Ch 7  ·  Q–
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Chapter 7 Exercise 7.9 Solutions

Integrals

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 7 Exercise 7.9

Class 12 Mathematics Exercise 7.9 NCERT Solutions Integrals Definite Integrals Indefinite Integrals Class 12 Mathematics Chapter 7 CBSE Board Exam JEE Main CUET Integration by Substitution Fundamental Theorem of Calculus Inverse Trigonometric Integrals Standard Integration Formulas Solved Examples
10 Questions
20–30 min Ideal time
Q1 Now at
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Q1
LONG ANSWER4 marks
Evaluate the following definite integral using substitution. \(\displaystyle \int\limits_0^1\frac{x}{x^2+1}\,dx\)
📘 Concept & Theory
Theory and Concept
›

This problem is based on the method of substitution, which simplifies an integral by replacing a complicated expression with a new variable. It is particularly useful when the integrand contains a function and its derivative.

The standard result used in this question is:

\[ \int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C \]

In the given integral, the denominator is \(x^2+1\), whose derivative is \(2x\). Since the numerator contains \(x\), it is proportional to the derivative of the denominator.

Therefore, we choose the substitution:

\[ t=x^2+1 \]

Important concept for definite integrals: When a substitution is made in a definite integral, we can either transform the limits into the new variable or substitute back into the original variable before applying the original limits. Both approaches give the same result.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the denominator as the expression suitable for substitution.

  2. Put \(t=x^2+1\) and differentiate to obtain \(dt=2x\,dx\).

  3. Express the numerator and denominator in terms of the new variable.

  4. Transform the original limits \(x=0\) and \(x=1\) into corresponding limits of \(t\).

  5. Evaluate the resulting logarithmic definite integral.

  6. Simplify the expression to obtain the final answer.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. Given —
    \[I=\int\limits_0^1\frac{x}{x^2+1}\,dx\]
  2. Apply substitution
  3. Let
    \[t=x^2+1\]
  4. Differentiating both sides with respect to \(x\), we get
    \[\frac{dt}{dx}=2x\]
    \[dt=2x\,dx\]
    \[x\,dx=\frac{dt}{2}\]
  5. Change the limits of integration
  6. When \(x=0\),
    \[t=0^2+1=1\]
  7. When \(x=1\),
    \[t=1^2+1=2\]
  8. Thus, the new limits are \(t=1\) and \(t=2\).
  9. Substitute in the integral
  10. Using \(t=x^2+1\) and \(x\,dx=\frac{dt}{2}\), we obtain
    \[ \begin{aligned} I&=\int\limits_0^1\frac{x}{x^2+1}\,dx\\[6pt] &=\frac{1}{2}\int\limits_1^2\frac{dt}{t} \end{aligned} \]
  11. Integrate using the standard logarithmic formula
  12. We know that
    \[\int\frac{1}{t}\,dt=\log|t|+C\]
  13. Therefore,
    \[ \begin{aligned} I&=\frac{1}{2}\int\limits_1^2\frac{dt}{t}\\[6pt] &=\frac{1}{2}\left[\log|t|\right]_1^2\\[6pt] &=\frac{1}{2}\left[\log 2-\log 1\right] \end{aligned} \]
  14. Simplify the result
  15. Since \(\log 1=0\),
    \[ \begin{aligned} I&=\frac{1}{2}[\log 2-0]\\[6pt] &=\frac{1}{2}\log 2 \end{aligned} \]
  16. Hence, the required value is
    \[ \boxed{\displaystyle\int\limits_0^1\frac{x}{x^2+1}\,dx =\frac{1}{2}\log 2} \]
💡 Answer
Final Answer
›
\[\boxed{\displaystyle\frac{1}{2}\log 2}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the fundamental application of substitution in evaluating definite integrals.
  • It reinforces the standard logarithmic integration formula and its application to rational functions.
  • Correctly changing the limits after substitution is an important step in presenting a complete board-examination solution.
  • Writing the differentiation and substitution steps explicitly helps secure marks for the mathematical procedure.

JEE Main and Other Competitive Entrance Examinations:

  • The problem develops the ability to recognise integrands of the form \(\frac{f'(x)}{f(x)}\).
  • This recognition is useful for solving logarithmic integrals quickly and accurately.
  • The same substitution technique frequently appears in questions involving rational functions, definite integrals and properties of logarithms.
  • Understanding the transformation of integration limits helps avoid errors in objective-type questions.

CUET and Other Entrance Examinations:

  • Tests direct application of standard integration formulas.
  • Provides practice in evaluating definite integrals with elementary substitutions.
  • Strengthens conceptual understanding required for MCQs and numerical-answer questions.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Substitution: Choose \(t=x^2+1\) when the integrand contains \(x^2+1\) in the denominator and \(x\) in the numerator.

  2. Differentiation: The derivative of \(x^2+1\) is \(2x\), giving \(x\,dx=\frac{dt}{2}\).

  3. Change of limits: The original limits \(x=0\) and \(x=1\) become \(t=1\) and \(t=2\), respectively.

  4. Standard formula: Remember that \(\displaystyle\int\frac{1}{t}\,dt=\log|t|+C\).

  5. Definite integration: Always use the transformed limits when integrating with respect to the substituted variable.

  6. Logarithmic property: Since \(\log 1=0\), the final expression simplifies directly.

  7. Exam strategy: Recognising the relationship between the numerator and the derivative of the denominator makes this type of integral straightforward.
↑ Top
1 / 10  ·  10%
Q2 →
Q2
LONG ANSWER4 marks
Evaluate the following integral using substitution \[I=\int\limits_0^{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\,d\phi\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the method of substitution in trigonometric integrals. The key idea is to select a substitution that simplifies both the radical expression and the trigonometric factor.

