interval,

\[ \int\limits_{-\pi/2}^{\pi/2}\sin^2 x\,dx = 2\int\limits_0^{\pi/2}\sin^2 x\,dx \]

Concept 3: Power-Reduction Identity

Using the double-angle identity

\[ \cos 2x=1-2\sin^2 x \]

we obtain

\[ \sin^2 x=\frac{1-\cos 2x}{2} \]

This identity converts the square of a trigonometric function into a simpler expression that can be integrated directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Establish that \(\sin^2 x\) is an even function.
Apply the even-function property to reduce the symmetric interval.
Use the power-reduction identity for \(\sin^2 x\).
Integrate the resulting expression and substitute the limits.
Simplify to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Assume the given integral to be \(I\)
\[I=\int\limits_{-\pi/2}^{\pi/2}\sin^2 x\,dx\]
Verify that the integrand is an even function
Consider
\[f(x)=\sin^2 x\]
Replacing \(x\) with \(-x\),
\[\begin{aligned}f(-x)&=\sin^2(-x)\\ &=(-\sin x)^2\\ &=\sin^2 x\\ &=f(x)\end{aligned}\]
Hence, \(f(x)=\sin^2 x\) is an even function
Apply the property of definite integrals for even functions
We know that for an even function,
\[\int\limits_{-a}^{a}f(x)\,dx=2\int\limits_0^a f(x)\,dx\]
Therefore,
\[\begin{aligned}I&=2\int\limits_0^{\pi/2}\sin^2 x\,dx\end{aligned}\]
Apply the trigonometric power-reduction identity
We know that
\[\sin^2 x=\frac{1-\cos 2x}{2}\]
Substituting this identity,
\[\begin{aligned}I&=2\int\limits_0^{\pi/2}\frac{1-\cos 2x}{2}\,dx\\[6pt] &=\int\limits_0^{\pi/2}(1-\cos 2x)\,dx\end{aligned}\]
Integrate and apply the limits
\[\begin{aligned}I&=\int\limits_0^{\pi/2}1\,dx-\int\limits_0^{\pi/2}\cos 2x\,dx\\[6pt] &=\left[x-\frac{\sin 2x}{2}\right]_0^{\pi/2}\\[6pt] &=\left[\frac{\pi}{2}-\frac{\sin\pi}{2}\right]-\left[0-\frac{\sin 0}{2}\right]\\[6pt] &=\frac{\pi}{2}-0\\[6pt] &=\frac{\pi}{2}\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_{-\pi/2}^{\pi/2}\sin^2 x\,dx=\frac{\pi}{2}}\]

🎯 Exam Significance

Exam Significance

›

Even-function property: The question tests the recognition and application of symmetry in definite integrals.
Trigonometric identities: It reinforces the double-angle formula and power-reduction identity.
Definite integration: Students practise evaluating an integral by applying limits correctly.
Conceptual clarity: It demonstrates how symmetry can reduce the integration interval and simplify calculations.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Symmetry-based evaluation: Recognising even and odd functions helps simplify many definite integrals.
Trigonometric integration: The power-reduction identity is useful for integrating higher powers of trigonometric functions.
Time-efficient calculations: Reducing the interval before integration can make calculations shorter.
Function properties: The problem develops the ability to identify parity and use it strategically in calculus.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Even-function condition: \(f(-x)=f(x)\).
Symmetry property: \(\displaystyle\int\limits_{-a}^{a}f(x)\,dx=2\int\limits_0^a f(x)\,dx\), when \(f(x)\) is even.
Trigonometric identity: \(\displaystyle\sin^2 x=\frac{1-\cos 2x}{2}\).
Standard result: \(\displaystyle\int\limits_0^{\pi/2}\sin^2 x\,dx=\frac{\pi}{4}\).
Final evaluation: \(\displaystyle\int\limits_{-\pi/2}^{\pi/2}\sin^2 x\,dx=\frac{\pi}{2}\).
Exam strategy: Whenever the limits of integration are symmetric about zero, first check whether the integrand is even or odd.

📘 Concept & Theory

Theory and Concept

›

We use the symmetry property of definite integrals:

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

For the interval \([0,\pi]\), this property gives:

\[ \int\limits_0^\pi f(x)\,dx = \int\limits_0^\pi f(\pi-x)\,dx \]

