Integrals

📘 Concept & Theory

Theory and Concept

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This problem involves integrating the square of a trigonometric function whose argument is a linear expression in \(x\). The most effective approach is to use the power-reduction identity for \(\sin^2\theta\), followed by the standard integration formula for cosine.

1. Power-reduction identity

The square of the sine function can be expressed in terms of the cosine of twice the angle:

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

2. Standard integration formulas

\[ \int 1\,dx=x+C \]

\[ \int \cos(ax+b)\,dx =\frac{\sin(ax+b)}{a}+C,\qquad a\ne 0 \]

3. Integration by substitution

When the argument of a trigonometric function is a linear expression, such as \(ax+b\), substitution can simplify the integral. Letting \(t=ax+b\) gives \(dt=a\,dx\), and therefore \(dx=\frac{dt}{a}\).

Important observation: The factor \(2\) in the double-angle identity must be applied to the entire argument \(2x+5\). Thus, \(2(2x+5)=4x+10\).

🗺️ Solution Roadmap

Step-by-step Plan

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Apply the identity \(\sin^2\theta=\frac{1-\cos 2\theta}{2}\).
Substitute \(\theta=2x+5\) and simplify the double-angle expression.
Separate the integral into a constant term and a cosine term.
Integrate the constant term directly.
Integrate the cosine term using substitution or the standard formula.
Combine the results and include the constant of integration \(C\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given —
\[I=\int\sin^2(2x+5)\,dx\]
Using
\[\sin^2\theta=\frac{1-\cos 2\theta}{2}\]
Let \(\theta=2x+5\) We obtain
\[\sin^2(2x+5)=\frac{1-\cos\left[2(2x+5)\right]}{2}\]
Simplify the argument of the cosine function.
Since
\[2(2x+5)=4x+10\]
the integral becomes
\[I=\int\frac{1-\cos(4x+10)}{2}\,dx\]
Separate the integral
Using the linearity of integration, we get
\[I=\frac12\int 1\,dx-\frac12\int\cos(4x+10)\,dx\]
Integrate the constant term
Since the integral of \(1\) with respect to \(x\) is \(x\),
\[\frac12\int 1\,dx=\frac{x}{2}\]
Therefore,
\[I=\frac{x}{2}-\frac12\int\cos(4x+10)\,dx\]
Apply substitution to the remaining integral.
Let
\[t=4x+10\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=4\]
Hence,
\[dt=4\,dx\]

\[dx=\frac{dt}{4}\]
Substitute into the integral
Using \(t=4x+10\) and \(dx=\frac{dt}{4}\),
\[I=\frac{x}{2}-\frac12\int\cos t\,\frac{dt}{4}\]
Taking the constant factor outside the integral,
\[I=\frac{x}{2}-\frac18\int\cos t\,dt\]
Integrate the cosine function
Using the standard formula
\[\int\cos t\,dt=\sin t+C\]
we obtain
\[I=\frac{x}{2}-\frac18\sin t+C\]
Substitute back \(t=4x+10\)
I=\frac{x}{2}-\frac18\sin(4x+10)+C

💡 Answer

Final Answer

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\[ \boxed{ \int\sin^2(2x+5)\,dx =\frac{x}{2}-\frac18\sin(4x+10)+C } \]

🎯 Exam Significance

Exam Significance

›

Application of identities: Tests the ability to transform a squared trigonometric function into a form suitable for integration.
Integration techniques: Reinforces the linearity of integration, standard formulas, and substitution.
Stepwise presentation: Writing the identity, substitution, integration, and final result separately makes the mathematical reasoning clear and helps avoid losing marks for omitted steps.
Verification: Differentiating the answer provides an independent check of the result.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Develops fluency in trigonometric identities and integration of composite functions, techniques that also appear in more complex integrals.
Time-efficient calculation: Recognising the power-reduction identity avoids unnecessary expansion or more complicated methods.
Chain-rule awareness: The factor \(4\) arising from differentiating \(4x+10\) illustrates why the coefficient of the inner linear expression matters.
Error prevention: The problem tests careful handling of numerical factors, negative signs, and the constant of integration.

🚫 Common Mistakes

Common Mistakes

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Forgetting the factor \(\frac12\) in the power-reduction identity.
Writing \(\cos(2x+10)\) instead of \(\cos(4x+10)\) after applying the double-angle identity.
Omitting the factor \(\frac14\) when integrating \(\cos(4x+10)\).
Incorrectly handling the negative sign before the cosine integral.
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

For squared sine functions, consider the identity
\[ \sin^2\theta=\frac{1-\cos 2\theta}{2} \]
Apply the double-angle operation to the entire argument:
\[ 2(2x+5)=4x+10 \]
Keep the factor \(\frac12\) outside the cosine integral when splitting the expression.
For a linear argument \(ax+b\), use
\[ \int\cos(ax+b)\,dx=\frac{\sin(ax+b)}{a}+C \]
Remember to include the constant of integration \(C\) in every indefinite integral.
Verify an indefinite integral by differentiating the proposed antiderivative.

📘 Concept & Theory

Theory and Concept

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This problem involves integrating the product of two trigonometric functions with different angles. Direct integration of the product is not straightforward, so we use a product-to-sum identity to convert the product into a sum or difference of sine functions.

Product-to-sum identity:

\[ \sin A\cos B =\frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right] \]

For the given integral, we take \(A=3x\) and \(B=4x\). Therefore,

\[ \sin 3x\cos 4x =\frac12\left[\sin(7x)+\sin(-x)\right] \]

Since sine is an odd function,

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

Hence, the product simplifies to

\[ \sin 3x\cos 4x =\frac12(\sin 7x-\sin x) \]

We can now integrate the two terms separately using the standard formula

\[ \int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C, \qquad a\ne 0 \]

🗺️ Solution Roadmap

Step-by-step Plan

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Identify the product of the sine and cosine functions.
Apply the product-to-sum identity with \(A=3x\) and \(B=4x\).
Simplify the resulting expression using \(\sin(-x)=-\sin x\).
Separate the integral into two simpler sine integrals.
Integrate each term, carefully accounting for the coefficients \(7\) and \(1\) in their arguments.
Combine the results and add the constant of integration.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Given —
\[I=\int\sin 3x\cos 4x\,dx\]
Apply the product-to-sum identity
Using
\[ \sin A\cos B =\frac12\left[\sin(A+B)+\sin(A-B)\right] \]
and substituting \(A=3x\) and \(B=4x\), we obtain
\[ I=\frac12\int \left[\sin(3x+4x)+\sin(3x-4x)\right]dx \]
Simplify the angles
Since
\[3x+4x=7x\]

\[3x-4x=-x\]
the integral becomes
\[I=\frac12\int\left[\sin 7x+\sin(-x)\right]dx\]
Use the odd-function property of sine.
\[\sin(-x)=-\sin x\]
Therefore,
\[I=\frac12\int(\sin 7x-\sin x)\,dx\]
Separate the integral
By the linearity of integration,
\[ I=\frac12\int\sin 7x\,dx -\frac12\int\sin x\,dx \]
Evaluate the first integral.
Consider
\[I_1=\int\sin 7x\,dx\]
Let
\[t=7x\]
Differentiating with respect to \(x\),
\[\frac{dt}{dx}=7\]

\[dt=7\,dx\]

\[dx=\frac{dt}{7}\]
Substituting into the integral,
\[I_1=\int\sin t\,\frac{dt}{7}\]
Taking the constant factor outside,
\[I_1=\frac17\int\sin t\,dt\]
Since
\[\int\sin t\,dt=-\cos t+C\]
we obtain
\[I_1=-\frac17\cos t+C\]
Substituting \(t=7x\) back,
\[\int\sin 7x\,dx=-\frac17\cos 7x+C\]
Consequently, the first term in the original integral is
\[ \frac12\int\sin 7x\,dx =-\frac1{14}\cos 7x \]
Evaluate the second integral
Using the standard formula,
\[\int\sin x\,dx=-\cos x+C\]
Therefore,
\[ -\frac12\int\sin x\,dx =-\frac12(-\cos x) \]
Simplifying the two negative signs,
\[-\frac12\int\sin x\,dx=\frac12\cos x\]
Combine the results
Substituting both evaluated integrals into the expression for \(I\),
\[I=-\frac1{14}\cos 7x+\frac12\cos x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\sin 3x\cos 4x\,dx=-\frac1{14}\cos 7x+\frac12\cos x+C}\]

🎯 Exam Significance

Exam Significance

›

Trigonometric identities: Tests the correct application of product-to-sum transformations.
Integration formulas: Reinforces the standard integral of \(\sin ax\) and the importance of the coefficient \(a\).
Stepwise presentation: Showing the identity, angle simplification, separation of integrals, and final result makes the reasoning clear and helps avoid errors.
Sign management: Demonstrates why \(\sin(-x)=-\sin x\) and why the negative sign in the second integral produces a positive cosine term.
Verification skills: Differentiation confirms that the proposed antiderivative is correct.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Product-to-sum identities are useful for simplifying trigonometric integrals and recognising equivalent forms.
Efficient problem-solving: Converting a product into a sum reduces the problem to standard integration formulas.
Coefficient awareness: The argument \(7x\) introduces a factor of \(\frac17\) during integration, a frequent source of errors in timed examinations.
Algebraic precision: Correct handling of negative angles and signs helps build accuracy in multi-step problems.
Transferable technique: The same identity can be applied to integrals involving other combinations of sine and cosine functions.

🚫 Common Mistakes

Common Mistakes

›

Using an incorrect product-to-sum identity or omitting the factor \(\frac12\).
Writing \(3x-4x=x\) instead of \(3x-4x=-x\).
Forgetting that \(\sin(-x)=-\sin x\).
Writing \(-\frac17\cos 7x\) for the first integrated term without multiplying by the outside factor \(\frac12\).
Giving the second term as \(-\frac12\cos x\) instead of \(+\frac12\cos x\).
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use the product-to-sum identity:
\[ \sin A\cos B =\frac12[\sin(A+B)+\sin(A-B)] \]
Remember that sine is an odd function:
\[ \sin(-x)=-\sin x \]
For a nonzero constant \(a\),
\[ \int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C \]
When integrating \(\sin 7x\), divide by \(7\), the coefficient of \(x\).
Keep track of the negative sign when integrating \(-\sin x\).
Include the constant of integration \(C\) in the final result.
Differentiate the final answer whenever possible to check its correctness.

📘 Concept & Theory

Theory and Concept

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This problem involves integrating the product of three cosine functions with different angles. Direct integration of this product is difficult, so we simplify the integrand by repeatedly applying product-to-sum identities and the double-angle identity.

1. Product-to-sum identity

\[ \cos A\cos B =\frac12\left[\cos(A+B)+\cos(A-B)\right] \]

2. Double-angle identity

\[ \cos 2A=2\cos^2 A-1 \]

Rearranging this identity gives

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

3. Standard integration formula

\[ \int\cos(ax)\,dx=\frac{\sin(ax)}{a}+C, \qquad a\ne 0 \]

These identities allow us to transform a product of cosine functions into a sum of individual cosine terms, each of which can be integrated directly.

Important observation: After expanding the product, the constant term must be integrated as \(\int 1\,dx=x+C\). It is not multiplied by \(\frac14\) unless that factor is explicitly outside the entire term.