The given integrand contains \(\sqrt{\sin\phi}\) and an odd power of \(\cos\phi\). We use the substitution:

\[ \sin\phi=t^2 \]

Differentiating both sides,

\[ \cos\phi\,d\phi=2t\,dt \]

This substitution is particularly useful because it transforms the square root into a simple algebraic expression:

\[ \sqrt{\sin\phi}=t \]

We also use the fundamental trigonometric identity:

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

Therefore,

\[ \cos^2\phi=1-\sin^2\phi=1-t^4 \]

Consequently, the original trigonometric integral reduces to a polynomial integral, which can be evaluated using the power rule of integration.

Important observation: Since the power of cosine is odd, we can separate one factor of \(\cos\phi\) and combine it with \(d\phi\). The remaining even power can then be expressed in terms of \(\sin\phi\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the square root of \(\sin\phi\) as a suitable expression for substitution.

  2. Put \(\sin\phi=t^2\) and differentiate to obtain \(\cos\phi\,d\phi=2t\,dt\).

  3. Separate \(\cos^5\phi\) into \(\cos^4\phi\cdot\cos\phi\).

  4. Express \(\cos^4\phi\) in terms of \(t\) using the identity \(\cos^2\phi=1-\sin^2\phi\).

  5. Transform the original limits into the corresponding limits of \(t\).

  6. Expand the polynomial and integrate each term separately.

  7. Substitute the transformed limits and simplify the final result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Given —
    \[I=\int\limits_0^{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\,d\phi\]
  2. Apply substitution
  3. Let
    \[\sin\phi=t^2\]
  4. Differentiating both sides with respect to \(\phi\),
    \[\cos\phi\,d\phi=2t\,dt\]
  5. Also
    \[\sqrt{\sin\phi}=t\]
  6. Express the remaining cosine factors in terms of \(t\).
  7. Using the identity \sin^2\phi+\cos^2\phi=1
  8. we get
    \[ \begin{aligned} \cos^2\phi&=1-\sin^2\phi\\ &=1-(t^2)^2\\ &=1-t^4 \end{aligned} \]
  9. Squaring both sides,
    \[\cos^4\phi=(1-t^4)^2\]
  10. Change the limits of integration
  11. When \(\phi=0\),
    \[ \begin{aligned} \sin 0&=t^2\\ 0&=t^2\\ t&=0 \end{aligned} \]
  12. When \(\phi=\frac{\pi}{2}\),
    \[ \begin{aligned} \sin\frac{\pi}{2}&=t^2\\ 1&=t^2\\ t&=1 \end{aligned} \]

  13. Since \(t=\sqrt{\sin\phi}\) and \(0\leq\phi\leq\frac{\pi}{2}\), we take the non-negative values of \(t\).

    Thus, the new limits are \(t=0\) and \(t=1\).

  14. Transform the given integral
  15. We have
    \[ \begin{aligned} I&=\int\limits_0^{\frac{\pi}{2}} \sqrt{\sin\phi}\cos^5\phi\,d\phi\\[6pt] &=\int\limits_0^{\frac{\pi}{2}} \sqrt{\sin\phi}\cos^4\phi \left(\cos\phi\,d\phi\right) \end{aligned} \]
  16. Substituting the expressions obtained above,
    \[ \begin{aligned} I&=\int\limits_0^1 t(1-t^4)^2(2t)\,dt\\[6pt] &=2\int\limits_0^1t^2(1-t^4)^2\,dt \end{aligned} \]
  17. Expand the polynomial
  18. Using the algebraic identity \((a-b)^2=a^2-2ab+b^2\),
  19. \[(1-t^4)^2=1-2t^4+t^8\]
  20. Therefore,
    \[ \begin{aligned} I&=2\int\limits_0^1t^2(1-2t^4+t^8)\,dt\\[6pt] &=2\int\limits_0^1(t^2-2t^6+t^{10})\,dt \end{aligned} \]
  21. Integrate each term separately
  22. Applying the power rule of integration,
    \[\begin{aligned}I&=2\left[\frac{t^3}{3}-\frac{2t^7}{7}+\frac{t^{11}}{11}\right]_0^1\\[8pt] &=2\left[\left(\frac{t^3}{3}-\frac{2t^7}{7}+\frac{t^{11}}{11}\right)\right]_0^1\end{aligned}\]
  23. Apply the limits
    \[\begin{aligned}I&=2\left[\left(\frac{1}{3}-\frac{2}{7}+\frac{1}{11}\right)-\left(0-0+0\right)\right]\\[8pt] &=2\left(\frac{1}{3}-\frac{2}{7}+\frac{1}{11}\right)\end{aligned}\]
  24. Taking the LCM of \(3,7,11\), which is \(231\),
    \[\begin{aligned}I&=2\left(\frac{77-66+21}{231}\right)\\[6pt] &=2\left(\frac{32}{231}\right)\\[6pt] &=\frac{64}{231}\end{aligned}\]
  25. Hence, the required value is
    \[\boxed{\displaystyle\int\limits_0^{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\,d\phi=\frac{64}{231}}\]
💡 Answer
Final Answer
›
\[\boxed{\displaystyle\frac{64}{231}}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the application of substitution in trigonometric definite integrals.
  • It evaluates the student's understanding of trigonometric identities and their role in simplifying integrands.
  • Correctly separating \(\cos^5\phi\) into \(\cos^4\phi\cdot\cos\phi\) demonstrates an important integration technique.
  • Changing the limits and showing the polynomial integration steps are essential for a complete mathematical solution.