Also,

\[ \sin(\pi-x)=\sin x \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the substitution \(x\mapsto\pi-x\).
Add the original and transformed integrals.
Reduce the problem to a simpler definite integral.
Evaluate the remaining integral using the half-angle substitution.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given —
\[I=\int\limits_0^\pi\frac{x}{1+\sin x}\,dx\]
Apply the symmetry property
Replacing \(x\) by \(\pi-x\), we obtain:
\[ \begin{aligned} I &=\int\limits_0^\pi \frac{\pi-x}{1+\sin(\pi-x)}\,dx\\[6pt] &=\int\limits_0^\pi \frac{\pi-x}{1+\sin x}\,dx \end{aligned} \]
Add the two expressions
Adding the original and transformed integrals:
\[ \begin{aligned} 2I &=\int\limits_0^\pi\frac{x}{1+\sin x}\,dx +\int\limits_0^\pi\frac{\pi-x}{1+\sin x}\,dx\\[6pt] &=\int\limits_0^\pi \frac{x+\pi-x}{1+\sin x}\,dx\\[6pt] &=\pi\int\limits_0^\pi\frac{dx}{1+\sin x} \end{aligned} \]
Therefore,
\[I=\frac{\pi}{2}\int\limits_0^\pi\frac{dx}{1+\sin x}\]
Evaluate the remaining integral
Let
\[J=\int\limits_0^\pi\frac{dx}{1+\sin x}\]
Using the substitution:
\[t=\tan\frac{x}{2}\]
We know that:
\[\sin x=\frac{2t}{1+t^2},\quad dx=\frac{2\,dt}{1+t^2}\]
The limits change as follows:
\[ \begin{aligned} x=0&\implies t=0\\ x=\pi&\implies t\to\infty \end{aligned} \]
Substituting these values:
\[ \begin{aligned} J &=\int\limits_0^\infty \frac{\frac{2\,dt}{1+t^2}} {1+\frac{2t}{1+t^2}}\\[6pt] &=2\int\limits_0^\infty \frac{dt}{1+t^2+2t}\\[6pt] &=2\int\limits_0^\infty\frac{dt}{(1+t)^2}\\[6pt] &=2\left[-\frac{1}{1+t}\right]_0^\infty\\[6pt] &=2\left(0-(-1)\right)\\[6pt] &=2 \end{aligned} \]
Calculate the required integral
We have:
\[I=\frac{\pi}{2}J\]
Substituting \(J=2\):
\[ \begin{aligned} I&=\frac{\pi}{2}\times2\\[4pt] &=\boxed{\pi} \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^\pi\frac{x}{1+\sin x}\,dx=\pi}\]

🎯 Exam Significance

Exam Significance

›

Tests the symmetry property of definite integrals.
Demonstrates how symmetry can simplify an integral containing a linear factor.
Applies the useful Weierstrass substitution \(t=\tan(x/2)\) for trigonometric integrals.
Develops an efficient approach for CBSE, JEE Main, JEE Advanced and CUET examinations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

For an integral over \([0,\pi]\), replacing \(x\) by \(\pi-x\) is often useful.
The identity \(\sin(\pi-x)=\sin x\) preserves the denominator.
Adding the original and transformed integrals eliminates the variable from the numerator.
The substitution \(t=\tan(x/2)\) converts the trigonometric integral into a rational integral.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the application of the symmetry property of definite integrals for odd functions. The integrand is an odd power of the sine function, and its values at opposite points of the interval are equal in magnitude but opposite in sign. Consequently, the positive and negative contributions cancel each other.

Concept 1: Odd Functions

A function \(f(x)\) is called an odd function if

\[ f(-x)=-f(x) \]

Since

\[ \sin(-x)=-\sin x \]

we obtain

\[ \begin{aligned} \sin^7(-x) &=(-\sin x)^7\\ &=-\sin^7 x \end{aligned} \]

Therefore, \(\sin^7 x\) is an odd function.

Concept 2: Integral of an Odd Function

If \(f(x)\) is an odd function and is integrable over the symmetric interval \([-a,a]\), then

\[ \boxed{ \int\limits_{-a}^{a}f(x)\,dx=0 } \]

This property follows because the contributions from the two halves of the interval cancel each other.

Concept 3: Symmetry Property of Definite Integrals

For a suitable integrable function,

\[ \int\limits_{-a}^{a}f(x)\,dx = \int\limits_0^a\left[f(x)+f(-x)\right]dx \]

For an odd function, \(f(x)+f(-x)=0\), so the integral vanishes.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Check the parity of the integrand by replacing \(x\) with \(-x\).
Establish that \(\sin^7 x\) is an odd function.
Apply the property of definite integrals for odd functions over symmetric limits.
Obtain the final value directly.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Assume the given integral to be \(I\)
\[I=\int\limits_{-\pi/2}^{\pi/2}\sin^7 x\,dx\]
Examine the parity of the integrand
Let
\[f(x)=\sin^7 x\]
Replacing \(x\) with \(-x\),
\[\begin{aligned}f(-x)&=\sin^7(-x)\\[4pt] &=(-\sin x)^7\\[4pt] &=-\sin^7 x\\[4pt] &=-f(x)\end{aligned}\]
Hence, \(f(x)=\sin^7 x\) is an odd function
Apply the property of definite integrals for odd functions
We know that if \(f(x)\) is odd, then
\[\int\limits_{-a}^{a}f(x)\,dx=0\]
Here,
\[a=\frac{\pi}{2}\]
Therefore,
\[\begin{aligned}I&=\int\limits_{-\pi/2}^{\pi/2}\sin^7 x\,dx\\[6pt] &=0\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_{-\pi/2}^{\pi/2}\sin^7 x\,dx=0}\]