🗺️ Solution Roadmap

Step-by-step Plan

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Apply the product-to-sum identity to \(\cos 2x\cos 4x\).
Multiply the resulting expression by \(\cos 6x\).
Apply the product-to-sum identity to \(\cos 2x\cos 6x\).
Use the identity for \(\cos^2 6x\) to simplify the squared cosine term.
Combine all terms and carefully retain their numerical coefficients.
Integrate each term separately.
Verify the final result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Given —
\[I=\int\cos 2x\cos 4x\cos 6x\,dx\]
Apply the product-to-sum identity to the first two factors.
Using
\[\cos A\cos B=\frac12[\cos(A+B)+\cos(A-B)]\]
with \(A=2x\) and \(B=4x\), we get
\[\cos 2x\cos 4x=\frac12[\cos(2x+4x)+\cos(2x-4x)]\]
Simplifying the angles,
\[\cos 2x\cos 4x=\frac12[\cos 6x+\cos(-2x)]\]
Since cosine is an even function, \(\cos(-2x)=\cos 2x\). Therefore,
\[\cos 2x\cos 4x=\frac12(\cos 6x+\cos 2x)\]
Substitute this expression into the original integral.
\[I=\int\frac12(\cos 6x+\cos 2x)\cos 6x\,dx\]
Taking the constant factor outside the integral and distributing \(\cos 6x\),
\[I=\frac12\int[\cos^2 6x+\cos 2x\cos 6x]\,dx\]
Expand the product \(\cos 2x\cos 6x\).
Applying the product-to-sum identity again, with \(A=2x\) and \(B=6x\),
\[\cos 2x\cos 6x=\frac12[\cos(2x+6x)+\cos(2x-6x)]\]
Simplifying,
\[\cos 2x\cos 6x=\frac12[\cos 8x+\cos(-4x)]\]
Since \(\cos(-4x)=\cos 4x\),
\[\cos 2x\cos 6x=\frac12(\cos 8x+\cos 4x)\]
Substituting this result,
\[I=\frac12\int\left[\cos^2 6x+\frac12(\cos 8x+\cos 4x)\right]dx\]
Simplify the squared cosine term.
Using the double-angle identity,
\[\cos 2A=2\cos^2 A-1\]
Rearranging,
\[\cos^2 A=\frac{1+\cos 2A}{2}\]
Putting \(A=6x\), we obtain
\[\cos^2 6x=\frac{1+\cos 12x}{2}\]
Substituting into the integral,
\[I=\frac12\int\left[\frac{1+\cos 12x}{2}+\frac12(\cos 8x+\cos 4x)\right]dx\]
Distribute the factor \(\frac12\) carefully.
Multiplying each term inside the brackets by \(\frac12\),
\[I=\int\left[\frac14+\frac14\cos 12x+\frac14\cos 8x+\frac14\cos 4x\right]dx\]
Separating the integral into individual terms,
\[I=\frac14\int dx+\frac14\int\cos 12x\,dx+\frac14\int\cos 8x\,dx+\frac14\int\cos 4x\,dx\]
Integrate each term separately
For the constant term,
\[\frac14\int dx=\frac{x}{4}\]
For the second term, using the standard formula,
\[\frac14\int\cos 12x\,dx=\frac14\cdot\frac{\sin 12x}{12}=\frac1{48}\sin 12x\]
For the third term,
\[\frac14\int\cos 8x\,dx=\frac14\cdot\frac{\sin 8x}{8}=\frac1{32}\sin 8x\]
For the fourth term,
\[\frac14\int\cos 4x\,dx=\frac14\cdot\frac{\sin 4x}{4}=\frac1{16}\sin 4x\]
Combine all the integrated terms
\[I=\frac{x}{4}+\frac1{48}\sin 12x+\frac1{32}\sin 8x+\frac1{16}\sin 4x+C\]

💡 Answer

Final Answer

›

\[\boxed{\begin{aligned}\int\cos 2x\cos 4x\cos 6x\,dx={}&\frac{x}{4}+\frac1{48}\sin 12x\\[6pt] &+\frac1{32}\sin 8x+\frac1{16}\sin 4x+C\end{aligned}}\]

🎯 Exam Significance

Exam Significance

›

Identity-based integration: Demonstrates how product-to-sum and double-angle identities simplify a product of three trigonometric functions.
Logical presentation: Applying one identity at a time makes the derivation easier to follow and supports stepwise marking.
Coefficient accuracy: Reinforces the need to retain every numerical factor when expanding expressions and integrating.
Standard integration formulas: Tests the ability to integrate cosine functions with linear arguments.
Checking the answer: Differentiation confirms the validity of the final antiderivative.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Builds fluency in simplifying products of multiple trigonometric functions, a useful technique in integration problems.
Efficient simplification: Converting the product into a sum reduces a complicated integral to standard forms.
Algebraic precision: The repeated transformations test accuracy in angle addition, angle subtraction, and coefficient handling.
Pattern recognition: Recognising squared cosine terms allows the use of double-angle identities without unnecessary expansion.
Error control under time pressure: Keeping the constant term and the factors \(\frac12\) and \(\frac14\) explicit helps prevent common mistakes.

🚫 Common Mistakes

Common Mistakes

›

Omitting the factor \(\frac12\) in the product-to-sum identity.
Forgetting that \(\cos(-\theta)=\cos\theta\).
Applying the double-angle identity incorrectly to \(\cos^2 6x\).
Distributing the outside factor \(\frac12\) incorrectly in Step 6.
Writing \(x\) instead of \(\frac{x}{4}\) for the integral of the constant term \(\frac14\).
Forgetting to divide by \(12\), \(8\), or \(4\) when integrating the respective cosine terms.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use the product-to-sum identity:
\[ \cos A\cos B =\frac12[\cos(A+B)+\cos(A-B)] \]
Remember that cosine is an even function:
\[ \cos(-\theta)=\cos\theta \]
Convert the squared cosine term using:
\[ \cos^2 A=\frac{1+\cos 2A}{2} \]
Integrate the constant term correctly:
\[ \int\frac14\,dx=\frac{x}{4} \]
For any nonzero constant \(a\),
\[ \int\cos(ax)\,dx=\frac{\sin(ax)}{a}+C \]
Keep track of all numerical factors when applying identities and integrating.
Verify the final expression by differentiation whenever possible.

📘 Concept & Theory

Theory and Concept

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This problem involves integrating the cube of a sine function whose argument is a linear expression in \(x\). Direct integration of the cube can be avoided by using a trigonometric identity to express \(\sin^3\theta\) in terms of \(\sin\theta\) and \(\sin 3\theta\).

1. Triple-angle identity for sine

The triple-angle identity is

\[ \sin 3\theta=3\sin\theta-4\sin^3\theta \]

Rearranging to express the cube of sine,

\[ 4\sin^3\theta=3\sin\theta-\sin 3\theta \]

\[ \sin^3\theta=\frac14(3\sin\theta-\sin 3\theta) \]

2. Standard integration formula

\[ \int\sin(ax+b)\,dx =-\frac{\cos(ax+b)}{a}+C, \qquad a\ne 0 \]

The coefficient \(a\) in the argument must be accounted for when integrating. In this question, the two sine terms have arguments \(2x+1\) and \(6x+3\), so their respective coefficients of \(x\) are \(2\) and \(6\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Apply the triple-angle identity to express \(\sin^3\theta\) as a linear combination of sine functions.
Substitute \(\theta=2x+1\) into the identity.
Expand the triple angle \(3(2x+1)\).
Separate the integral into two simpler sine integrals.
Integrate each term, dividing by the coefficient of \(x\) in its argument.
Combine the terms and include the constant of integration.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 26 steps

Given —
\[I=\int\sin^3(2x+1)\,dx\]
\[\sin 3\theta=3\sin\theta-4\sin^3\theta\]
Rearranging,
\[4\sin^3\theta=3\sin\theta-\sin 3\theta\]
Dividing both sides by \(4\),
\[\sin^3\theta=\frac14(3\sin\theta-\sin 3\theta)\]
Apply the identity to the given integral.
Putting \(\theta=2x+1\), we obtain
\[\sin^3(2x+1)=\frac14\left[3\sin(2x+1)-\sin\big(3(2x+1)\big)\right]\]
Substituting this expression into the integral,
\[I=\frac14\int\left[3\sin(2x+1)-\sin\big(3(2x+1)\big)\right]dx\]
Simplify the triple-angle argument
Expanding the expression inside the second sine function,
\[3(2x+1)=6x+3\]
Therefore,
\[I=\frac14\int[3\sin(2x+1)-\sin(6x+3)]\,dx\]
Separate the integral
Using the linearity of integration,
\[I=\frac34\int\sin(2x+1)\,dx-\frac14\int\sin(6x+3)\,dx\]
Integrate the first term.
Consider
\[I_1=\int\sin(2x+1)\,dx\]
Let
\[t=2x+1\]
Differentiating both sides,
\[\frac{dt}{dx}=2\]

\[dt=2\,dx\]

\[dx=\frac{dt}{2}\]
Substituting into the integral,
\[I_1=\int\sin t\,\frac{dt}{2}\]
Taking the constant factor outside,
\[I_1=\frac12\int\sin t\,dt\]
Since
\[\int\sin t\,dt=-\cos t+C\]
we get
\[I_1=-\frac12\cos t+C\]
Substituting \(t=2x+1\) back,
\[\int\sin(2x+1)\,dx=-\frac12\cos(2x+1)+C\]
Multiplying by the coefficient \(\frac34\), the first term becomes
\[\frac34\int\sin(2x+1)\,dx=-\frac34\cdot\frac12\cos(2x+1)\]

\[=-\frac38\cos(2x+1)\]
Integrate the second term
Consider
\[I_2=\int\sin(6x+3)\,dx\]
Let
\[t=6x+3\]
Differentiating
\[\frac{dt}{dx}=6\]

\[dt=6\,dx\]

\[dx=\frac{dt}{6}\]
Substituting into the integral,
\[I_2=\int\sin t\,\frac{dt}{6}\]
Taking the constant factor outside,
\[I_2=\frac16\int\sin t\,dt\]
Integrating,
\[I_2=-\frac16\cos t+C\]
Substituting \(t=6x+3\) back,
\[\int\sin(6x+3)\,dx=-\frac16\cos(6x+3)+C\]
Recall that the second integral in the original expression has a negative coefficient of \(-\frac14\). Therefore,
\[-\frac14\int\sin(6x+3)\,dx=-\frac14\left[-\frac16\cos(6x+3)\right]\]
The two negative signs cancel, giving
\[-\frac14\int\sin(6x+3)\,dx=\frac1{24}\cos(6x+3)\]
Combine the results
Combining the first and second terms, we obtain
\[I=-\frac38\cos(2x+1)+\frac1{24}\cos(6x+3)+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\sin^3(2x+1)\,dx=-\frac38\cos(2x+1)+\frac1{24}\cos(6x+3)+C}\]

🎯 Exam Significance

Exam Significance

›

Application of trigonometric identities: Tests the use of the triple-angle identity to simplify a cubic trigonometric expression.
Integration by substitution: Reinforces the treatment of linear arguments such as \(2x+1\) and \(6x+3\).
Stepwise presentation: Demonstrating the identity, expansion, separation, and integration makes the mathematical process clear and supports stepwise marking.
Sign accuracy: The negative sign in front of the second sine integral combines with the negative sign produced by integration to give a positive cosine term.
Verification: Differentiation confirms that the final expression is an antiderivative of the given function.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Builds familiarity with trigonometric identities used to simplify powers of sine and cosine.
Efficient integration: Replacing \(\sin^3\theta\) with a combination of \(\sin\theta\) and \(\sin 3\theta\) reduces the problem to standard integrals.
Chain-rule awareness: The coefficients \(2\) and \(6\) in the arguments demonstrate why the inner derivative must be accounted for.
Sign management: Careful treatment of negative coefficients helps prevent errors in multi-step calculations.
Transferable strategy: The same principle can be adapted to higher powers of trigonometric functions when suitable identities are available.

🚫 Common Mistakes

Common Mistakes

›

Rearranging the triple-angle identity incorrectly or omitting the factor \(\frac14\).
Writing \(3(2x+1)=6x+1\) instead of \(6x+3\).
Forgetting to divide by \(2\) when integrating \(\sin(2x+1)\).
Forgetting to divide by \(6\) when integrating \(\sin(6x+3)\).
Giving a negative coefficient for \(\cos(6x+3)\) instead of the correct positive coefficient \(\frac1{24}\).
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Remember the triple-angle identity:
\[ \sin 3\theta=3\sin\theta-4\sin^3\theta \]
To express the cube of sine, rearrange it as:
\[ \sin^3\theta=\frac14(3\sin\theta-\sin 3\theta) \]
Expand the entire triple-angle argument correctly:
\[ 3(2x+1)=6x+3 \]
For \(a\ne0\), use the formula:
\[ \int\sin(ax+b)\,dx=-\frac{\cos(ax+b)}{a}+C \]
When integrating a negative sine term, carefully account for the two negative signs.
Include the constant of integration \(C\) in the final answer.
Differentiate the final expression to verify the result.