JEE Main and Other Competitive Entrance Examinations:

  • The problem develops pattern recognition for trigonometric integrals involving odd powers of sine or cosine.
  • The substitution technique is useful in integrals containing fractional powers of trigonometric functions.
  • It builds familiarity with converting trigonometric expressions into algebraic polynomials.
  • The problem also strengthens algebraic expansion, definite-limit evaluation and rational-number simplification skills.

CUET and Other Entrance Examinations:

  • Tests the direct application of substitution and standard integration formulas.
  • Provides practice in evaluating definite integrals involving trigonometric functions and fractional powers.
  • Improves accuracy in numerical-answer and multiple-choice questions.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Choice of substitution: When a square root of a trigonometric function appears, consider a substitution that eliminates the radical. Here, \(\sin\phi=t^2\) converts \(\sqrt{\sin\phi}\) into \(t\).

  2. Odd power of cosine: Separate one factor of \(\cos\phi\) and combine it with \(d\phi\) to facilitate substitution.

  3. Trigonometric identity: Use \(\cos^2\phi=1-\sin^2\phi\) to express the remaining cosine factors in terms of the substituted variable.

  4. Correct differential: From \(\sin\phi=t^2\), we obtain \(\cos\phi\,d\phi=2t\,dt\). The factor \(2\) must not be omitted.

  5. Transformed limits: The original limits \(0\) and \(\frac{\pi}{2}\) become \(0\) and \(1\), respectively.

  6. Polynomial integration: After substitution, apply the power rule separately to every term.

  7. Common error to avoid: Do not forget the factor \(2\) introduced by differentiation. Also, ensure that the transformed limits are used consistently.

  8. Exam strategy: Recognising odd trigonometric powers and selecting an appropriate substitution can significantly simplify definite integrals.
← Q1
2 / 10  ·  20%
Q3 →
Q3
LONG ANSWER4 marks
Evaluate the following integral using substitution \[I=\int\limits_0^1\sin^{-1}\left(\frac{2x}{1+x^2}\right)\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the application of inverse trigonometric identities and substitution in definite integration. The key is to recognise that the expression inside the inverse sine function can be simplified using the double-angle identity for sine.

Recall the trigonometric identity:

\[ \sin 2\theta=\frac{2\tan\theta}{1+\tan^2\theta} \]

Putting \(x=\tan\theta\), we obtain

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

Consequently, the inverse sine expression can be simplified by using the principal-value property of the inverse sine function.

We also need the standard identity:

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

provided that

\[ -\frac{\pi}{2}\leq\alpha\leq\frac{\pi}{2} \]

Since the integration interval is \(0\leq x\leq1\), the substitution \(x=\tan\theta\) gives \(0\leq\theta\leq\frac{\pi}{4}\). Therefore, \(2\theta\) lies within the principal-value interval of the inverse sine function.

This allows us to simplify the inverse trigonometric expression before evaluating the integral.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Recognise the expression \(\frac{2x}{1+x^2}\) as the double-angle expression for sine.