🎯 Exam Significance

Exam Significance

›

Odd-function property: The question tests the ability to identify odd functions and apply their integral property.
Trigonometric functions: It reinforces the parity of sine and the behaviour of odd powers of trigonometric functions.
Symmetric limits: Students learn to recognise when the limits of integration allow direct simplification.
Conceptual efficiency: The integral can be evaluated without finding an antiderivative.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Parity-based evaluation: Identifying whether an integrand is even or odd is an important shortcut in definite integration.
Higher powers of trigonometric functions: The result applies to every positive odd integer power of sine over a symmetric interval.
Time-efficient problem solving: Recognising odd symmetry avoids unnecessary algebraic expansion and integration.
Function analysis: The question develops the ability to use functional properties to simplify calculus problems.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Odd-function condition: \(f(-x)=-f(x)\).
Sine parity: \(\sin(-x)=-\sin x\).
Odd powers of sine: \(\sin^{2n+1}x\) is an odd function for every non-negative integer \(n\).
Integral property: \(\displaystyle\int\limits_{-a}^{a}f(x)\,dx=0\), when \(f(x)\) is odd and integrable.
Final evaluation: \(\displaystyle\int\limits_{-\pi/2}^{\pi/2}\sin^7 x\,dx=0\).
Exam strategy: Whenever the limits are symmetric about zero, check the parity of the integrand before attempting integration.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the use of symmetry and periodicity in evaluating definite integrals. The integrand is an odd power of the cosine function, and its values over the two halves of a complete period cancel each other.

Concept 1: Symmetry Property of Definite Integrals

For a suitable integrable function \(f(x)\),

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx \]

Concept 2: Cosine Function and Its Shift Property

We know that

\[ \cos(\pi+x)=-\cos x \]

Consequently,

\[ \cos^5(\pi+x)=-\cos^5x \]

Thus, the contributions from the intervals \([0,\pi]\) and \([\pi,2\pi]\) are equal in magnitude and opposite in sign.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Split the interval \([0,2\pi]\) into two equal intervals.
Apply the substitution \(x=\pi+t\) to the second integral.
Use the identity \(\cos(\pi+t)=-\cos t\).
Show that the two integrals cancel each other.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Assume the given integral to be \(I\).
\[I=\int\limits_0^{2\pi}\cos^5x\,dx\]
Split the integral into two equal intervals
\[I=\int\limits_0^\pi\cos^5x\,dx+\int\limits_\pi^{2\pi}\cos^5x\,dx\]
Evaluate the second integral using substitution
In the second integral, put
\[x=\pi+t\]
Then
\[dx=dt\]
Changing the limits:
- When \(x=\pi\), \(t=0\)
- When \(x=2\pi\), \(t=\pi\)
Therefore,
\[ \begin{aligned} \int\limits_\pi^{2\pi}\cos^5x\,dx &=\int\limits_0^\pi\cos^5(\pi+t)\,dt\\[6pt] &=\int\limits_0^\pi(-\cos t)^5\,dt\\[6pt] &=-\int\limits_0^\pi\cos^5t\,dt \end{aligned} \]
Substitute this result into the original integral
\[ \begin{aligned} I &=\int\limits_0^\pi\cos^5x\,dx -\int\limits_0^\pi\cos^5t\,dt\\[6pt] &=\int\limits_0^\pi\cos^5x\,dx -\int\limits_0^\pi\cos^5x\,dx\\[6pt] &=0 \end{aligned} \]
Hence, the positive and negative contributions over one complete period cancel each other.

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{2\pi}\cos^5x\,dx=0}\]

🎯 Exam Significance

Exam Significance

›

Periodicity: Recognising the behaviour of trigonometric functions over a complete period helps simplify definite integrals.
Odd powers of cosine: An odd power changes sign under a shift of \(\pi\).
Symmetry-based shortcuts: Such properties can eliminate lengthy integration calculations.
Substitution techniques: The transformation \(x=\pi+t\) is a useful tool for exploiting trigonometric symmetry.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Cosine shift identity: \(\cos(\pi+x)=-\cos x\).
Odd power behaviour: \(\cos^5(\pi+x)=-\cos^5x\).
Interval symmetry: The integrals over \([0,\pi]\) and \([\pi,2\pi]\) cancel.
Efficient evaluation: Definite integrals can often be evaluated using symmetry without finding an antiderivative.
Final result: The integral over one complete period of \(\cos^5x\) is zero.

📘 Concept & Theory

Theory and Concept

›

This problem demonstrates the application of the symmetry property of definite integrals. The limits of integration are complementary angles, and the substitution \(x\rightarrow \frac{\pi}{2}-x\) interchanges the sine and cosine functions.

Concept 1: Symmetry Property of Definite Integrals

For a suitable integrable function \(f(x)\),

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx \]

In particular, when the limits are \(0\) and \(\frac{\pi}{2}\),

\[ \int\limits_0^{\pi/2}f(x)\,dx = \int\limits_0^{\pi/2} f\left(\frac{\pi}{2}-x\right)dx \]

Concept 2: Complementary Angle Identities

We know that

\[ \sin\left(\frac{\pi}{2}-x\right)=\cos x \]

\[ \cos\left(\frac{\pi}{2}-x\right)=\sin x \]

Also,

\[ \sin\left(\frac{\pi}{2}-x\right) \cos\left(\frac{\pi}{2}-x\right) =\sin x\cos x \]