📘 Concept & Theory

Theory and Concept

›

This problem involves integrating the product of the cubes of sine and cosine. The key idea is to combine the two trigonometric functions using the double-angle identity and then apply the triple-angle identity to simplify the resulting cubic sine function.

1. Double-angle identity

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

Rearranging,

\[ \sin x\cos x=\frac12\sin 2x \]

Therefore,

\[ \sin^3x\cos^3x =(\sin x\cos x)^3 =\frac18\sin^3 2x \]

2. Triple-angle identity

We know that

\[ \sin 3\theta=3\sin\theta-4\sin^3\theta \]

Rearranging,

\[ \sin^3\theta=\frac14(3\sin\theta-\sin 3\theta) \]

Putting \(\theta=2x\),

\[ \sin^3 2x=\frac14(3\sin 2x-\sin 6x) \]

3. Standard integration formula

\[ \int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C, \qquad a\ne0 \]

These identities convert the original product into a linear combination of sine functions, which can be integrated directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Express the product as \((\sin x\cos x)^3\).
Apply the double-angle identity to replace \(\sin x\cos x\) with \(\frac12\sin 2x\).
Take the cube of the expression and extract the constant factor \(\frac18\).
Apply the triple-angle identity to expand \(\sin^3 2x\).
Substitute the identity into the integral and separate the terms.
Integrate each sine term, accounting for the coefficients \(2\) and \(6\) in their arguments.
Combine the results and include the constant of integration.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 17 steps

Given —
\[I=\int\sin^3x\cos^3x\,dx\]
Combine the sine and cosine factors
Using the algebraic identity \(a^3b^3=(ab)^3\),
\[I=\int(\sin x\cos x)^3\,dx\]
\sin 2x=2\sin x\cos x
Dividing both sides by \(2\),
\[\sin x\cos x=\frac12\sin 2x\]
Substituting this expression into the integral,
\[I=\int\left(\frac12\sin 2x\right)^3\,dx\]
Simplify the cube
Since \(\left(\frac12\right)^3=\frac18\),
\[I=\frac18\int\sin^3 2x\,dx\]
Apply the triple-angle identity
We know that
\[\sin^3\theta=\frac14(3\sin\theta-\sin 3\theta)\]
Putting \(\theta=2x\),
\[\sin^3 2x=\frac14(3\sin 2x-\sin 6x)\]
Substituting into the integral,
\[I=\frac18\int\frac14(3\sin 2x-\sin 6x)\,dx\]
Multiplying the constant factors,
\[I=\frac1{32}\int(3\sin 2x-\sin 6x)\,dx\]
Separate the integral
Using the linearity of integration,
\[I=\frac3{32}\int\sin 2x\,dx-\frac1{32}\int\sin 6x\,dx\]
Integrate the first term
Using the standard formula,
\[\int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C\]
For \(a=2\),
\[\int\sin 2x\,dx=-\frac12\cos 2x\]
Therefore,
\[\frac3{32}\int\sin 2x\,dx=\frac3{32}\left(-\frac12\cos 2x\right)\]
Simplifying,
\[\frac3{32}\int\sin 2x\,dx=-\frac3{64}\cos 2x\]
Integrate the second term
For \(a=6\),
\[\int\sin 6x\,dx=-\frac16\cos 6x\]
Therefore,
\[-\frac1{32}\int\sin 6x\,dx=-\frac1{32}\left(-\frac16\cos 6x\right)\]
The two negative signs cancel, giving
\[-\frac1{32}\int\sin 6x\,dx=\frac1{192}\cos 6x\]
Combine the results
Combining both integrated terms and adding the constant of integration,
\[I=-\frac3{64}\cos 2x+\frac1{192}\cos 6x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\sin^3x\cos^3x\,dx=-\frac3{64}\cos 2x+\frac1{192}\cos 6x+C}\]

🎯 Exam Significance

Exam Significance

›

Trigonometric identities: Tests the application of both double-angle and triple-angle identities in one problem.
Stepwise derivation: Shows how a product of powers can be transformed into simpler trigonometric terms before integration.
Standard integration formulas: Reinforces the need to divide by the coefficient of \(x\) in each sine argument.
Sign and coefficient accuracy: Helps prevent mistakes when multiplying constants and integrating negative terms.
Verification: Demonstrates how differentiation can be used to check the correctness of an indefinite integral.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Builds fluency in combining multiple trigonometric identities to simplify integrals.
Efficient problem-solving: Transforming the original product into a single cubic sine function reduces the complexity of the calculation.
Identity recognition: Helps identify when a double-angle identity should be applied before a power-reduction or triple-angle identity.
Calculation accuracy: The factors \(\frac18\), \(\frac14\), \(\frac12\), and \(\frac16\) illustrate the importance of tracking coefficients carefully.
Transferable techniques: The same strategy can help simplify other products of trigonometric powers.

🚫 Common Mistakes

Common Mistakes

›

Forgetting to cube the factor \(\frac12\), which gives \(\frac18\).
Applying the triple-angle identity incorrectly to \(\sin^3 2x\).
Forgetting the factor \(\frac14\) in the triple-angle identity.
Dropping the outside factor \(\frac1{32}\) from either term after separating the integral.
Forgetting to divide by \(2\) or \(6\) while integrating \(\sin 2x\) or \(\sin 6x\).
Giving the coefficient of \(\cos 6x\) a negative sign instead of the correct positive sign.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use the double-angle identity:
\[ \sin 2x=2\sin x\cos x \]
Consequently,
\[ \sin^3x\cos^3x=\frac18\sin^3 2x \]
Use the triple-angle identity in the form:
\[ \sin^3\theta=\frac14(3\sin\theta-\sin 3\theta) \]
Keep all constant factors outside the integral and multiply them carefully.
For \(a\ne0\),
\[ \int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C \]
Pay attention to the negative sign before the second sine integral, which produces a positive cosine term after integration.
Verify the final answer by differentiating it.

📘 Concept & Theory

Theory and Concept

›

This problem involves integrating the product of three sine functions with different angles. The product-to-sum identities allow us to convert products of trigonometric functions into sums or differences of simpler functions, which can then be integrated directly.

1. Product-to-sum identity for two sine functions

\[ \sin A\sin B =\frac12[\cos(A-B)-\cos(A+B)] \]

2. Product-to-sum identity for cosine and sine

\[ \cos A\sin B =\frac12[\sin(A+B)-\sin(A-B)] \]

3. Double-angle identity

\[ 2\sin A\cos A=\sin 2A \]

4. Standard integration formula

\[ \int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C, \qquad a\ne0 \]

In this problem, we first simplify \(\sin x\sin 2x\), then multiply the result by \(\sin 3x\). The resulting products can be converted into sine terms using the identities above.

Important observation: Since cosine is an even function, \(\cos(-x)=\cos x\). However, sine is an odd function, so \(\sin(-x)=-\sin x\). These properties must be applied carefully when simplifying the angles.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Apply the product-to-sum identity to \(\sin x\sin 2x\).
Use the even-function property of cosine to simplify the expression.
Multiply the result by \(\sin 3x\).
Apply the product-to-sum identity to \(\cos x\sin 3x\).
Use the double-angle identity to simplify \(\cos 3x\sin 3x\).
Combine the resulting sine terms and integrate each term separately.
Verify the final expression by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Given —
\[I=\int\sin x\sin 2x\sin 3x\,dx\]
Apply the product-to-sum identity
\[\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]\]
Putting \(A=x\) and \(B=2x\),
\[\sin x\sin 2x=\frac12[\cos(x-2x)-\cos(x+2x)]\]
Simplifying the angles,
\[\sin x\sin 2x=\frac12[\cos(-x)-\cos 3x]\]
Since cosine is an even function,
\[\cos(-x)=\cos x\]
Therefore,
\[\sin x\sin 2x=\frac12(\cos x-\cos 3x)\]
Substitute into the original integral
Replacing \(\sin x\sin 2x\) with the expression obtained above,
\[I=\frac12\int(\cos x-\cos 3x)\sin 3x\,dx\]
Distributing \(\sin 3x\),
\[I=\frac12\int[\cos x\sin 3x-\cos 3x\sin 3x]\,dx\]
Simplify \(\cos x\sin 3x\).
Using the identity
\[\cos A\sin B=\frac12[\sin(A+B)-\sin(A-B)]\]
put \(A=x\) and \(B=3x\). Then
\[\cos x\sin 3x=\frac12[\sin(x+3x)-\sin(x-3x)]\]
Simplifying,
\[\cos x\sin 3x=\frac12[\sin 4x-\sin(-2x)]\]
Since sine is an odd function,
\[\sin(-2x)=-\sin 2x\]
Hence,
\[\cos x\sin 3x=\frac12(\sin 4x+\sin 2x)\]
Simplify \(\cos 3x\sin 3x\).
Using the double-angle identity
\[2\sin A\cos A=\sin 2A\]
with \(A=3x\),
\[2\sin 3x\cos 3x=\sin 6x\]
Dividing both sides by \(2\),
\[\cos 3x\sin 3x=\frac12\sin 6x\]
Substitute both identities into the integral.
We have
\[I=\frac12\int\left[\frac12(\sin 4x+\sin 2x)-\frac12\sin 6x\right]dx\]
Distributing the outside factor \(\frac12\),
\[I=\frac14\int(\sin 4x+\sin 2x-\sin 6x)\,dx\]
Separating the integral into individual terms,
\[I=\frac14\int\sin 4x\,dx+\frac14\int\sin 2x\,dx-\frac14\int\sin 6x\,dx\]
Integrate each term separately.
For the first term,
\[\int\sin 4x\,dx=-\frac14\cos 4x\]
Therefore,
\[\frac14\int\sin 4x\,dx=\frac14\left(-\frac14\cos 4x\right)=-\frac1{16}\cos 4x\]
For the second term,
\[\int\sin 2x\,dx=-\frac12\cos 2x\]
Therefore,
\[\frac14\int\sin 2x\,dx=\frac14\left(-\frac12\cos 2x\right)=-\frac18\cos 2x\]
For the third term,
\[\int\sin 6x\,dx=-\frac16\cos 6x\]
Therefore,
\[-\frac14\int\sin 6x\,dx=-\frac14\left(-\frac16\cos 6x\right)=\frac1{24}\cos 6x\]
Combine the results
Adding all three integrated terms and including the constant of integration,
\[I=-\frac1{16}\cos 4x-\frac18\cos 2x+\frac1{24}\cos 6x+C\]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \int\sin x\sin 2x\sin 3x\,dx ={}&-\frac1{16}\cos 4x\\[6pt] &-\frac18\cos 2x +\frac1{24}\cos 6x+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

Product-to-sum identities: Tests the ability to transform products of trigonometric functions into sums and differences.
Angle simplification: Reinforces the even and odd properties of cosine and sine.
Standard integration formulas: Requires accurate integration of sine functions with different coefficients of \(x\).
Stepwise presentation: Showing each identity and simplification makes the solution easier to follow and supports stepwise marking.
Verification: Differentiation helps identify sign and numerical-factor errors.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Develops fluency in combining product-to-sum and double-angle identities.
Efficient simplification: Converts a product of three sine functions into a sum of simpler sine terms.
Algebraic accuracy: Trains careful handling of negative angles, signs, and numerical coefficients.
Method selection: Reinforces the decision to simplify a trigonometric product before integrating.
Error detection: Differentiation is a useful independent check for multi-step integration problems.

🚫 Common Mistakes

Common Mistakes

›

Using an incorrect sign in the product-to-sum identity for \(\sin A\sin B\).
Forgetting that \(\sin(-2x)=-\sin 2x\).
Omitting the factor \(\frac12\) when applying either product-to-sum identity.
Forgetting to divide by \(4\), \(2\), or \(6\) when integrating the corresponding sine terms.
Combining terms without carefully checking the signs.
Failing to verify the final answer by differentiation.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use the product-to-sum identity:
\[ \sin A\sin B =\frac12[\cos(A-B)-\cos(A+B)] \]
Remember the parity properties:
\[ \cos(-x)=\cos x,\qquad \sin(-x)=-\sin x \]
Use the identity:
\[ \cos A\sin B =\frac12[\sin(A+B)-\sin(A-B)] \]
Apply the double-angle formula:
\[ 2\sin A\cos A=\sin 2A \]
For \(a\ne0\), remember:
\[ \int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C \]
Track all factors when separating and integrating terms.
Verify the final antiderivative by differentiating it.