  2. Put \(x=\tan\theta\) to convert the inverse sine expression into a simpler angular expression.

  3. Transform the differential using \(dx=\sec^2\theta\,d\theta\).

  4. Change the integration limits according to the substitution.

  5. Simplify the resulting trigonometric integral using \(\sec^2\theta=1+\tan^2\theta\).

  6. Integrate using integration by parts.

  7. Apply the transformed limits and simplify the final answer.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Given —
    \[I=\int\limits_0^1\sin^{-1}\left(\frac{2x}{1+x^2}\right)\,dx\]
  2. Apply substitution
  3. Let
    \[x=\tan\theta\]
  4. Differentiating both sides,
    \[dx=\sec^2\theta\,d\theta\]
  5. Also,
    \[ \begin{aligned} \frac{2x}{1+x^2} &=\frac{2\tan\theta}{1+\tan^2\theta}\\[6pt] &=\frac{2\tan\theta}{\sec^2\theta}\\[6pt] &=2\sin\theta\cos\theta\\[6pt] &=\sin 2\theta \end{aligned} \]
  6. Therefore, the inverse sine expression becomes
    \[ \begin{aligned} \sin^{-1}\left(\frac{2x}{1+x^2}\right) &=\sin^{-1}(\sin 2\theta)\\ &=2\theta \end{aligned} \]
    Here, \(0\leq\theta\leq\frac{\pi}{4}\), so \(2\theta\) belongs to the principal-value interval of \(\sin^{-1}\).
  7. Change the limits of integration
  8. When \(x=0\),
    \[ \begin{aligned} 0&=\tan\theta\\ \theta&=0 \end{aligned} \]
  9. When \(x=1\),
    \[ \begin{aligned} 1&=\tan\theta\\ \theta&=\frac{\pi}{4} \end{aligned} \]
  10. Thus, the new limits are \(0\) and \(\frac{\pi}{4}\).
  11. Transform the given integral
  12. Substituting the expressions obtained above,
    \[ \begin{aligned} I&=\int\limits_0^1 \sin^{-1}\left(\frac{2x}{1+x^2}\right)\,dx\\[6pt] &=\int\limits_0^{\frac{\pi}{4}}2\theta\sec^2\theta\,d\theta \end{aligned} \]
  13. Apply integration by parts
  14. We know that
    \[\int u\,dv=uv-\int v\,du\]
  15. Taking
    \[u=2\theta,\qquad dv=\sec^2\theta\,d\theta\]
  16. Therefore,
    \[du=2\,d\theta\]
    \[v=\tan\theta\]
  17. Applying integration by parts,
    \[ \begin{aligned} I&=\left[2\theta\tan\theta\right]_0^{\frac{\pi}{4}} -2\int\limits_0^{\frac{\pi}{4}}\tan\theta\,d\theta \end{aligned} \]
  18. Evaluate the remaining integral
  19. Using the standard result
    \[\int\tan\theta\,d\theta=-\log|\cos\theta|+C\]
  20. we get
    \[ \begin{aligned} I&=\left[2\theta\tan\theta\right]_0^{\frac{\pi}{4}} +2\left[\log|\cos\theta|\right]_0^{\frac{\pi}{4}} \end{aligned} \]
  21. Apply the limits
    \[\begin{aligned}I={}&\left(2\cdot\frac{\pi}{4}\cdot\tan\frac{\pi}{4}-2\cdot0\cdot\tan0\right)\\[6pt] &+2\left(\log\left|\cos\frac{\pi}{4}\right|-\log|\cos0|\right)\end{aligned}\]
  22. Using
    \[ \tan\frac{\pi}{4}=1,\qquad \cos\frac{\pi}{4}=\frac{1}{\sqrt2}, \qquad \cos0=1 \]
  23. we obtain
    \[ \begin{aligned} I&=\frac{\pi}{2} +2\left[ \log\left(\frac{1}{\sqrt2}\right)-\log1 \right]\\[6pt] &=\frac{\pi}{2} +2\log\left(\frac{1}{\sqrt2}\right) \end{aligned} \]
  24. Since
    \[\log\left(\frac{1}{\sqrt2}\right)=-\frac{1}{2}\log2\]
  25. we get
    \[ \begin{aligned} I&=\frac{\pi}{2} +2\left(-\frac{1}{2}\log2\right)\\[6pt] &=\frac{\pi}{2}-\log2 \end{aligned} \]
  26. Hence, the required value is
    \[ \boxed{\displaystyle \int\limits_0^1\sin^{-1}\left(\frac{2x}{1+x^2}\right)\,dx =\frac{\pi}{2}-\log2 } \]
💡 Answer
Final Answer
›
\[\boxed{\displaystyle\frac{\pi}{2}-\log2}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the application of inverse trigonometric identities in definite integration.
  • It reinforces the double-angle identity for sine and the principal-value property of the inverse sine function.
  • It combines substitution with integration by parts, testing the ability to apply two integration techniques in a single problem.
  • Showing the transformed limits and all integration-by-parts steps is important for a complete board-examination solution.

JEE Main and Other Competitive Entrance Examinations:

  • Recognition of \(\frac{2x}{1+x^2}\) as a double-angle trigonometric expression is an important problem-solving skill.
  • The question tests the relationship between inverse trigonometric functions and trigonometric substitutions.
  • It develops proficiency in combining substitution and integration by parts.
  • The principal-value restriction is particularly important because ignoring it can lead to incorrect simplification of inverse trigonometric expressions.

CUET and Other Entrance Examinations:

  • Tests the direct application of standard inverse trigonometric identities.
  • Strengthens understanding of definite integration involving inverse trigonometric functions.
  • Provides practice in evaluating logarithmic and trigonometric expressions at definite limits.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Recognising identities: The expression \(\frac{2x}{1+x^2}\) can be transformed into \(\sin2\theta\) by putting \(x=\tan\theta\).

  2. Inverse trigonometric property: Always check the principal-value interval before using \(\sin^{-1}(\sin\alpha)=\alpha\).

  3. Substitution: From \(x=\tan\theta\), we obtain \(dx=\sec^2\theta\,d\theta\).

  4. Changing limits: The original limits \(x=0\) and \(x=1\) become \(\theta=0\) and \(\theta=\frac{\pi}{4}\).

  5. Integration by parts: The transformed integral contains the product of \(\theta\) and \(\sec^2\theta\), making integration by parts the appropriate technique.

  6. Standard logarithmic integral: Remember that \(\int\tan\theta\,d\theta=-\log|\cos\theta|+C\).

  7. Common error to avoid: Do not simplify \(\sin^{-1}(\sin2\theta)\) without checking whether \(2\theta\) lies in the principal-value interval.

  8. Exam strategy: Look for trigonometric identities hidden inside algebraic expressions before attempting direct integration.
← Q2
3 / 10  ·  30%
Q4 →
Q4
LONG ANSWER4 marks
Evaluate the following definite integral using substitution \[I=\int\limits_0^2 x\sqrt{x+2}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem is based on the method of substitution involving fractional powers and algebraic expressions. The objective is to transform the radical expression into a simpler algebraic form so that the resulting integral can be evaluated using the power rule of integration.

Since the integrand contains \(\sqrt{x+2}\), we choose the substitution:

\[ x+2=t^2 \]

Consequently,

\[ x=t^2-2 \]

and differentiating gives

\[ dx=2t\,dt \]

This substitution eliminates the square root because

\[ \sqrt{x+2}=t \]

The original integral is therefore transformed into a polynomial integral in \(t\).