Thus, the denominator remains unchanged, while the numerator changes its sign.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Apply the substitution \(x\rightarrow\frac{\pi}{2}-x\).
Use the complementary angle identities to simplify the transformed integral.
Observe that the transformed integral is equal to \(-I\).
Combine both expressions to determine the value of \(I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Assume the given integral to be \(I\).
\[I=\int\limits_0^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx\]
Apply the symmetry property
Using the substitution
\[x=\frac{\pi}{2}-t\]
we get
\[dx=-dt\]
The limits change as follows:
- When \(x=0\), \(t=\frac{\pi}{2}\).
- When \(x=\frac{\pi}{2}\), \(t=0\).
Therefore, reversing the limits after substitution,
\[ \begin{aligned} I &=\int\limits_0^{\pi/2} \frac{ \sin\left(\frac{\pi}{2}-t\right) -\cos\left(\frac{\pi}{2}-t\right) }{ 1+ \sin\left(\frac{\pi}{2}-t\right) \cos\left(\frac{\pi}{2}-t\right) }\,dt\\[10pt] &=\int\limits_0^{\pi/2} \frac{\cos t-\sin t}{1+\cos t\sin t}\,dt \end{aligned} \]
Since
\[\cos t-\sin t=-(\sin t-\cos t)\]
we obtain
\[\begin{aligned}I&=-\int\limits_0^{\pi/2}\frac{\sin t-\cos t}{1+\sin t\cos t}\,dt\\[8pt]&=-I\end{aligned}\]
Determine the value of the integral
\[I=-I\]

\[2I=0\]
Hence,
\[\boxed{I=0}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx=0}\]

🎯 Exam Significance

Exam Significance

›

Symmetry property: Tests the application of the transformation \(x\rightarrow\frac{\pi}{2}-x\).
Complementary angles: Reinforces the relationship between sine and cosine functions.
Definite integral evaluation: Demonstrates how symmetry can simplify an integral without direct integration.
Logical reasoning: Develops the ability to recognise when an integral is equal to its own negative.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Symmetry-based shortcuts: The substitution \(x\rightarrow\frac{\pi}{2}-x\) is frequently useful in trigonometric definite integrals.
Algebraic structure: Recognising an unchanged denominator and sign-reversed numerator makes evaluation immediate.
Efficient problem solving: Avoids unnecessary differentiation, substitution or antiderivative calculations.
Conceptual understanding: Reinforces the relationship between complementary-angle identities and definite integral properties.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Symmetry transformation: \(x\rightarrow\frac{\pi}{2}-x\).
Complementary angle identities: \(\sin(\frac{\pi}{2}-x)=\cos x\) and \(\cos(\frac{\pi}{2}-x)=\sin x\).
Denominator: \(1+\sin x\cos x\) remains unchanged under the transformation.
Numerator: \(\sin x-\cos x\) changes to its negative.
Final result: The integral is equal to its own negative and therefore has value zero.

📘 Concept & Theory

Theory and Concept

›

This problem involves the application of the symmetry property of definite integrals, logarithmic identities and trigonometric half-angle identities.

Concept 1: Symmetry Property of Definite Integrals

For a suitable integrable function \(f(x)\),

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx \]

For the interval \([0,\pi]\), this becomes

\[ \int\limits_0^\pi f(x)\,dx = \int\limits_0^\pi f(\pi-x)\,dx \]

Concept 2: Trigonometric Half-Angle Identity

We know that

\[ 1+\cos x=2\cos^2\frac{x}{2} \]

Therefore,

\[ \log(1+\cos x) = \log 2+2\log\cos\frac{x}{2} \]

Concept 3: Logarithmic Integral

The logarithmic singularity at an endpoint does not necessarily make an improper integral divergent. In this problem, \(\log(1+\cos x)\) behaves like \(2\log(\pi-x)\) near \(x=\pi\), and this singularity is integrable.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Apply the symmetry property using \(x\rightarrow\pi-x\).
Add the original and transformed integrals.
Use the identity \(1+\cos x=2\cos^2\frac{x}{2}\).
Evaluate the resulting logarithmic integral using the standard result \(\displaystyle\int\limits_0^{\pi/2}\log\cos t\,dt=-\frac{\pi}{2}\log 2\).
Obtain the final value.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Assume the given integral to be \(I\)
\[I=\int\limits_0^\pi\log(1+\cos x)\,dx\]
Apply the symmetry property of definite integrals
Replacing \(x\) with \(\pi-x\),
\[\begin{aligned}I&=\int\limits_0^\pi\log\left(1+\cos(\pi-x)\right)\,dx\\[6pt] &=\int\limits_0^\pi\log(1-\cos x)\,dx\end{aligned}\]
Here, we have used
\[\cos(\pi-x)=-\cos x\]
Add the original and transformed integrals
\[ \begin{aligned} 2I &=\int\limits_0^\pi\log(1+\cos x)\,dx\\ &\quad+\int\limits_0^\pi\log(1-\cos x)\,dx \end{aligned} \]
Using the logarithmic identity \(\log a+\log b=\log(ab)\),
\[ \begin{aligned} 2I &=\int\limits_0^\pi \log\left[(1+\cos x)(1-\cos x)\right]dx\\[6pt] &=\int\limits_0^\pi\log(1-\cos^2x)\,dx\\[6pt] &=\int\limits_0^\pi\log(\sin^2x)\,dx\\[6pt] &=2\int\limits_0^\pi\log\sin x\,dx \end{aligned} \]
Dividing both sides by 2,
\[I=\int\limits_0^\pi\log\sin x\,dx\]
Evaluate the standard logarithmic integral
Using the symmetry property,
\[\int\limits_0^\pi\log\sin x\,dx=2\int\limits_0^{\pi/2}\log\sin x\,dx\]
We use the standard result
\[\int\limits_0^{\pi/2}\log\sin x\,dx=-\frac{\pi}{2}\log 2\]
Therefore,
\[\begin{aligned}I&=2\left(-\frac{\pi}{2}\log 2\right)\\[6pt] &=-\pi\log 2\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^\pi\log(1+\cos x)\,dx=-\pi\log 2}\]