📘 Concept & Theory

Theory and Concept

›

This problem involves integrating the product of two sine functions with different angles. The product-to-sum identity transforms the product into a difference of cosine functions, allowing us to integrate each term using a standard formula.

1. Product-to-sum identity

\[ \sin A\sin B =\frac12[\cos(A-B)-\cos(A+B)] \]

2. Even-function property of cosine

\[ \cos(-\theta)=\cos\theta \]

3. Standard integration formula

\[ \int\cos(ax)\,dx=\frac{\sin(ax)}{a}+C, \qquad a\ne0 \]

In this problem, the product-to-sum identity is applied with \(A=4x\) and \(B=8x\). The resulting angles are \(4x-8x=-4x\) and \(4x+8x=12x\). Since cosine is an even function, \(\cos(-4x)=\cos 4x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Apply the product-to-sum identity to \(\sin 4x\sin 8x\).
Simplify the negative angle using the even-function property of cosine.
Separate the integral into two cosine integrals.
Integrate both terms, dividing by the respective coefficients \(4\) and \(12\).
Combine the terms and add the constant of integration.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Given —
\[I=\int\sin 4x\sin 8x\,dx\]
Apply the product-to-sum identity
We know that
\[\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]\]
Putting \(A=4x\) and \(B=8x\),
\[\sin 4x\sin 8x=\frac12[\cos(4x-8x)-\cos(4x+8x)]\]
Simplifying the angles,
\[\sin 4x\sin 8x=\frac12[\cos(-4x)-\cos 12x]\]
Simplify the negative angle.
Since cosine is an even function,
\[\cos(-4x)=\cos 4x\]
Therefore,
\[\sin 4x\sin 8x=\frac12(\cos 4x-\cos 12x)\]
Substitute into the integral.
\[I=\frac12\int(\cos 4x-\cos 12x)\,dx\]
Separating the integral,
\[I=\frac12\int\cos 4x\,dx-\frac12\int\cos 12x\,dx\]
Integrate the first term.
Using the standard formula
\[\int\cos(ax)\,dx=\frac{\sin(ax)}{a}+C\]
for \(a=4\),
\[\int\cos 4x\,dx=\frac14\sin 4x\]
Hence,
\[\frac12\int\cos 4x\,dx=\frac12\cdot\frac14\sin 4x=\frac18\sin 4x\]
Integrate the second term.
For \(a=12\),
\[\int\cos 12x\,dx=\frac1{12}\sin 12x\]
Therefore,
\[-\frac12\int\cos 12x\,dx=-\frac12\cdot\frac1{12}\sin 12x=-\frac1{24}\sin 12x\]
Combine the results
\[I=\frac18\sin 4x-\frac1{24}\sin 12x+C\]

💡 Answer

Final Answer

›

\[ \boxed{ \int\sin 4x\sin 8x\,dx =\frac18\sin 4x-\frac1{24}\sin 12x+C } \]

🎯 Exam Significance

Exam Significance

›

Trigonometric identities: Tests the application of the product-to-sum identity to simplify a product of sine functions.
Angle simplification: Reinforces the property \(\cos(-\theta)=\cos\theta\).
Standard integration: Requires the correct integration of cosine functions with linear arguments.
Stepwise presentation: Writing each transformation explicitly makes the solution easier to follow and supports stepwise marking.
Verification: Differentiation provides a direct way to check the final result.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Builds fluency in simplifying trigonometric products before integration.
Time-efficient calculation: Converting a product into a difference of cosines reduces the problem to two standard integrals.
Coefficient accuracy: Reinforces the need to divide by \(4\) and \(12\) when integrating the respective cosine terms.
Sign management: Helps prevent mistakes involving the subtraction sign in the product-to-sum identity.
Transferable technique: The same approach applies to many integrals involving products of sine functions with different frequencies.

🚫 Common Mistakes

Common Mistakes

›

Using a plus sign instead of a minus sign in the product-to-sum identity for \(\sin A\sin B\).
Forgetting that \(\cos(-4x)=\cos 4x\).
Omitting the factor \(\frac12\) when expanding the product.
Forgetting to divide by \(4\) or \(12\) while integrating the cosine terms.
Changing the negative sign before the second term into a positive sign.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use the product-to-sum identity:
\[ \sin A\sin B =\frac12[\cos(A-B)-\cos(A+B)] \]
Remember that cosine is an even function:
\[ \cos(-\theta)=\cos\theta \]
For \(a\ne0\), use the standard formula:
\[ \int\cos(ax)\,dx=\frac{\sin(ax)}{a}+C \]
Retain the factor \(\frac12\) from the product-to-sum identity throughout the integration.
Keep track of the negative sign before the second cosine integral.
Include the constant of integration \(C\).
Verify the antiderivative by differentiating the final expression.

📘 Concept & Theory

Theory and Concept

›

This problem involves integrating a rational expression containing trigonometric functions. The numerator and denominator can be simplified by multiplying them by a suitable expression and using the Pythagorean identity. The resulting expression can then be integrated using standard trigonometric formulas.

1. Pythagorean identity

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

Therefore,

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

2. Reciprocal and quotient identities

\[ \operatorname{cosec}x=\frac1{\sin x} \]

\[ \cot x=\frac{\cos x}{\sin x} \]

Consequently,

\[ \frac{\cos x}{\sin^2x} =\operatorname{cosec}x\cot x \]

3. Standard integration formulas

\[ \int\operatorname{cosec}^2x\,dx=-\cot x+C \]

\[ \int\operatorname{cosec}x\cot x\,dx =-\operatorname{cosec}x+C \]

\[ \int 1\,dx=x+C \]

Important correction: In the supplied working, the derivative formula for \(\operatorname{cosec}x\) was applied incorrectly. Since

\[ \frac{d}{dx}(\operatorname{cosec}x) =-\operatorname{cosec}x\cot x, \]

the integral of \(-2\operatorname{cosec}x\cot x\) is \(+2\operatorname{cosec}x\), not \(-2\operatorname{cosec}x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Multiply the numerator and denominator by \(1-\cos x\).
Expand the numerator and use \(1-\cos^2x=\sin^2x\) in the denominator.
Rewrite the resulting expression using cosecant and cotangent.
Integrate each term separately using standard formulas.
Combine the terms and include the constant of integration.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Given —
\[I=\int\frac{1-\cos x}{1+\cos x}\,dx\]
Rationalise the trigonometric expression.
Multiply both the numerator and denominator by \(1-\cos x\):
\[I=\int\frac{1-\cos x}{1+\cos x}\cdot\frac{1-\cos x}{1-\cos x}\,dx\]
Combining the fractions,
\[I=\int\frac{(1-\cos x)^2}{(1+\cos x)(1-\cos x)}\,dx\]
Expand the numerator and simplify the denominator.
Expanding the numerator,
\[(1-\cos x)^2=1-2\cos x+\cos^2x\]
Using the difference of two squares in the denominator,
\[(1+\cos x)(1-\cos x)=1-\cos^2x\]
Therefore,
\[I=\int\frac{1-2\cos x+\cos^2x}{1-\cos^2x}\,dx\]
Using the identity \(1-\cos^2x=\sin^2x\),
\[I=\int\frac{1-2\cos x+\cos^2x}{\sin^2x}\,dx\]
Rewrite the numerator to simplify the fraction.
Since \(1=\sin^2x+\cos^2x\), we have
\[1+\cos^2x-2\cos x=2\cos^2x+\sin^2x-2\cos x\]
Alternatively, to obtain the simplest decomposition, write
\[1-2\cos x+\cos^2x=2-2\cos x-\sin^2x\]
This follows from \(\cos^2x=1-\sin^2x\). Hence,
\[I=\int\frac{2-2\cos x-\sin^2x}{\sin^2x}\,dx\]
Separating the terms,
\[I=\int\left(\frac{2}{\sin^2x}-\frac{2\cos x}{\sin^2x}-1\right)dx\]
Express the terms using cosecant and cotangent.
Using
\[\operatorname{cosec}x=\frac1{\sin x},\quad\cot x=\frac{\cos x}{\sin x},\]
we obtain
\[\frac1{\sin^2x}=\operatorname{cosec}^2x\]
and
\[\frac{\cos x}{\sin^2x}=\operatorname{cosec}x\cot x\]
Therefore, the integral becomes
\[I=\int\left(2\operatorname{cosec}^2x-2\operatorname{cosec}x\cot x-1\right)dx\]
Separate the integral.
\[I=2\int\operatorname{cosec}^2x\,dx-2\int\operatorname{cosec}x\cot x\,dx-\int1\,dx\]
Integrate each term.
Using the standard formula
\[\int\operatorname{cosec}^2x\,dx=-\cot x+C,\]
the first term gives
\[2\int\operatorname{cosec}^2x\,dx=-2\cot x\]
Next, since
\[\frac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x,\]
we have
\[\int\operatorname{cosec}x\cot x\,dx=-\operatorname{cosec}x+C\]
Therefore, the second term gives
\[-2\int\operatorname{cosec}x\cot x\,dx=-2(-\operatorname{cosec}x)=2\operatorname{cosec}x\]
Finally,
\[-\int1\,dx=-x\]
Combine the results
\[I=-2\cot x+2\operatorname{cosec}x-x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{1-\cos x}{1+\cos x}\,dx=2\operatorname{cosec}x-2\cot x-x+C}\]

🎯 Exam Significance

Exam Significance

›

Trigonometric identities: Tests the use of the Pythagorean identity to simplify a rational trigonometric expression.
Algebraic manipulation: Reinforces multiplying the numerator and denominator by a suitable expression to simplify the denominator.
Standard integrals: Requires knowledge of the integrals of \(\operatorname{cosec}^2x\) and \(\operatorname{cosec}x\cot x\).
Sign accuracy: Highlights the importance of correctly using the derivative of cosecant.
Stepwise reasoning: Showing the simplification and integration separately helps make the solution clear and supports stepwise marking.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Develops the ability to simplify rational expressions involving trigonometric functions.
Method selection: Demonstrates when algebraic manipulation is more efficient than direct substitution.
Formula recall: Reinforces the derivative and integral relationships involving cosecant and cotangent.
Error prevention: Careful handling of signs and common denominators is essential for accuracy under time pressure.
Verification skills: Differentiation provides a reliable check for a multistep integration problem.

🚫 Common Mistakes

Common Mistakes

›

Expanding \((1-\cos x)^2\) incorrectly.
Using \(1-\cos^2x=\cos^2x\) instead of the correct identity \(1-\cos^2x=\sin^2x\).
Confusing \(\operatorname{cosec}x\cot x\) with \(\operatorname{cosec}^2x\).
Writing \(\int\operatorname{cosec}x\cot x\,dx=+\operatorname{cosec}x+C\), instead of the correct negative sign.
Forgetting that the negative coefficient of the cosecant-cotangent integral changes the sign of its antiderivative.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Multiplying the numerator and denominator by \(1-\cos x\) helps simplify the rational expression.
Use the Pythagorean identity:
\[ 1-\cos^2x=\sin^2x \]
Remember:
\[ \frac{\cos x}{\sin^2x} =\operatorname{cosec}x\cot x \]
Use the standard formulas:
\[ \int\operatorname{cosec}^2x\,dx=-\cot x+C \]

\[ \int\operatorname{cosec}x\cot x\,dx =-\operatorname{cosec}x+C \]
Keep the negative signs associated with each term throughout the calculation.
Verify the final answer by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem involves integrating a rational trigonometric expression. The numerator and denominator can be simplified using half-angle identities, after which the integral can be expressed in terms of the secant and tangent functions.

1. Half-angle identities

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

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

These identities allow us to rewrite the original fraction in terms of functions of \(\frac{x}{2}\).