Important observation: When making a substitution in a definite integral, the original limits must also be transformed. Moreover, the expression \(x\) in the numerator must be replaced by \(t^2-2\); it cannot be omitted during substitution.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the expression inside the square root as a suitable substitution.

  2. Put \(x+2=t^2\) and obtain \(x=t^2-2\).

  3. Differentiate to obtain \(dx=2t\,dt\).

  4. Transform the original limits \(x=0\) and \(x=2\) into the corresponding limits of \(t\).

  5. Substitute all expressions into the original integral.

  6. Expand the resulting polynomial and integrate term by term.

  7. Apply the transformed limits and simplify the final answer.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Given —
    \[I=\int\limits_0^2 x\sqrt{x+2}\,dx\]
  2. Apply substitution
  3. Let
    \[x+2=t^2\]
  4. Therefore,
    \[x=t^2-2\]
  5. Differentiating both sides with respect to \(t\),
    \[dx=2t\,dt\]
  6. Also,
    \[\sqrt{x+2}=t\]
  7. Change the limits of integration
  8. When \(x=0\),
    \[ \begin{aligned} 0+2&=t^2\\ t^2&=2\\ t&=\sqrt{2} \end{aligned} \]
  9. When \(x=2\),
    \[ \begin{aligned} 2+2&=t^2\\ t^2&=4\\ t&=2 \end{aligned} \]
  10. Since \(t=\sqrt{x+2}\) and \(x\) lies between \(0\) and \(2\), we take the positive values of \(t\).
  11. Thus, the new limits are \(t=\sqrt{2}\) and \(t=2\).
  12. Substitute in the given integral
  13. Using
    \[x=t^2-2,\quad \sqrt{x+2}=t,\quad dx=2t\,dt\]
  14. we obtain
    \[ \begin{aligned} I&=\int\limits_0^2x\sqrt{x+2}\,dx\\[6pt] &=\int\limits_{\sqrt{2}}^2(t^2-2)\cdot t\cdot 2t\,dt\\[6pt] &=2\int\limits_{\sqrt{2}}^2(t^2-2)t^2\,dt\\[6pt] &=2\int\limits_{\sqrt{2}}^2(t^4-2t^2)\,dt \end{aligned} \]
  15. Integrate term by term
  16. Applying the power rule of integration,
    \[ \begin{aligned} I&=2\int\limits_{\sqrt{2}}^2(t^4-2t^2)\,dt\\[6pt] &=2\left[ \frac{t^5}{5}-\frac{2t^3}{3} \right]_{\sqrt{2}}^2\\[6pt] &=\left[ \frac{2t^5}{5}-\frac{4t^3}{3} \right]_{\sqrt{2}}^2 \end{aligned} \]
  17. Apply the upper limit
  18. Putting \(t=2\),
    \[ \begin{aligned} \frac{2(2)^5}{5}-\frac{4(2)^3}{3} &=\frac{64}{5}-\frac{32}{3}\\[6pt] &=\frac{192-160}{15}\\[6pt] &=\frac{32}{15} \end{aligned} \]
  19. Apply the lower limit
  20. Putting \(t=\sqrt{2}\),
    \[ \begin{aligned} \frac{2(\sqrt{2})^5}{5} -\frac{4(\sqrt{2})^3}{3} &=\frac{2\cdot4\sqrt{2}}{5} -\frac{4\cdot2\sqrt{2}}{3}\\[6pt] &=\frac{8\sqrt{2}}{5} -\frac{8\sqrt{2}}{3}\\[6pt] &=\frac{24\sqrt{2}-40\sqrt{2}}{15}\\[6pt] &=-\frac{16\sqrt{2}}{15} \end{aligned} \]
  21. Subtract the lower-limit value from the upper-limit value
    \[ \begin{aligned} I&=\frac{32}{15} -\left(-\frac{16\sqrt{2}}{15}\right)\\[6pt] &=\frac{32}{15}+\frac{16\sqrt{2}}{15}\\[6pt] &=\frac{16}{15}(2+\sqrt{2}) \end{aligned} \]
  22. Hence, the required value is
    \[ \boxed{\displaystyle \int\limits_0^2x\sqrt{x+2}\,dx =\frac{16}{15}(2+\sqrt{2}) } \]
💡 Answer
Final Answer
›
\[\boxed{\displaystyle\frac{16}{15}(2+\sqrt{2})}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the application of substitution in definite integrals involving algebraic expressions and square roots.
  • It reinforces the relationship between a radical expression and an appropriate substitution.
  • It assesses the ability to transform both the integrand and the limits correctly.
  • Showing the complete polynomial expansion, integration and limit evaluation helps demonstrate the mathematical procedure required for stepwise marking.

JEE Main and Other Competitive Entrance Examinations:

  • The problem develops recognition of substitutions that convert radical expressions into polynomial functions.
  • It strengthens algebraic manipulation involving fractional powers and definite integration.
  • It provides practice in evaluating expressions containing irrational numbers at definite limits.
  • The same technique is useful in more advanced integrals involving shifted variables and fractional powers.

CUET and Other Entrance Examinations:

  • Tests the direct application of substitution and the power rule of integration.
  • Provides practice in transforming definite integration limits.
  • Improves accuracy in numerical-answer and multiple-choice questions involving radicals.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Choice of substitution: For an expression involving \(\sqrt{x+2}\), the substitution \(x+2=t^2\) eliminates the radical.