🎯 Exam Significance

Exam Significance

›

Symmetry property: Demonstrates how the substitution \(x\rightarrow\pi-x\) transforms a trigonometric logarithmic integral.
Logarithmic identities: Reinforces the use of logarithmic addition and multiplication properties.
Trigonometric identities: Connects complementary-angle identities with the fundamental identity \(\sin^2x+\cos^2x=1\).
Standard integrals: Develops familiarity with important logarithmic trigonometric integrals.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Symmetry-based evaluation: Recognising transformations of the integration interval helps simplify complicated integrands.
Logarithmic transformations: Combining logarithms can convert a difficult expression into a standard integral.
Improper integrals: The problem illustrates that an integrable logarithmic singularity at an endpoint does not prevent a finite result.
Standard integral application: The result is useful in evaluating other trigonometric integrals involving logarithms.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Symmetry transformation: \(x\rightarrow\pi-x\).
Complementary cosine identity: \(\cos(\pi-x)=-\cos x\).
Logarithmic combination: \(\log(1+\cos x)+\log(1-\cos x)=2\log\sin x\).
Standard result: \(\displaystyle\int\limits_0^{\pi/2}\log\sin x\,dx=-\frac{\pi}{2}\log 2\).
Final evaluation: \(\displaystyle\int\limits_0^\pi\log(1+\cos x)\,dx=-\pi\log 2\).

📘 Concept & Theory

Theory and Concept

›

This problem demonstrates the application of the symmetry property of definite integrals. The numerator contains \(\sqrt{x}\), while the denominator contains the symmetric expression \(\sqrt{x}+\sqrt{a-x}\). Replacing \(x\) with \(a-x\) interchanges the two square-root terms in the numerator and denominator.

Concept 1: Symmetry Property of Definite Integrals

For a suitable integrable function \(f(x)\),

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx \]

For the interval \([0,a]\), the property becomes

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

Concept 2: Symmetric Expressions

Under the transformation \(x\rightarrow a-x\),

\[ \sqrt{x}\longleftrightarrow\sqrt{a-x} \]

Therefore, the transformed numerator becomes \(\sqrt{a-x}\), while the denominator remains unchanged.

Concept 3: Complementary Fractions

For non-negative quantities \(u\) and \(v\),

\[ \frac{u}{u+v}+\frac{v}{u+v}=1 \]

This identity is the central step in evaluating the given integral.

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the given integral to be \(I\).
Apply the symmetry property using \(x\rightarrow a-x\).
Write the transformed integral with numerator \(\sqrt{a-x}\).
Add the original and transformed integrals.
Simplify the numerator and evaluate the resulting constant integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Assume the given integral to be \(I\).
\[I=\int\limits_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx\]
Apply the symmetry property of definite integrals
Using the substitution
\[x=a-t\]
we get
\[dx=-dt\]
The limits change as follows:
- When \(x=0\), \(t=a\).
- When \(x=a\), \(t=0\).
Reversing the limits,
\[ \begin{aligned} I &=\int\limits_0^a \frac{\sqrt{a-t}} {\sqrt{a-t}+\sqrt{a-(a-t)}}\,dt\\[8pt] &=\int\limits_0^a \frac{\sqrt{a-t}} {\sqrt{a-t}+\sqrt{t}}\,dt \end{aligned} \]
Renaming the dummy variable \(t\) as \(x\),
\[I=\int\limits_0^a\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx\]
Add the original and transformed integrals
\[ \begin{aligned} 2I &=\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx\\ &\quad+ \int\limits_0^a \frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx \end{aligned} \]
Combining the two integrals,
\[ \begin{aligned} 2I &=\int\limits_0^a \left[ \frac{\sqrt{x}+\sqrt{a-x}} {\sqrt{x}+\sqrt{a-x}} \right]dx\\[8pt] &=\int\limits_0^a 1\,dx\\[6pt] &=\left[x\right]_0^a\\[6pt] &=a \end{aligned} \]
Determine the value of the integral
\[2I=a\]
Therefore,
\[\boxed{I=\frac{a}{2}}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx=\frac{a}{2}}\]

Note: The result is valid for \(a>0\), as required for the given real-valued square roots over the interval \([0,a]\).