2. Pythagorean identity

\[ 1+\tan^2\theta=\sec^2\theta \]

Therefore,

\[ \tan^2\theta=\sec^2\theta-1 \]

3. Standard integration formula

\[ \int\sec^2(ax)\,dx=\frac1a\tan(ax)+C, \qquad a\ne0 \]

In particular, using the chain rule for the half-angle argument,

\[ \int\sec^2\frac{x}{2}\,dx =2\tan\frac{x}{2}+C \]

Important observation: When integrating \(\sec^2\frac{x}{2}\), the derivative of \(\frac{x}{2}\) is \(\frac12\). Therefore, the integral introduces a factor of \(2\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Apply the half-angle identities to the numerator and denominator.
Simplify the fraction into a combination of \(\sec^2\frac{x}{2}\) and \(\tan^2\frac{x}{2}\).
Replace \(\tan^2\frac{x}{2}\) using the Pythagorean identity.
Combine like terms and separate the integral.
Integrate using the standard formula for \(\sec^2\frac{x}{2}\).
Write the final answer and verify it by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Given —
\[I=\int\frac{\cos x}{1+\cos x}\,dx\]
Apply the half-angle identities.
We know that
\[\cos x=1-2\sin^2\frac{x}{2}\]
and
\[1+\cos x=2\cos^2\frac{x}{2}\]
Substituting these identities into the integral,
\[I=\int\frac{1-2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}\,dx\]
Split the fraction into two terms.
Dividing each term in the numerator by the denominator,
\[I=\int\left[\frac{1}{2\cos^2\frac{x}{2}}-\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}\right]dx\]
Simplifying the second fraction,
\[I=\int\left[\frac12\sec^2\frac{x}{2}-\tan^2\frac{x}{2}\right]dx\]
Apply the Pythagorean identity
\[1+\tan^2\theta=\sec^2\theta\]
Therefore,
\[\tan^2\theta=\sec^2\theta-1\]
Putting \(\theta=\frac{x}{2}\),
\[\tan^2\frac{x}{2}=\sec^2\frac{x}{2}-1\]
Substituting into the integral,
\[I=\int\left[\frac12\sec^2\frac{x}{2}-\left(\sec^2\frac{x}{2}-1\right)\right]dx\]
Simplify the expression
Removing the brackets,
\[I=\int\left[\frac12\sec^2\frac{x}{2}-\sec^2\frac{x}{2}+1\right]dx\]
Combining the coefficients of the secant-squared term,
\[I=\int\left[-\frac12\sec^2\frac{x}{2}+1\right]dx\]
Separating the integral,
\[I=-\frac12\int\sec^2\frac{x}{2}\,dx+\int1\,dx\]
Integrate the secant-squared term.
Let
\[t=\frac{x}{2}\]
Differentiating,
\[dt=\frac12\,dx\]
Therefore,
\[dx=2\,dt\]
Now,
\[\int\sec^2\frac{x}{2}\,dx =2\int\sec^2t\,dt\]
Using
\[\int\sec^2t\,dt=\tan t+C\]
we obtain
\[\int\sec^2\frac{x}{2}\,dx =2\tan t+C\]
Substituting \(t=\frac{x}{2}\) back,
\[\int\sec^2\frac{x}{2}\,dx=2\tan\frac{x}{2}+C\]
Hence,
\[-\frac12\int\sec^2\frac{x}{2}\,dx=-\frac12\left(2\tan\frac{x}{2}\right)\]
Simplifying,
\[-\frac12\int\sec^2\frac{x}{2}\,dx=-\tan\frac{x}{2}\]
Integrate the constant term.
\[\int1\,dx=x\]
Combine the results.
Combining both terms and including the constant of integration,
\[I=x-\tan\frac{x}{2}+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\cos x}{1+\cos x}\,dx=x-\tan\frac{x}{2}+C}\]

🎯 Exam Significance

Exam Significance

›

Half-angle identities: Tests the use of half-angle formulas to simplify rational trigonometric expressions.
Algebraic simplification: Demonstrates how a complicated fraction can be reduced to standard trigonometric functions.
Standard integrals: Reinforces the integration of \(\sec^2\frac{x}{2}\) and the importance of the chain-rule factor.
Stepwise presentation: Showing each identity and simplification clearly supports a logical, marks-oriented solution.
Verification: Differentiating the final expression confirms the result and helps identify coefficient errors.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Develops fluency in transforming rational trigonometric expressions using half-angle identities.
Method selection: Shows when half-angle identities can simplify an integral more effectively than direct substitution.
Chain-rule accuracy: Highlights the factor introduced by the inner function \(\frac{x}{2}\).
Time-efficient calculation: Converting the fraction into a combination of secant-squared and constant terms reduces the integration to standard formulas.
Error prevention: Careful handling of the factor \(\frac12\) is essential for obtaining the correct coefficient of the tangent term.

🚫 Common Mistakes

Common Mistakes

›

Using \(1+\cos x=2\sin^2\frac{x}{2}\) instead of the correct identity \(1+\cos x=2\cos^2\frac{x}{2}\).
Forgetting to square the half-angle trigonometric functions when applying the identities.
Applying \(\tan^2\theta=\sec^2\theta-1\) with an incorrect sign.
Forgetting the factor \(2\) in \(\int\sec^2\frac{x}{2}\,dx\).
Incorrectly simplifying the coefficient of the tangent term.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Use the half-angle identities:
\[ 1+\cos x=2\cos^2\frac{x}{2} \]

\[ \cos x=1-2\sin^2\frac{x}{2} \]
Remember the Pythagorean identity:
\[ \tan^2\theta=\sec^2\theta-1 \]
For the half-angle argument,
\[ \int\sec^2\frac{x}{2}\,dx=2\tan\frac{x}{2}+C \]
Apply the chain rule carefully when integrating trigonometric functions with arguments such as \(\frac{x}{2}\).
Keep constant factors outside the integral until the final simplification.
Verify the final result by differentiation.

📘 Concept & Theory

Theory and Concept

›

To integrate an even power of a trigonometric function, we use the power-reduction identity derived from the double-angle formula:

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

Squaring this identity gives an expression for \(\sin^4 x\). We then use the identity

\[ \cos^2 2x=\frac{1+\cos 4x}{2} \]

This converts the fourth power into a sum of simpler trigonometric terms that can be integrated directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite \(\sin^4 x\) as \((\sin^2 x)^2\).
Apply the power-reduction identity for \(\sin^2 x\).
Expand the square and simplify using the identity for \(\cos^2 2x\).
Integrate each term separately and add the constant of integration.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Express the fourth power as a square
\[ I=\int \sin^4 x\,dx =\int (\sin^2 x)^2\,dx \]
Apply the double-angle identity.
\[\sin^2 x=\frac{1-\cos 2x}{2}\]
we obtain
\[I=\int \left(\frac{1-\cos 2x}{2}\right)^2 dx\]
Expand the square
\[I=\frac14\int (1-2\cos 2x+\cos^2 2x)\,dx\]
Distributing the factor \(\frac14\),
\[I=\frac14\int dx-\frac12\int\cos 2x\,dx+\frac14\int\cos^2 2x\,dx\]
Reduce the remaining squared cosine term. Using the identity
\[\cos^2 2x=\frac{1+\cos 4x}{2}\]
we get
\[I=\frac14\int dx-\frac12\int\cos 2x\,dx +\frac18\int(1+\cos 4x)\,dx\]
Integrate each term
\[\int dx=x,\quad\int\cos 2x\,dx=\frac12\sin 2x\]
and
\[\int\cos 4x\,dx=\frac14\sin 4x\]
Substituting these results,
\[I=\frac{x}{4}-\frac12\left(\frac12\sin 2x\right)+\frac18\left(x+\frac14\sin 4x\right)+C\]
Simplifying,
\[I=\frac{x}{4}-\frac14\sin 2x+\frac{x}{8}+\frac1{32}\sin 4x+C\]
Combining the terms involving \(x\),
\[I=\frac{3x}{8}-\frac14\sin 2x+\frac1{32}\sin 4x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\sin^4 x\,dx=\frac{3x}{8}-\frac14\sin 2x+\frac1{32}\sin 4x+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests trigonometric identities, algebraic simplification, and integration of trigonometric powers.
JEE Main and other entrance examinations: Power-reduction identities are useful for simplifying trigonometric integrals and evaluating related expressions.
Common error to avoid: The identity \(\sin^2 x=\frac{1-\cos 2x}{2}\) must be squared as a complete fraction. Also, the coefficients in \(\int\cos ax\,dx=\frac{\sin ax}{a}+C\) must be handled correctly.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\sin^2 x=\frac{1-\cos 2x}{2}\) to reduce even powers of sine.
For \(\sin^4 x\), applying the identity twice produces terms involving \(1\), \(\cos 2x\), and \(\cos 4x\).
Remember the equivalent expansion:
\[ \sin^4 x=\frac38-\frac12\cos 2x+\frac18\cos 4x \]
Always include the constant of integration \(C\) in an indefinite integral.

📘 Concept & Theory

Theory and Concept

›

To integrate an even power of cosine, we use the power-reduction identity obtained from the double-angle formula:

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

Squaring this identity gives an expression for \(\cos^4 x\). We then apply the identity

\[ \cos^2 2x=\frac{1+\cos 4x}{2} \]

This reduces the fourth power to a sum of simpler terms that can be integrated individually.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write \(\cos^4 x\) as \((\cos^2 x)^2\).
Apply the power-reduction identity for \(\cos^2 x\).
Expand the square and simplify \(\cos^2 2x\).
Integrate each term separately, taking care with the coefficients.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Express the fourth power as a square
\[I=\int \cos^4 x\,dx=\int (\cos^2 x)^2\,dx\]
Apply the double-angle identity.
We know that
\[\cos^2 x=\frac{1+\cos 2x}{2}\]
Therefore,
\[I=\int \left(\frac{1+\cos 2x}{2}\right)^2 dx\]
Expand the square
\[I=\frac14\int (1+2\cos 2x+\cos^2 2x)\,dx\]
Reduce the squared cosine term
Using
\[\cos^2 2x=\frac{1+\cos 4x}{2}\]
we obtain
\[I=\frac14\int\left(1+2\cos 2x+\frac{1+\cos 4x}{2}\right)dx\]
Combining the terms inside the bracket,
\[I=\frac14\int\left(\frac32+2\cos 2x+\frac12\cos 4x\right)dx\]
Equivalently,
\[I=\int\left(\frac38+\frac12\cos 2x+\frac18\cos 4x\right)dx\]
Integrate each term
Using the standard formulas
\[\int dx=x+C\]

\[\int\cos 2x\,dx=\frac12\sin 2x+C\]
and
\[\int\cos 4x\,dx=\frac14\sin 4x+C\]
we get
\[I=\frac38x+\frac12\left(\frac12\sin 2x\right)+\frac18\left(\frac14\sin 4x\right)+C\]
Simplifying,
\[I=\frac{3x}{8}+\frac14\sin 2x+\frac1{32}\sin 4x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\cos^4 x\,dx=\frac{3x}{8}+\frac14\sin 2x+\frac1{32}\sin 4x+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the use of trigonometric identities, algebraic expansion, and integration of multiple-angle functions.
JEE Main and other entrance examinations: Power-reduction identities are useful for evaluating trigonometric integrals and simplifying expressions involving even powers.
Common mistake to avoid: The integral of \(\cos ax\) is \(\frac{\sin ax}{a}+C\), not \(-\frac{\sin ax}{a}+C\). The negative sign applies to the integral of \(\sin ax\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\cos^2 x=\frac{1+\cos 2x}{2}\) to reduce even powers of cosine.
Apply the identity a second time to simplify \(\cos^2 2x\).
Remember the useful expansion:
\[ \cos^4 x=\frac38+\frac12\cos 2x+\frac18\cos 4x \]
Integrate each multiple-angle term carefully and include the constant \(C\).