  2. Express every variable: Remember that \(x=t^2-2\), not simply \(t^2\). Omitting the \(-2\) changes the entire integral.

  3. Differential transformation: From \(x+2=t^2\), we obtain \(dx=2t\,dt\).

  4. Change of limits: The original limits \(0\) and \(2\) become \(\sqrt{2}\) and \(2\), respectively.

  5. Polynomial conversion: The transformed integrand becomes \(2(t^4-2t^2)\), which can be integrated using the power rule.

  6. Definite integration: Always subtract the value at the lower limit from the value at the upper limit.

  7. Common error to avoid: Do not retain the original limits after changing the integration variable. Also, take care while simplifying powers such as \((\sqrt{2})^5=4\sqrt{2}\).

  8. Exam strategy: For integrals involving square roots of linear expressions, consider substitutions that convert the radical into a simple variable.
← Q3
4 / 10  ·  40%
Q5 →
Q5
LONG ANSWER4 marks
Evaluate the following definite integral using substitution \[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sin x}{1+\cos^2x}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem is based on the method of substitution in trigonometric definite integrals. The key is to identify a function whose derivative is also present in the integrand.

In the given integral, the denominator contains \(\cos^2x\), while the numerator contains \(\sin x\). Since the derivative of \(\cos x\) is \(-\sin x\), we choose:

\[ t=\cos x \]

Differentiating both sides,

\[ dt=-\sin x\,dx \]

Therefore,

\[ \sin x\,dx=-dt \]

This substitution transforms the trigonometric integral into a rational algebraic integral:

\[ -\int\frac{dt}{1+t^2} \]

We can evaluate this integral using the standard formula:

\[ \int\frac{dt}{1+t^2}=\tan^{-1}t+C \]

Important observation: The negative sign introduced during differentiation must be retained. The integration limits must also be changed according to the new variable.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify \(\cos x\) as the suitable expression for substitution.

  2. Put \(t=\cos x\) and differentiate to obtain \(\sin x\,dx=-dt\).

  3. Transform the original limits \(x=0\) and \(x=\frac{\pi}{2}\) into the corresponding limits of \(t\).

  4. Rewrite the integral in terms of \(t\).

  5. Apply the standard inverse tangent integration formula.

  6. Evaluate the resulting expression at the transformed limits.

  7. Simplify to obtain the final answer.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Given —
    \[I=\int\limits_0^{\frac{\pi}{2}}\frac{\sin x}{1+\cos^2x}\,dx\]
  2. Apply substitution
  3. Let
    \[t=\cos x\]
  4. Differentiating both sides with respect to \(x\),
    \[\frac{dt}{dx}=-\sin x\]
  5. Therefore,
    \[dt=-\sin x\,dx\]
    \[\sin x\,dx=-dt\]
  6. Change the limits of integration
  7. When \(x=0\),
    \[ \begin{aligned} t&=\cos 0\\ &=1 \end{aligned} \]
  8. When \(x=\frac{\pi}{2}\),
    \[ \begin{aligned} t&=\cos\frac{\pi}{2}\\ &=0 \end{aligned} \]
  9. Thus, the new limits are \(t=1\) and \(t=0\).
  10. Transform the given integral
  11. Substituting \(t=\cos x\) and \(\sin x\,dx=-dt\), we obtain
    \[ \begin{aligned} I&=\int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos^2x}\,dx\\[6pt] &=-\int\limits_1^0\frac{dt}{1+t^2} \end{aligned} \]
  12. Reversing the limits changes the sign of the integral:
    \[I=\int\limits_0^1\frac{dt}{1+t^2}\]
  13. Apply the standard integration formula
  14. \[\int\frac{dt}{1+t^2}=\tan^{-1}t+C\]
  15. Therefore,
    \[ \begin{aligned} I&=\int\limits_0^1\frac{dt}{1+t^2}\\[6pt] &=\left[\tan^{-1}t\right]_0^1 \end{aligned} \]
  16. Apply the limits.
    \[ \begin{aligned} I&=\tan^{-1}1-\tan^{-1}0\\[6pt] &=\frac{\pi}{4}-0\\[6pt] &=\frac{\pi}{4} \end{aligned} \]
  17. Hence, the required value is
    \[ \boxed{\displaystyle \int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos^2x}\,dx =\frac{\pi}{4} } \]
💡 Answer
Final Answer
›
\[\boxed{\displaystyle\frac{\pi}{4}}\]
🎯 Exam Significance
Exam Significance
›

CBSE Board Examination:

  • This problem tests the fundamental application of substitution in definite trigonometric integrals.
  • It reinforces the relationship between a trigonometric function and its derivative.
  • It tests the ability to identify and apply the standard inverse tangent integration formula.
  • Correctly handling the negative sign and changing the limits are important steps in obtaining full procedural credit.

JEE Main and Other Competitive Entrance Examinations:

  • The problem develops recognition of integrals involving a function and its derivative.
  • It strengthens familiarity with standard inverse trigonometric integrals.
  • The transformation of trigonometric integrals into rational integrals is a frequently useful problem-solving technique.
  • It reinforces accuracy in handling reversed integration limits and negative differentials.

CUET and Other Entrance Examinations:

  • Tests direct application of substitution and standard integration formulas.
  • Provides practice in evaluating definite integrals involving trigonometric functions.
  • Builds confidence in solving objective questions involving inverse trigonometric functions.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Choice of substitution: When the numerator contains \(\sin x\) and the denominator contains \(\cos x\), consider putting \(t=\cos x\).