🎯 Exam Significance

Exam Significance

›

Symmetry property: Demonstrates the application of \(x\rightarrow a-x\) in definite integration.
Radical expressions: Develops the ability to simplify integrals involving square roots of complementary expressions.
Algebraic simplification: Shows how adding two equivalent integrals can eliminate a complicated fraction.
Conceptual approach: Evaluates the integral without requiring direct antiderivative calculation.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Symmetric integrands: Recognising complementary expressions is an important technique in definite integration.
Complementary fractions: The identity \(\frac{u}{u+v}+\frac{v}{u+v}=1\) provides an immediate simplification.
Parameter-dependent integrals: The result illustrates how a parameter \(a\) determines the value of a definite integral.
Time-efficient problem solving: A suitable symmetry transformation avoids lengthy integration techniques.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Symmetry transformation: \(x\rightarrow a-x\).
Complementary square roots: \(\sqrt{x}\) and \(\sqrt{a-x}\) interchange under the transformation.
Denominator: \(\sqrt{x}+\sqrt{a-x}\) remains unchanged.
Complementary fractions: The sum of the original and transformed integrands is \(1\).
Final result: \(\displaystyle I=\frac{a}{2}\).

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the evaluation of a definite integral involving an absolute value function. The absolute value expression changes its sign at \(x=1\), so the integral must be considered over two separate intervals.

Concept 1: Absolute Value Function

The absolute value of a real number is defined as

\[ |x-1|= \begin{cases} 1-x, & x < 1\\ x-1, & x\geq 1 \end{cases} \]

Thus, the expression inside the absolute value is negative on \([0,1)\) and non-negative on \([1,4]\).

Concept 2: Splitting a Definite Integral

If \(a

\[ \int\limits_a^b f(x)\,dx = \int\limits_a^c f(x)\,dx+\int\limits_c^b f(x)\,dx \]

We use this property to split the integral at the point where the expression inside the absolute value changes sign.

Concept 3: Geometrical Interpretation

The integral of \(|x-1|\) represents the area between the graph \(y=|x-1|\) and the \(x\)-axis over the interval \([0,4]\). The graph consists of two straight-line segments meeting at the vertex \((1,0)\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the point where \(x-1=0\).
Express the absolute value function separately over \([0,1]\) and \([1,4]\).
Split the original integral at \(x=1\).
Evaluate the two resulting integrals.
Add their values to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Identify the critical point
The expression inside the absolute value becomes zero when
\[x-1=0\]

\[x=1\]
Therefore,
\[ |x-1|= \begin{cases} 1-x, & 0\leq x < 1\\ x-1, & 1\leq x\leq 4 \end{cases} \]
Split the integral at \(x=1\)
\[ \begin{aligned} I &=\int\limits_0^1|x-1|\,dx+\int\limits_1^4|x-1|\,dx\\[6pt] &=\int\limits_0^1(1-x)\,dx+\int\limits_1^4(x-1)\,dx \end{aligned} \]
Evaluate the first integral
\[ \begin{aligned} I_1 &=\int\limits_0^1(1-x)\,dx\\[6pt] &=\left[x-\frac{x^2}{2}\right]_0^1\\[6pt] &=\left(1-\frac12\right)-0\\[6pt] &=\frac12 \end{aligned} \]
Evaluate the second integral
\[ \begin{aligned} I_2 &=\int\limits_1^4(x-1)\,dx\\[6pt] &=\left[\frac{x^2}{2}-x\right]_1^4\\[6pt] &=\left(\frac{16}{2}-4\right) -\left(\frac12-1\right)\\[6pt] &=(8-4)-\left(-\frac12\right)\\[6pt] &=4+\frac12\\[6pt] &=\frac92 \end{aligned} \]
Add the two contributions
\[ \begin{aligned} I &=I_1+I_2\\[6pt] &=\frac12+\frac92\\[6pt] &=\frac{10}{2}\\[6pt] &=5 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^4|x-1|\,dx=5}\]

🎯 Exam Significance

Exam Significance

›

Absolute value functions: Tests the ability to express a modulus function as a piecewise-defined function.
Properties of definite integrals: Reinforces the technique of splitting an integral at a critical point.
Geometrical interpretation: Connects definite integration with the area under a curve.
Algebraic accuracy: Develops proficiency in evaluating polynomial integrals over different intervals.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Piecewise functions: Recognising sign changes is essential when evaluating integrals involving modulus expressions.
Area calculations: The problem illustrates how a definite integral of a non-negative function represents geometrical area.
Critical-point analysis: Identifying the zero of the expression inside the modulus is the first step in solving such problems.
Multiple approaches: The integral can be evaluated algebraically or verified using geometrical area.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Critical point: \(x=1\), where the expression inside the modulus changes sign.
Piecewise representation: \(|x-1|=1-x\) for \(0\leq x<1\), and \(|x-1|=x-1\) for \(1\leq x\leq4\).
Integral splitting: Split the interval at the critical point before integrating.
Geometrical interpretation: The integral represents the sum of the areas of two triangles.
Final result: \(\displaystyle I=5\).

📘 Concept & Theory

Theory and Concept

›

Concept 1: Symmetry Property of Definite Integrals

For a suitable integrable function \(F(x)\),

\[ \int\limits_0^a F(x)\,dx = \int\limits_0^a F(a-x)\,dx \]

This property is obtained by substituting \(x=a-t\) and reversing the limits of integration.

Concept 2: Symmetric Function

The condition

\[ f(x)=f(a-x) \]

means that \(f(x)\) is symmetric about \(x=\frac a2\). Its value remains unchanged when \(x\) is replaced by \(a-x\).