📘 Concept & Theory

Theory and Concept

›

This problem can be simplified using the fundamental trigonometric identity

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

Rearranging gives

\[ \sin^2 x=1-\cos^2 x=(1-\cos x)(1+\cos x) \]

The numerator therefore contains the factor \(1+\cos x\), which cancels with the denominator. This algebraic simplification is more direct than expanding the expression or introducing a double-angle identity.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write \(\sin^2 x\) as \(1-\cos^2 x\).
Factor the difference of two squares.
Cancel the common factor \(1+\cos x\).
Integrate the resulting expression term by term.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int\frac{\sin^2 x}{1+\cos x}\,dx\]
Apply the Pythagorean identity
Since
\[\sin^2 x=1-\cos^2 x\]
we obtain
\[I=\int\frac{1-\cos^2 x}{1+\cos x}\,dx\]
Factor the numerator
Using the algebraic identity \(a^2-b^2=(a-b)(a+b)\),
\[1-\cos^2 x=(1-\cos x)(1+\cos x)\]
Therefore,
\[I=\int\frac{(1-\cos x)(1+\cos x)} {1+\cos x}\,dx\]
Cancel the common factor \(1+\cos x\), wherever it is nonzero:
\[I=\int(1-\cos x)\,dx\]
Integrate term by term
\[I=\int 1\,dx-\int\cos x\,dx\]
Using
\[\int 1\,dx=x\]
and
\[\int\cos x\,dx=\sin x\]
we get
\[I=x-\sin x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\sin^2 x}{1+\cos x}\,dx=x-\sin x+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This question tests the application of trigonometric identities and algebraic factorisation before integration.
JEE Main and other entrance examinations: Recognising a factorisation pattern can reduce a complicated-looking rational trigonometric expression to a simple integral.
Common mistake to avoid: Do not cancel terms across addition or subtraction. Cancellation is valid here only after factoring \(1-\cos^2 x\) as \((1-\cos x)(1+\cos x)\).
Efficiency tip: Simplify the integrand algebraically before applying more complicated identities. The factorisation method is shorter and less error-prone than expanding the expression.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\sin^2 x=1-\cos^2 x\) to transform the numerator.
Recognise the difference of squares:
\[ 1-\cos^2 x=(1-\cos x)(1+\cos x) \]
Cancel common factors only after factorisation.
The simplified integral is \(\int(1-\cos x)\,dx\), giving \(x-\sin x+C\).

📘 Concept & Theory

Theory and Concept

›

This problem uses the double-angle identity and the algebraic identity for the difference of two squares.

The double-angle identity is

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

Similarly,

\[ \cos 2\alpha=2\cos^2\alpha-1 \]

Subtracting these expressions gives

\[ \cos 2x-\cos 2\alpha =2(\cos^2 x-\cos^2\alpha) \]

Now apply the difference-of-squares identity:

\[ a^2-b^2=(a-b)(a+b) \]

This factorisation produces the factor \(\cos x-\cos\alpha\), which cancels with the denominator and simplifies the integration.

🗺️ Solution Roadmap

Step-by-step Plan

›

Express both cosine double angles in terms of squared cosines.
Subtract and factor the numerator using the difference of squares.
Cancel the common factor \(\cos x-\cos\alpha\).
Integrate the resulting expression term by term.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[I=\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx\]
Apply the double-angle identity
We know that
\[\cos 2x=2\cos^2 x-1\]
and
\[\cos 2\alpha=2\cos^2\alpha-1\]
Substituting these expressions,
\[I=\int\frac{(2\cos^2 x-1)-(2\cos^2\alpha-1)}{\cos x-\cos\alpha}\,dx\]
Simplify the numerator
Expanding the numerator,
\[I=\int\frac{2\cos^2 x-1-2\cos^2\alpha+1}{\cos x-\cos\alpha}\,dx\]
The constant terms cancel:
\[I=\int\frac{2\cos^2 x-2\cos^2\alpha}{\cos x-\cos\alpha}\,dx\]
Taking \(2\) common,
\[I=2\int\frac{\cos^2 x-\cos^2\alpha}{\cos x-\cos\alpha}\,dx\]
Factor the numerator
Using the difference-of-squares identity,
\[\cos^2 x-\cos^2\alpha=(\cos x-\cos\alpha)(\cos x+\cos\alpha)\]
Therefore,
\[I=2\int\frac{(\cos x-\cos\alpha)(\cos x+\cos\alpha)}{\cos x-\cos\alpha}\,dx\]
Cancel the common factor \(\cos x-\cos\alpha\), wherever it is nonzero:
\[I=2\int(\cos x+\cos\alpha)\,dx\]
Integrate term by term.
Since \(\alpha\) is a constant, \(\cos\alpha\) is also constant with respect to \(x\). Thus,
\[I=2\left(\int\cos x\,dx+\int\cos\alpha\,dx\right)\]
Using
\[\int\cos x\,dx=\sin x\]
and
\[\int\cos\alpha\,dx=x\cos\alpha\]
we obtain
\[I=2(\sin x+x\cos\alpha)+C\]
Hence,
\[I=2\sin x+2x\cos\alpha+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx=2\sin x+2x\cos\alpha+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This question assesses the use of double-angle identities, algebraic factorisation, and integration with respect to a variable while treating a parameter as constant.
JEE Main and other entrance examinations: Recognising a difference of squares can simplify a rational trigonometric expression quickly.
Important concept: The parameter \(\alpha\) is constant with respect to \(x\), so \(\cos\alpha\) behaves as a constant during integration.
Common mistake to avoid: Do not forget the factor \(2\) produced when subtracting the double-angle expressions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\cos 2x=2\cos^2 x-1\) to rewrite the numerator.
Factor \(\cos^2 x-\cos^2\alpha\) as \((\cos x-\cos\alpha)(\cos x+\cos\alpha)\).
After cancellation, integrate \(2\cos x+2\cos\alpha\).
Always distinguish the integration variable \(x\) from the constant parameter \(\alpha\).

📘 Concept & Theory

Theory and Concept

›

This integral can be simplified using the double-angle identity:

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

Since

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

we can express the denominator as

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

Recognising a perfect square gives

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

The numerator, \(\cos x-\sin x\), is the derivative of \(\sin x+\cos x\) with a negative sign. This observation leads to a direct substitution.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite the denominator as \((\sin x+\cos x)^2\).
Express the numerator in terms of the derivative of \(\sin x+\cos x\).
Use substitution to simplify the integral.
Integrate and verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int\frac{\cos x-\sin x}{1+\sin 2x}\,dx\]
Simplify the denominator.
Using \(\sin 2x=2\sin x\cos x\),
\[1+\sin 2x=1+2\sin x\cos x\]
Since \(1=\sin^2 x+\cos^2 x\),
\[1+\sin 2x=\sin^2 x+\cos^2 x+2\sin x\cos x\]
Therefore,
\[1+\sin 2x=(\sin x+\cos x)^2\]
Substituting this into the integral,
\[I=\int\frac{\cos x-\sin x}{(\sin x+\cos x)^2}\,dx\]
Apply substitution
Let
\[t=\sin x+\cos x\]
Differentiating with respect to \(x\),
\[\frac{dt}{dx}=\cos x-\sin x\]
Hence,
\[dt=(\cos x-\sin x)\,dx\]
The integral becomes
\[I=\int\frac{dt}{t^2}\]
Integrate
Writing the integrand in index form,
\[I=\int t^{-2}\,dt\]
Using the power rule for integration,
\[I=\frac{t^{-1}}{-1}+C\]
Therefore,
\[I=-\frac1t+C\]
Substitute back
Since \(t=\sin x+\cos x\),
\[I=-\frac1{\sin x+\cos x}+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\cos x-\sin x}{1+\sin 2x}\,dx=-\frac1{\sin x+\cos x}+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests double-angle identities, perfect-square recognition, substitution, and the power rule of integration.
JEE Main and other entrance examinations: Recognising a denominator as a perfect square can reveal a direct substitution and substantially shorten the calculation.
Important observation: The numerator is the derivative of \(\sin x+\cos x\), which is the key to solving the integral efficiently.
Common mistake to avoid: Remember the negative sign in \(\int t^{-2}\,dt=-t^{-1}+C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(1+\sin 2x=(\sin x+\cos x)^2\).
Choose \(t=\sin x+\cos x\), so that \(dt=(\cos x-\sin x)\,dx\).
Apply the power rule carefully when integrating \(t^{-2}\).
The final result is \(-1/(\sin x+\cos x)+C\).

📘 Concept & Theory

Theory and Concept

›

This integral involves powers of \(\tan 2x\) and the factor \(\sec 2x\). We use the trigonometric identity

\[ \tan^2\theta=\sec^2\theta-1 \]

to rewrite \(\tan^3 2x\) as \(\tan^2 2x\cdot\tan 2x\). This allows us to express the integral in terms of \(\sec 2x\) and its derivative.

The key differentiation formula is

\[ \frac{d}{dx}(\sec 2x)=2\sec 2x\tan 2x \]

The factor \(2\) arises from the chain rule because the angle is \(2x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate one factor of \(\tan 2x\).
Replace \(\tan^2 2x\) with \(\sec^2 2x-1\).
Choose \(u=\sec 2x\) for substitution.
Integrate the resulting polynomial expression.
Substitute back and verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given —
\[I=\int\tan^2 2x\cdot\tan 2x\sec 2x\,dx\]
Apply the trigonometric identity
Using
\[\tan^2 2x=\sec^2 2x-1\]
we obtain
\[I=\int(\sec^2 2x-1)\tan 2x\sec 2x\,dx\]
Use substitution
Let
\[u=\sec 2x\]
Differentiating both sides with respect to \(x\),
\[\frac{du}{dx}=2\sec 2x\tan 2x\]
Therefore,
\[du=2\sec 2x\tan 2x\,dx\]
or
\[\sec 2x\tan 2x\,dx=\frac12\,du\]
Transform the integral
Substituting \(u=\sec 2x\), we get
\[I=\frac12\int(u^2-1)\,du\]
Expanding,
\[I=\frac12\left(\int u^2\,du-\int 1\,du\right)\]
Integrate term by term.
Using the power rule,
\[\int u^2\,du=\frac{u^3}{3}\]
and
\[\int 1\,du=u\]
Hence,
\[I=\frac12\left(\frac{u^3}{3}-u\right)+C\]
Simplifying,
\[I=\frac{u^3}{6}-\frac{u}{2}+C\]
Substitute back
Since \(u=\sec 2x\),
\[I=\frac{\sec^3 2x}{6}-\frac{\sec 2x}{2}+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\tan^3 2x\sec 2x\,dx=\frac16\sec^3 2x-\frac12\sec 2x+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests trigonometric identities, substitution, the power rule of integration, and the chain rule.

JEE Main and other entrance examinations: Recognising the derivative of \(\sec 2x\) helps transform a trigonometric integral into a simple polynomial integral.

Common mistake to avoid: Do not omit the factor \(2\) in \(d(\sec 2x)=2\sec 2x\tan 2x\,dx\). Missing it changes the coefficients in the final answer.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\tan^2\theta=\sec^2\theta-1\) to reduce the power of tangent.
Choose \(u=\sec 2x\) because its derivative contains \(\sec 2x\tan 2x\).
Account for the chain-rule factor \(2\) when differentiating functions of \(2x\).
Always verify a trigonometric integral by differentiation when possible.

📘 Concept & Theory

Theory and Concept

›

To integrate an even power of tangent, we use the fundamental trigonometric identity

\[ 1+\tan^2 x=\sec^2 x \]

Rearranging gives

\[ \tan^2 x=\sec^2 x-1 \]

We can express \(\tan^4 x\) as \(\tan^2 x\cdot\tan^2 x\), replace one factor of \(\tan^2 x\) using the identity, and then integrate the resulting expression using standard integration formulas.