  2. Differential transformation: From \(t=\cos x\), we obtain \(\sin x\,dx=-dt\).

  3. Change of limits: The original limits \(0\) and \(\frac{\pi}{2}\) become \(1\) and \(0\), respectively.

  4. Standard formula: Remember that \(\displaystyle\int\frac{dt}{1+t^2}=\tan^{-1}t+C\).

  5. Reversing limits: Reversing the integration limits changes the sign of the integral.

  6. Common error to avoid: Do not omit the negative sign in \(dt=-\sin x\,dx\). Alternatively, reverse the transformed limits to obtain a positive integral.

  7. Exam strategy: Look for the derivative of the denominator's inner function in the numerator. Such integrals often reduce directly to standard forms.
← Q4
5 / 10  ·  50%
Q6 →
Q6
LONG ANSWER4 marks
Evaluate the following definite integral using substitution \[I=\int\limits_0^2\frac{dx}{x+4-x^2}\]
📘 Concept & Theory
Theory and Concept
›

This problem is based on the method of substitution in rational algebraic integrals. The denominator is a quadratic expression that can be simplified by completing the square or by factorisation.

First, consider the quadratic expression:

\[ x+4-x^2 \]

Rearranging and completing the square,

\[ \begin{aligned} x+4-x^2 &=\frac{17}{4}-\left(x-\frac12\right)^2 \end{aligned} \]

This suggests a trigonometric substitution involving the tangent function. Alternatively, we can use a substitution that converts the denominator into a product of linear factors.

Factorising the denominator,

\[ \begin{aligned} x+4-x^2 &=(2-x)(x+2)+x\\ &\text{or, equivalently,}\\ &=\left(\frac{1+\sqrt{17}}{2}-x\right) \left(x+\frac{\sqrt{17}-1}{2}\right) \end{aligned} \]

We can therefore use the standard logarithmic integration result:

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

The definite integral can then be evaluated by applying the given limits.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Factorise the quadratic expression in the denominator.

  2. Express the denominator as a product of two linear factors.

  3. Use partial fractions to separate the rational expression.

  4. Integrate the resulting terms using logarithmic integration.

  5. Apply the original limits \(0\) and \(2\).

  6. Simplify the logarithmic expression to obtain the final answer.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  23 steps
  1. Given —
    \[I=\int\limits_0^2\frac{dx}{x+4-x^2}\]
  2. Factorise the denominator.
  3. We have
    \[x+4-x^2=-(x^2-x-4)\]
  4. Solving the quadratic equation
    \[x^2-x-4=0\]
  5. Using the quadratic formula,
    \[ \begin{aligned} x&=\frac{1\pm\sqrt{1+16}}{2}\\[6pt] &=\frac{1\pm\sqrt{17}}{2} \end{aligned} \]
  6. Thus,
    \[\begin{aligned}x+4-x^2&=\left(\frac{1+\sqrt{17}}{2}-x\right)\left(x+\frac{\sqrt{17}-1}{2}\right)\end{aligned}\]
  7. Apply partial fractions
  8. Therefore,
    \[\begin{aligned}I=\int\limits_0^2\frac{dx}{\left(\frac{1+\sqrt{17}}{2}-x\right)\left(x+\frac{\sqrt{17}-1}{2}\right)}\end{aligned}\]
  9. Let
    \[a=\frac{1+\sqrt{17}}{2},\quad b=\frac{\sqrt{17}-1}{2}\]
  10. Then
    \[\begin{aligned}I&=\int\limits_0^2\frac{dx}{(a-x)(x+b)}\end{aligned}\]
  11. Resolving into partial fractions,
    \[\frac{1}{(a-x)(x+b)}=\frac{A}{a-x}+\frac{B}{x+b}\]
  12. Multiplying both sides by \((a-x)(x+b)\),
    \[1=A(x+b)+B(a-x)\]
  13. Comparing coefficients of \(x\),
    \[A-B=0\]
    \[A=B\]
  14. Comparing constant terms,
    \[Ab+Ba=1\]
  15. Since \(A=B\),
    \[A(a+b)=1\]
  16. Now,
    \[ \begin{aligned} a+b &=\frac{1+\sqrt{17}}{2} +\frac{\sqrt{17}-1}{2}\\ &=\sqrt{17} \end{aligned} \]
  17. Hence,
    \[A=B=\frac{1}{\sqrt{17}}\]
  18. Therefore,
    \[\frac{1}{(a-x)(x+b)}=\frac{1}{\sqrt{17}}\left(\frac{1}{a-x}+\frac{1}{x+b}\right)\]
  19. Integrate using the logarithmic formula
    \[\begin{aligned}I&=\frac{1}{\sqrt{17}}\int\limits_0^2\left(\frac{1}{a-x}+\frac{1}{x+b}\right)dx\\[8pt] &=\frac{1}{\sqrt{17}}\left[-\log|a-x|+\log|x+b|\right]_0^2\\[8pt] &=\frac{1}{\sqrt{17}}\left[\log\left|\frac{x+b}{a-x}\right|\right]_0^2\end{aligned}\]
  20. Apply the upper limit
  21. Putting \(x=2\),
    \[ \begin{aligned} \frac{x+b}{a-x} &=\frac{2+\frac{\sqrt{17}-1}{2}} {\frac{1+\sqrt{17}}{2}-2}\\[8pt] &=\frac{\frac{3+\sqrt{17}}{2}} {\frac{\sqrt{17}-3}{2}}\\[8pt] &=\frac{3+\sqrt{17}}{\sqrt{17}-3} \end{aligned} \]
  22. Rationalising,
    \[ \begin{aligned} \frac{3+\sqrt{17}}{\sqrt{17}-3} &=\frac{(3+\sqrt{17})^2}{17-9}\\[6pt] &=\frac{26+6\sqrt{17}}{8}\\[6pt] &=\frac{13+3\sqrt{17}}{4} \end{aligned} \]
  23. Apply the lower limit
  24. Putting \(x=0\),
    \[\begin{aligned}\frac{x+b}{a-x}&=\frac{\frac{\sqrt{17}-1}{2}}{\frac{1+\sqrt{17}}{2}}\\[8pt] &=\frac{\sqrt{17}-1}{\sqrt{17}+1}\end{aligned}\]
  25. Rationalising,
    \[ \begin{aligned} \frac{\sqrt{17}-1}{\sqrt{17}+1} &=\frac{(\sqrt{17}-1)^2}{17-1}\\[6pt] &=\frac{18-2\sqrt{17}}{16}\\[6pt] &=\frac{9-\sqrt{17}}{8} \end{aligned} \]
  26. Subtract the lower-limit value from the upper-limit value
    \[ \begin{aligned} I&=\frac{1}{\sqrt{17}} \left[ \log\left(\frac{13+3\sqrt{17}}{4}\right) -\log\left(\frac{9-\sqrt{17}}{8}\right) \right]\\[8pt] &=\frac{1}{\sqrt{17}} \log\left[ \frac{13+3\sqrt{17}}{4} \cdot\frac{8}{9-\sqrt{17}} \right]\\[8pt] &=\frac{1}{\sqrt{17}} \log\left[ \frac{2(13+3\sqrt{17})}{9-\sqrt{17}} \right] \end{aligned} \]
  27. Rationalising the expression inside the logarithm,
    \[ \begin{aligned} \frac{2(13+3\sqrt{17})}{9-\sqrt{17}} &= \frac{2(13+3\sqrt{17})(9+\sqrt{17})}{81-17}\\[6pt] &= \frac{2(168+40\sqrt{17})}{64}\\[6pt] &=\frac{21+5\sqrt{17}}{4} \end{aligned} \]
  28. Thus,
    \[\boxed{I=\frac{1}{\sqrt{17}}\log\left(\frac{21+5\sqrt{17}}{4}\right)}\]
💡 Answer
Final Answer
›
\[\boxed{\displaystyle\frac{1}{\sqrt{17}}\log\left(\frac{21+5\sqrt{17}}{4}\right)}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the evaluation of definite integrals involving quadratic expressions in the denominator.
  • It reinforces factorisation of quadratic polynomials and the method of partial fractions.
  • It develops proficiency in integrating rational functions using logarithmic formulas.
  • It tests accuracy in applying definite limits and simplifying logarithmic expressions.