Concept 3: Complementary Function Values

The second condition

\[ g(x)+g(a-x)=4 \]

states that the sum of the values of \(g\) at two complementary points is always equal to 4.

Combining these two conditions allows the product integral to be simplified into an integral involving only \(f(x)\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Assume the left-hand side integral to be \(I\).
Apply the symmetry property by replacing \(x\) with \(a-x\).
Use the given conditions to simplify the transformed integral.
Add the original and transformed expressions.
Obtain the required identity.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Assume the given integral to be \(I\)
\[I=\int\limits_0^a f(x)g(x)\,dx\]
Apply the symmetry property of definite integrals
Replacing \(x\) with \(a-x\), we obtain
\[\begin{aligned}I&=\int\limits_0^a f(a-x)g(a-x)\,dx\end{aligned}\]
Using the given condition
\[f(a-x)=f(x)\]
we get
\[I=\int\limits_0^a f(x)g(a-x)\,dx\]
Add the original and transformed integrals
We have
\[I=\int\limits_0^a f(x)g(x)\,dx\]
and
\[I=\int\limits_0^a f(x)g(a-x)\,dx\]
Adding these two equations,
\[ \begin{aligned} 2I &=\int\limits_0^a f(x)g(x)\,dx\\ &\quad+\int\limits_0^a f(x)g(a-x)\,dx \end{aligned} \]
Combining the integrals,
\[\begin{aligned}2I&=\int\limits_0^af(x)\left[g(x)+g(a-x)\right]dx\end{aligned}\]
Apply the second given condition
Since
\[g(x)+g(a-x)=4\]
we obtain
\[ \begin{aligned} 2I &=\int\limits_0^a f(x)\cdot4\,dx\\[6pt] &=4\int\limits_0^a f(x)\,dx \end{aligned} \]
Dividing both sides by 2,
\[I=2\int\limits_0^a f(x)\,dx\]
Since \(I=\displaystyle\int\limits_0^a f(x)g(x)\,dx\), we have proved that
\[\boxed{\displaystyle\int\limits_0^a f(x)g(x)\,dx=2\int\limits_0^a f(x)\,dx}\]
Hence Proved

🎯 Exam Significance

Exam Significance

›

Properties of definite integrals: Tests the ability to apply the substitution \(x\rightarrow a-x\).
Symmetric functions: Reinforces the interpretation of \(f(x)=f(a-x)\).
Algebraic manipulation: Demonstrates how adding two equivalent integrals simplifies a complicated expression.
Proof-based questions: Develops the logical sequence required to establish mathematical identities.

For JEE Main, JEE Advanced, CUET and Other Competitive Entrance Examinations:

Functional symmetry: Recognising complementary function values is useful in evaluating definite integrals.
Integral transformations: The substitution \(x\rightarrow a-x\) is a powerful technique for simplifying integrals involving multiple functions.
Generalisation: The proof illustrates how functional conditions can determine the value of an integral without knowing explicit formulas for the functions.
Efficient reasoning: The result is established without calculating any antiderivative.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Symmetry of \(f(x)\): \(f(x)=f(a-x)\).
Complementary values of \(g(x)\): \(g(x)+g(a-x)=4\).
Integral transformation: \(\displaystyle\int\limits_0^a f(x)g(x)\,dx=\int\limits_0^a f(x)g(a-x)\,dx\).
Addition technique: Adding the two equivalent integrals produces the constant factor 4.
Required identity: \(\displaystyle\int\limits_0^a f(x)g(x)\,dx=2\int\limits_0^a f(x)\,dx\).

Q20

LONG ANSWER4 marks

Choose the Correct Answer by evaluating \[ I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^3+x\cos x+\tan^5x+1\right)\,dx \]

\(0\)
\(2\)
\(\pi\)
\(1\)

📘 Concept & Theory

Theory and Concept

›

This question tests the properties of odd and even functions over an interval symmetric about the origin.

Odd Function: A function \(f(x)\) is odd if

\[ f(-x)=-f(x) \]

For an odd function, its integral over symmetric limits is zero, provided the integral exists:

\[ \int\limits_{-a}^{a}f(x)\,dx=0 \]

Even Function: A function \(f(x)\) is even if

\[ f(-x)=f(x) \]

For an even function:

\[ \int\limits_{-a}^{a}f(x)\,dx=2\int\limits_0^a f(x)\,dx \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the odd and even terms in the integrand.
Use symmetry to evaluate the odd terms in the Cauchy principal-value sense.
Evaluate the constant term over the given interval.
Check the convergence of the improper integral because \(\tan x\) is undefined at the endpoints.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[ I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^3+x\cos x+\tan^5x+1\right)\,dx \]
Identify the parity of each term
(i) \(x^3\) is an odd function.
\[(-x)^3=-x^3\]
(ii) \(x\cos x\) is an odd function.
Since \(x\) is odd and \(\cos x\) is even, their product is odd
\[(-x)\cos(-x)=-x\cos x\]
(iii) \(\tan^5x\) is an odd function
\[\tan^5(-x)=-\tan^5x\]
(iv) \(1\) is an even function
\[1=1\]
Apply symmetry
In the Cauchy principal-value sense, the odd terms cancel over symmetric limits:
\[\operatorname{PV}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x^3\,dx=0\]

\[\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\cos x\,dx=0\]

\[\operatorname{PV}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\tan^5x\,dx=0\]
Evaluate the constant term
\[ \begin{aligned} \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1\,dx &=\left[x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\ &=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\\ &=\pi \end{aligned} \]
Check the endpoint behaviour
The term \(\tan^5x\) has singularities at \(x=\pm\frac{\pi}{2}\). Near \(x=\frac{\pi}{2}\), it behaves like a constant multiple of \((\frac{\pi}{2}-x)^{-5}\). Consequently, the corresponding one-sided improper integral diverges. The same issue occurs at the negative endpoint.