The useful formulas are

\[ \int \tan^2 x\,dx=\tan x-x+C \]

and

\[ \int \sec^2 x\,dx=\tan x+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite \(\tan^4 x\) as \(\tan^2 x(\sec^2 x-1)\).
Expand the expression and separate the integrals.
Use the identity \(\tan^2 x=\sec^2 x-1\) to simplify the terms.
Integrate each term separately.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Express the fourth power using the identity
\[I=\int\tan^4 x\,dx\]
Since
\[\tan^2 x=\sec^2 x-1\]
we can write
\[I=\int\tan^2 x(\sec^2 x-1)\,dx\]
Expand the expression
\[I=\int\tan^2 x\sec^2 x\,dx-\int\tan^2 x\,dx\]
Evaluate the first integral
For the first integral, let
\[u=\tan x\]
Differentiating,
\[du=\sec^2 x\,dx\]
Therefore,
\[\int\tan^2 x\sec^2 x\,dx=\int u^2\,du\]
Applying the power rule,
\[\int u^2\,du=\frac{u^3}{3}\]
Substituting \(u=\tan x\),
\[\int\tan^2 x\sec^2 x\,dx=\frac{\tan^3 x}{3}\]
Evaluate the second integral
Using \(\tan^2 x=\sec^2 x-1\),
\[\int\tan^2 x\,dx=\int(\sec^2 x-1)\,dx\]
Integrating each term,
\[\int\tan^2 x\,dx=\tan x-x\]
Combine the results
Recall that
\[I=\int\tan^2 x\sec^2 x\,dx -\int\tan^2 x\,dx\]
Substituting the evaluated integrals,
\[I=\frac{\tan^3 x}{3}-(\tan x-x)+C\]
Removing the brackets and simplifying,
\[I=\frac{\tan^3 x}{3}-\tan x+x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\tan^4 x\,dx=\frac{\tan^3 x}{3}-\tan x+x+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the use of trigonometric identities, algebraic manipulation, substitution, and standard integration formulas.
JEE Main and other entrance examinations: The identity \(\tan^2 x=\sec^2 x-1\) is an important technique for reducing higher powers of tangent.
Common mistake to avoid: When subtracting \(\int\tan^2 x\,dx\), remember to subtract both terms in \(\tan x-x\), giving \(-\tan x+x\).
Exam strategy: For even powers of tangent, converting a factor of \(\tan^2 x\) into \(\sec^2 x-1\) often makes the integral easier to evaluate.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(1+\tan^2 x=\sec^2 x\) to reduce higher powers of tangent.
The substitution \(u=\tan x\) is useful when the integrand contains \(\sec^2 x\,dx\).
Remember that \(\int\tan^2 x\,dx=\tan x-x+C\).
Always simplify carefully and verify the final result by differentiation.

📘 Concept & Theory

Theory and Concept

›

This integral can be simplified using the fundamental Pythagorean identity:

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

Therefore, the denominator of the given fraction is equal to \(1\). The integral reduces to the sum of two standard trigonometric integrals.

For the cubic terms, we use the identities

\[ \sin^3 x=\sin x(1-\cos^2 x) \]

and

\[ \cos^3 x=\cos x(1-\sin^2 x) \]

These identities allow us to evaluate each integral using substitution.

🗺️ Solution Roadmap

Step-by-step Plan

›

Simplify the denominator using \(\sin^2 x+\cos^2 x=1\).
Separate the integral into the sine-cube and cosine-cube terms.
Rewrite each cubic term using the Pythagorean identity.
Integrate each term using substitution.
Combine the results and verify by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 20 steps

Simplify the denominator
We know that
\[\sin^2 x+\cos^2 x=1\]
Therefore,
\[I=\int\frac{\sin^3 x+\cos^3 x}{1}\,dx\]
Hence,
\[I=\int(\sin^3 x+\cos^3 x)\,dx\]
Separate the integrals
By the linearity of integration,
\[I=\int\sin^3 x\,dx+\int\cos^3 x\,dx\]
Evaluate \(\int\sin^3 x\,dx\).
Write
\[\sin^3 x=\sin x\sin^2 x\]
Using \(\sin^2 x=1-\cos^2 x\),
\[\int\sin^3 x\,dx=\int\sin x(1-\cos^2 x)\,dx\]
Let
\[u=\cos x\]
Differentiating,
\[du=-\sin x\,dx\]
Thus, \(\sin x\,dx=-du\), and the integral becomes
\[\int\sin^3 x\,dx=-\int(1-u^2)\,du\]
Expanding and integrating,
\[=-\int 1\,du+\int u^2\,du\]

\[=-u+\frac{u^3}{3}+C\]
Substituting \(u=\cos x\), we obtain
\[\int\sin^3 x\,dx=-\cos x+\frac{\cos^3 x}{3}+C\]
Evaluate \(\int\cos^3 x\,dx\)
Write
\[\cos^3 x=\cos x\cos^2 x\]
Using \(\cos^2 x=1-\sin^2 x\),
\[\int\cos^3 x\,dx=\int\cos x(1-\sin^2 x)\,dx\]
Let
\[v=\sin x\]
Differentiating,
\[dv=\cos x\,dx\]
The integral becomes
\[\int\cos^3 x\,dx=\int(1-v^2)\,dv\]
Integrating term by term,
\[=v-\frac{v^3}{3}+C\]
Substituting \(v=\sin x\),
\[\int\cos^3 x\,dx=\sin x-\frac{\sin^3 x}{3}+C\]
Combine the results
Substituting both integrals into the original expression,
\[I=-\cos x+\frac{\cos^3 x}{3}+\sin x-\frac{\sin^3 x}{3}+C\]
Rearranging the terms,
\[I=\sin x-\cos x+\frac{\cos^3 x-\sin^3 x}{3}+C\]

💡 Answer

Final Answer

›

\[\boxed{I=\sin x-\cos x+\frac{\cos^3 x-\sin^3 x}{3}+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This question tests the Pythagorean identity, linearity of integration, and substitution for odd powers of sine and cosine.
JEE Main and other entrance examinations: Simplifying the denominator before integrating reduces the problem to two standard trigonometric integrals.
Common mistake to avoid: When substituting \(u=\cos x\), remember that \(du=-\sin x\,dx\), which introduces a negative sign.
Exam strategy: For odd powers of sine or cosine, separate one factor and express the remaining even power using the Pythagorean identity.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Always simplify \(\sin^2 x+\cos^2 x\) to \(1\) before integrating.
For \(\sin^3 x\), reserve one factor of \(\sin x\) and use \(u=\cos x\).
For \(\cos^3 x\), reserve one factor of \(\cos x\) and use \(v=\sin x\).
Combine the antiderivatives carefully and verify the result by differentiation.

📘 Concept & Theory

Theory and Concept

›

This integral can be simplified using the double-angle identity

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

We also use the Pythagorean identity

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

By simplifying the numerator first, we can rewrite the integrand in terms of \(\tan x\) and \(\sec x\). This makes it possible to apply standard integration formulas directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Replace \(\cos 2x\) with \(\cos^2 x-\sin^2 x\).
Simplify the numerator by combining like terms.
Divide each term by \(\cos^2 x\).
Use the identities involving \(\tan x\) and \(\sec x\).
Integrate and verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Given —
\[I=\int\frac{\cos 2x+2\sin^2 x}{\cos^2 x}\,dx\]
Apply the double-angle identity
Since
\[\cos 2x=\cos^2 x-\sin^2 x\]
substituting this identity gives
\[I=\int\frac{\cos^2 x-\sin^2 x+2\sin^2 x}{\cos^2 x}\,dx\]
Simplify the numerator
Combining the terms involving \(\sin^2 x\),
\[I=\int\frac{\cos^2 x+\sin^2 x}{\cos^2 x}\,dx\]
Using \(\sin^2 x+\cos^2 x=1\), we obtain
\[I=\int\frac{1}{\cos^2 x}\,dx\]
Express the integrand in standard form.
Since \(\sec x=\frac{1}{\cos x}\),
\[I=\int\sec^2 x\,dx\]
Integrate
Using the standard formula
\[\int\sec^2 x\,dx=\tan x+C\]
we obtain
\[I=\tan x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\cos 2x+2\sin^2 x}{\cos^2 x}\,dx=\tan x+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This question tests double-angle identities, the Pythagorean identity, algebraic simplification, and standard integration formulas.
JEE Main and other entrance examinations: Simplifying the numerator before integrating is an efficient strategy for reducing a complicated-looking trigonometric fraction.
Common mistake to avoid: Remember that \(\cos 2x=\cos^2 x-\sin^2 x\). Combining the \(-\sin^2 x\) term with \(+2\sin^2 x\) leaves \(+\sin^2 x\).
Exam strategy: Simplify trigonometric expressions algebraically before applying integration formulas.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\cos 2x=\cos^2 x-\sin^2 x\) to simplify the numerator.
Recognise that \(\cos^2 x+\sin^2 x=1\).
Reduce the given integral to \(\int\sec^2 x\,dx\).
Remember the standard result \(\int\sec^2 x\,dx=\tan x+C\).

📘 Concept & Theory

Theory and Concept

›

This integral can be simplified by expressing the integrand in terms of cosecant and tangent functions. We use the identities

\[ \csc x=\frac{1}{\sin x},\qquad \sec x=\frac{1}{\cos x} \]

and

\[ \sec^2 x=1+\tan^2 x \]

The key observation is that the derivative of \(\csc x\) is \(-\csc x\cot x\), while the derivative of \(\tan x\) is \(\sec^2 x\). A useful substitution is \(u=\cos x\), because \(du=-\sin x\,dx\); however, the integrand contains \(1/\sin x\), so rewriting the expression using \(\tan x\) and \(\sec x\) is more direct.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite the integrand using \(\tan x\) and \(\sec x\).
Separate the expression into terms involving \(\sec^2 x\).
Use a suitable substitution to evaluate the resulting integral.
Combine the terms and verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Given —
\[I=\int\frac{1}{\sin x\cos^3 x}\,dx\]
Multiply the numerator and denominator by \(\cos x\):
\[I=\int\frac{\cos x}{\sin x\cos^4 x}\,dx\]
Since \(\tan x=\frac{\sin x}{\cos x}\) and \(\sec x=\frac1{\cos x}\), we can write
\[\frac{1}{\sin x\cos^3 x}=\frac{\sec^4 x}{\tan x}\]
Thus,
\[I=\int\frac{\sec^4 x}{\tan x}\,dx\]
Use the identity \(\sec^2 x=1+\tan^2 x\).
Write
\[I=\int\frac{\sec^2 x\cdot\sec^2 x}{\tan x}\,dx\]
Since \(\sec^2 x=1+\tan^2 x\),
\[I=\int\frac{(1+\tan^2 x)\sec^2 x}{\tan x}\,dx\]
Separating the terms,
\[I=\int\left(\frac1{\tan x}+\tan x\right)\sec^2 x\,dx\]
Apply substitution
Let
\[u=\tan x\]
Then
\[du=\sec^2 x\,dx\]
The integral becomes
\[I=\int\left(\frac1u+u\right)\,du\]
Integrate term by term
\[I=\int\frac1u\,du+\int u\,du\]
Using the standard integration formulas,
\[\int\frac1u\,du=\ln|u|\]
and
\[\int u\,du=\frac{u^2}{2}\]
we obtain
\[I=\ln|u|+\frac{u^2}{2}+C\]
Substitute back
Since \(u=\tan x\),
\[I=\ln|\tan x|+\frac12\tan^2 x+C\]

💡 Answer

Final Answer

›

\[ \boxed{ \int\frac{1}{\sin x\cos^3 x}\,dx =\ln|\tan x|+\frac12\tan^2 x+C } \]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This question tests trigonometric identities, algebraic transformation, substitution, and logarithmic integration.
JEE Main and other entrance examinations: Recognising when to use \(u=\tan x\) can convert a complicated trigonometric integral into a straightforward algebraic integral.
Common mistake to avoid: Remember that \(\int\frac1u\,du=\ln|u|+C\), including the absolute-value signs.
Exam strategy: When a trigonometric expression contains powers of secant and tangent, check whether the substitution \(u=\tan x\), \(du=\sec^2 x\,dx\), is suitable.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Rewrite \(\frac{1}{\sin x\cos^3 x}\) as \(\frac{\sec^4 x}{\tan x}\).
Use \(\sec^2 x=1+\tan^2 x\) to simplify the integrand.
Choose \(u=\tan x\), so \(du=\sec^2 x\,dx\).
Remember to include \(\ln|\tan x|\) and the constant of integration \(C\).

📘 Concept & Theory

Theory and Concept

›

This integral can be simplified using the double-angle identity

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

Using the difference-of-squares identity, we can factorise the numerator as

\[ \cos 2x=(\cos x-\sin x)(\cos x+\sin x) \]

The denominator contains \((\cos x+\sin x)^2\). Cancelling a common factor reduces the integral to a simpler trigonometric expression.