Therefore, the ordinary improper integral does not converge. Its Cauchy principal value is:

\[ \operatorname{PV}(I)=0+0+0+\pi=\pi \]

💡 Answer

Final Answer

›

Option (C): \(\pi\), if the intended interpretation is the Cauchy principal value.

🎯 Exam Significance

Exam Significance

›

Tests the identification of odd and even functions.
Reinforces the use of symmetry over intervals of the form \([-a,a]\).
Highlights the distinction between symmetry-based cancellation and convergence of improper integrals.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

\(x^3\), \(x\cos x\), and \(\tan^5x\) are odd functions.
The constant function \(1\) is even.
Symmetry gives zero for the odd terms only when the relevant integral exists, or when explicitly using the Cauchy principal value.
Endpoint singularities must be checked before applying definite-integral properties.

📘 Concept & Theory

Theory and Concept

›

We use the symmetry property of definite integrals:

\[ \int\limits_0^a f(x)\,dx = \int\limits_0^a f(a-x)\,dx \]

In particular, for the interval \(\left[0,\frac{\pi}{2}\right]\), we can replace \(x\) by \(\frac{\pi}{2}-x\), using:

\[ \sin\left(\frac{\pi}{2}-x\right)=\cos x \]

\[ \cos\left(\frac{\pi}{2}-x\right)=\sin x \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the substitution \(x\mapsto\frac{\pi}{2}-x\).
Obtain a second expression for the same integral.
Add the two expressions and simplify the logarithms.
Determine the value of the integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 3 steps

Given —
\[ I=\int\limits_0^{\frac{\pi}{2}} \log\left(\frac{4+3\sin x}{4+3\cos x}\right)\,dx \]
Apply the symmetry property
Replacing \(x\) by \(\frac{\pi}{2}-x\), we get:
\[ \begin{aligned} I&=\int\limits_0^{\frac{\pi}{2}} \log\left( \frac{4+3\sin\left(\frac{\pi}{2}-x\right)} {4+3\cos\left(\frac{\pi}{2}-x\right)} \right)\,dx\\[6pt] &=\int\limits_0^{\frac{\pi}{2}} \log\left(\frac{4+3\cos x}{4+3\sin x}\right)\,dx \end{aligned} \]
Add the two expressions
Adding the original and transformed expressions:
\[ \begin{aligned} 2I &=\int\limits_0^{\frac{\pi}{2}} \left[ \log\left(\frac{4+3\sin x}{4+3\cos x}\right) + \log\left(\frac{4+3\cos x}{4+3\sin x}\right) \right]dx\\[6pt] &=\int\limits_0^{\frac{\pi}{2}} \log\left[ \frac{4+3\sin x}{4+3\cos x} \cdot \frac{4+3\cos x}{4+3\sin x} \right]dx\\[6pt] &=\int\limits_0^{\frac{\pi}{2}}\log 1\,dx\\[6pt] &=\int\limits_0^{\frac{\pi}{2}}0\,dx\\[6pt] &=0 \end{aligned} \]
Calculate the required value
\[2I=0\]

\[\boxed{I=0}\]

💡 Answer

Final Answer

›

Option (C):

\[\boxed{0}\]

🎯 Exam Significance

Exam Significance

›

Tests the symmetry property of definite integrals.
Demonstrates how logarithmic identities simplify definite integrals.
Illustrates an efficient approach to MCQs without evaluating the integral directly.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 3 points

For an integral over \(\left[0,\frac{\pi}{2}\right]\), the substitution \(x\mapsto\frac{\pi}{2}-x\) interchanges sine and cosine.
The logarithm of a reciprocal is the negative of the original logarithm.
When the transformed integrand is the negative of the original integrand, the integral is zero.

Integrals

Chapter Complete!

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INTEGRALS — Learning Resources

Frequently Asked Questions

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Integrals

Chapter Complete!

Recent posts

INTEGRALS — Learning Resources

Frequently Asked Questions

Q 01 What is covered in NCERT Class 12 Mathematics Chapter 7 Exercise 7.10?

Q 02 How many questions are included in Integrals Exercise 7.10?

Q 03 What are the important properties of definite integrals used in Exercise 7.10?

Q 04 How is the symmetry property of definite integrals applied?

Q 05 How can odd and even functions simplify definite integrals?

Q 06 Why is the substitution x ? p-x useful in definite integrals?

Q 07 Why is the substitution x ? p/2-x useful in trigonometric integrals?

Q 08 Are definite integrals important for the CBSE Class 12 Mathematics Board Exam?

Q 09 Is NCERT Integrals Exercise 7.10 useful for JEE Main and JEE Advanced?

Q 10 How should students prepare for Integrals Exercise 7.10?

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