🗺️ Solution Roadmap

Step-by-step Plan

›

Factorise \(\cos 2x\) using the difference of squares.
Cancel the common factor \(\cos x+\sin x\).
Use a substitution based on \(\cos x+\sin x\).
Integrate the resulting expression and verify the answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int\frac{\cos 2x}{(\cos x+\sin x)^2}\,dx\]
Apply the double-angle identity
We know that
\[\cos 2x=\cos^2 x-\sin^2 x\]
Factorising the difference of squares,
\[\cos 2x=(\cos x-\sin x)(\cos x+\sin x)\]
Substituting into the integral,
\[I=\int\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^2}\,dx\]
Simplify the integrand
Cancelling one factor of \(\cos x+\sin x\), wherever it is nonzero, gives
\[I=\int\frac{\cos x-\sin x}{\cos x+\sin x}\,dx\]
Apply substitution
Let
\[u=\cos x+\sin x\]
Differentiating with respect to \(x\),
\[\frac{du}{dx}=-\sin x+\cos x\]
Therefore,
\[du=(\cos x-\sin x)\,dx\]
The integral becomes
\[I=\int\frac{du}{u}\]
Integrate
Using the standard formula
\[\int\frac{du}{u}=\ln|u|+C\]
we obtain
\[I=\ln|u|+C\]
Substituting \(u=\cos x+\sin x\),
\[I=\ln|\cos x+\sin x|+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\cos 2x}{(\cos x+\sin x)^2}\,dx=\ln|\cos x+\sin x|+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests double-angle identities, algebraic factorisation, substitution, and logarithmic integration.
JEE Main and other entrance examinations: Recognising a difference of squares can reveal a direct substitution and simplify a rational trigonometric integral.
Common mistake to avoid: When differentiating \(\cos x+\sin x\), remember that its derivative is \(\cos x-\sin x\).
Exam strategy: Before choosing a substitution, check whether the numerator is the derivative of an expression appearing in the denominator.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\cos 2x=\cos^2 x-\sin^2 x\).
Factorise the numerator as \((\cos x-\sin x)(\cos x+\sin x)\).
Choose \(u=\cos x+\sin x\), so \(du=(\cos x-\sin x)\,dx\).
Apply \(\int\frac{du}{u}=\ln|u|+C\), including the absolute-value signs.

📘 Concept & Theory

Theory and Concept

›

This problem involves the inverse trigonometric function \(\sin^{-1}(\cos x)\). The key is to use the identity

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

However, the inverse sine function returns its principal value in the interval \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Therefore, the identity

\[ \sin^{-1}(\sin\theta)=\theta \]

is valid directly only when \(\theta\) lies in this principal interval.

For the standard textbook assumption \(0\leq x\leq\pi\), we have \(\frac{\pi}{2}-x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Thus,

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

The integral then reduces to a simple polynomial integral. For unrestricted real \(x\), the integrand is periodic and piecewise linear, so the answer must be handled piecewise.

🗺️ Solution Roadmap

Step-by-step Plan

›

Express \(\cos x\) as \(\sin\left(\frac{\pi}{2}-x\right)\).
Apply the principal-value property of inverse sine.
Integrate the resulting linear expression.
State the interval on which this simplification is valid.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Given —
\[I=\int\sin^{-1}(\cos x)\,dx\]
Use the complementary-angle identity
We know that
\[\cos x=\sin\left(\frac{\pi}{2}-x\right)\]
Therefore,
\[I=\int\sin^{-1}\left[\sin\left(\frac{\pi}{2}-x\right)\right]dx\]
For \(0\leq x\leq\pi\), the angle \(\frac{\pi}{2}-x\) lies in the principal range of inverse sine. Hence,
\[\sin^{-1}\left[\sin\left(\frac{\pi}{2}-x\right)\right]=\frac{\pi}{2}-x\]
Thus,
\[I=\int\left(\frac{\pi}{2}-x\right)dx\]
Integrate term by term
\[I=\frac{\pi}{2}\int dx-\int x\,dx\]
Using the standard formulas
\[\int dx=x\]
and
\[\int x\,dx=\frac{x^2}{2}\]
we obtain
\[I=\frac{\pi x}{2}-\frac{x^2}{2}+C\]

💡 Answer

Final Answer

›

For \(0\leq x\leq\pi\),

\[\boxed{\int\sin^{-1}(\cos x)\,dx=\frac{\pi x}{2}-\frac{x^2}{2}+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests complementary-angle identities, principal values of inverse trigonometric functions, and basic integration.
JEE Main and other entrance examinations: Understanding the principal-value range is essential when simplifying compositions of inverse trigonometric and trigonometric functions.
Common mistake to avoid: Do not assume that \(\sin^{-1}(\sin\theta)=\theta\) for every real \(\theta\). The equality holds directly only when \(\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\cos x=\sin\left(\frac{\pi}{2}-x\right)\).
Check the principal-value range before simplifying an inverse trigonometric composition.
On \(0\leq x\leq\pi\), the integrand simplifies to \(\frac{\pi}{2}-x\).
The corresponding antiderivative is \(\frac{\pi x}{2}-\frac{x^2}{2}+C\).

📘 Concept & Theory

Theory and Concept

›

This integral can be evaluated using the trigonometric identity for the difference of two tangent functions:

\[ \tan A-\tan B =\frac{\sin(A-B)}{\cos A\cos B} \]

Putting \(A=x-a\) and \(B=x-b\), we get

\[ \tan(x-a)-\tan(x-b) =\frac{\sin(b-a)} {\cos(x-a)\cos(x-b)} \]

When \(a\) and \(b\) are constants with \(\sin(b-a)\ne0\), this identity allows us to express the integrand as a constant multiple of a difference of tangent functions. Since

\[ \int\tan x\,dx=-\ln|\cos x|+C \]

the integral can be evaluated directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the tangent-difference identity.
Rearrange the identity to express the given integrand.
Integrate the two tangent terms separately.
Simplify the logarithmic expression and verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int\frac{1}{\cos(x-a)\cos(x-b)}\,dx\]
Apply the tangent-difference identity
We know that
\[\tan A-\tan B=\frac{\sin(A-B)}{\cos A\cos B}\]
Let \(A=x-a\) and \(B=x-b\). Then
\[\tan(x-a)-\tan(x-b)=\frac{\sin[(x-a)-(x-b)]}{\cos(x-a)\cos(x-b)}\]
Simplifying the angle in the numerator,
\[(x-a)-(x-b)=b-a\]
Therefore,
\[\tan(x-a)-\tan(x-b)=\frac{\sin(b-a)}{\cos(x-a)\cos(x-b)}\]
Express the integrand using tangent functions.
For \(\sin(b-a)\ne0\), rearranging gives
\[\frac{1}{\cos(x-a)\cos(x-b)}=\frac{\tan(x-a)-\tan(x-b)}{\sin(b-a)}\]
Hence,
\[I=\frac{1}{\sin(b-a)}\int[\tan(x-a)-\tan(x-b)]\,dx\]
Integrate term by term
Using
\[\int\tan u\,du=-\ln|\cos u|+C\]
and noting that the derivatives of \(x-a\) and \(x-b\) with respect to \(x\) are both \(1\), we obtain
\[I=\frac{1}{\sin(b-a)}\left[-\ln|\cos(x-a)|+\ln|\cos(x-b)|\right]+C\]
Combine the logarithms
Using the logarithmic identity \(\ln u-\ln v=\ln\left(\frac{u}{v}\right)\), we get
\[I=\frac{1}{\sin(b-a)}\ln\left|\frac{\cos(x-b)}{\cos(x-a)}\right|+C\]

💡 Answer

Final Answer

›

For \(\sin(b-a)\ne0\),

\[\boxed{\int\frac{dx}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(b-a)}\ln\left|\frac{\cos(x-b)}{\cos(x-a)}\right|+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests trigonometric identities, integration of tangent functions, and logarithmic simplification.
JEE Main and other entrance examinations: Recognising a tangent-difference identity is an efficient method for integrating products of shifted cosine factors in the denominator.
Common mistake to avoid: Keep the order of the angles consistent. The numerator is \(\sin(b-a)\), not \(\sin(a-b)\), when the tangent difference is written as \(\tan(x-a)-\tan(x-b)\).
Important condition: Handle \(\sin(b-a)=0\) separately rather than using the logarithmic formula.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(\tan A-\tan B=\frac{\sin(A-B)}{\cos A\cos B}\).
Convert the product of shifted cosines into a difference of tangent functions.
Remember \(\int\tan u\,du=-\ln|\cos u|+C\).
Check special parameter values before dividing by \(\sin(b-a)\).

📘 Concept & Theory

Theory and Concept

›

The key technique is to split the fraction into two simpler terms and use standard trigonometric identities.

We use the identities

\[ \frac{1}{\cos^2 x}=\sec^2 x,\qquad \frac{1}{\sin^2 x}=\csc^2 x \]

Recall the standard integrals

\[ \int\sec^2 x\,dx=\tan x+C \]

and

\[ \int\csc^2 x\,dx=-\cot x+C \]

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Split the fraction
\[I=\int\frac{\sin^2 x}{\sin^2 x\cos^2 x}\,dx-\int\frac{\cos^2 x}{\sin^2 x\cos^2 x}\,dx\]
Cancelling common factors,
\[I=\int\frac{1}{\cos^2 x}\,dx-\int\frac{1}{\sin^2 x}\,dx\]
Express in standard trigonometric form
\[I=\int\sec^2 x\,dx-\int\csc^2 x\,dx\]
Integrate each term
\[I=\tan x-(-\cot x)+C\]
Therefore,
\[I=\tan x+\cot x+C\]

💡 Answer

Final Answer

›

Correct Answer is (A)

\[\boxed{\tan x+\cot x+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: Tests algebraic simplification and standard trigonometric integration formulas.
JEE Main and other entrance examinations: Splitting a rational trigonometric expression into simpler terms is an efficient integration technique.
Common mistake to avoid: The integral of \(\csc^2 x\) is \(-\cot x+C\). Subtracting this integral produces a positive \(\cot x\) term.

Q24

LONG ANSWER4 marks

Choose the correct answer \[\begin{aligned}&\int \dfrac{e^{x}\left( 1+x\right) }{\cos ^{2}\left( e^{x}x\right) }dx\\[6pt] \left( A\right)\quad &-\cot \left( ex^{x}\right) +C\\ \left( B\right)\quad &\tan \left( ex^{x}\right) +C\\ \left( C\right)\quad &\tan \left( e^{x}\right) +C\\ \left( D\right)\quad &\cot \left( e^{x}\right) +C\end{aligned}\]

📘 Concept & Theory

Theory and Concept

›

This integral is designed for the method of substitution. We use the chain rule in reverse and the standard formula

\[ \int\sec^2 u\,du=\tan u+C \]

The key observation is that

\[ \frac{d}{dx}(xe^x)=e^x+xe^x=e^x(1+x) \]

Therefore, the numerator is the derivative of \(xe^x\), which appears as the argument of the cosine in the denominator.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given —
\[I=\int e^x(1+x)\sec^2(xe^x)\,dx\]
Apply substitution
Let
\[u=xe^x\]
Differentiating with respect to \(x\), using the product rule,
\[\frac{du}{dx}=xe^x+e^x=e^x(1+x)\]
Hence,
\[du=e^x(1+x)\,dx\]
Transform the integral
Substituting \(u=xe^x\) and \(du=e^x(1+x)\,dx\),
\[I=\int\sec^2 u\,du\]
Integrate
Using the standard formula,
\[I=\tan u+C\]
Substituting \(u=xe^x\) back,
\[I=\tan(xe^x)+C\]

💡 Answer

Final Answer

›

Correct Answer is (B)

\[\boxed{\tan(xe^x)+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: Tests substitution, the product rule, and standard trigonometric integration.
JEE Main and other entrance examinations: Recognising the derivative of a composite argument is essential for solving integrals efficiently.
Common mistake to avoid: Differentiate \(xe^x\) using the product rule; its derivative is \(e^x(1+x)\), not merely \(e^x\).

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

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Integrals

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

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INTEGRALS — Learning Resources

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