JEE Main and Other Competitive Entrance Examinations:

  • Factorisation of quadratic expressions is an important preliminary skill in rational integration.
  • The problem develops the ability to recognise and apply partial fractions efficiently.
  • It strengthens algebraic manipulation involving surds and logarithms.
  • The same technique is useful in more advanced problems involving rational functions and definite integrals.

CUET and Other Entrance Examinations:

  • Tests standard logarithmic integration and partial-fraction decomposition.
  • Provides practice in evaluating definite integrals with irrational roots.
  • Improves accuracy in numerical-answer and multiple-choice questions.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Quadratic factorisation: Before integrating a rational function, examine whether its denominator can be factorised.

  2. Roots of the quadratic: The roots of \(x^2-x-4=0\) are \(\frac{1+\sqrt{17}}{2}\) and \(\frac{1-\sqrt{17}}{2}\).

  3. Partial fractions: A rational expression with two distinct linear factors in its denominator can be decomposed into simpler fractions.

  4. Logarithmic integration: Remember that \(\displaystyle\int\frac{dx}{a-x}=-\log|a-x|+C\) and \(\displaystyle\int\frac{dx}{x+b}=\log|x+b|+C\).

  5. Definite limits: Evaluate the antiderivative at the upper limit and subtract its value at the lower limit.

  6. Surd simplification: Rationalisation and careful simplification help express the final answer in a compact form.

  7. Common error to avoid: The negative sign in the integral of \(\frac{1}{a-x}\) must be retained. Also, do not interchange the upper and lower limits.

  8. Exam strategy: For rational functions with quadratic denominators, check factorisation first and then select an appropriate integration technique.
← Q5
6 / 10  ·  60%
Q7 →
Q7
LONG ANSWER4 marks
Evaluate the following definite integral using substitution \[I=\int\limits_{-1}^{1}\frac{dx}{x^2+2x+5}\]
📘 Concept & Theory
Theory and Concept
›

This problem is based on the method of substitution using completion of the square. The denominator is a quadratic expression that can be transformed into the standard form associated with the inverse tangent function.

The first step is to complete the square in the denominator:

\[ \begin{aligned} x^2+2x+5 &=x^2+2x+1+4\\ &=(x+1)^2+2^2 \end{aligned} \]

This resembles the standard inverse tangent integral:

\[ \int\frac{dt}{t^2+a^2} =\frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)+C \]

Since the quadratic expression contains \((x+1)^2+4\), we choose the substitution:

\[ t=x+1 \]

Consequently,

\[ dt=dx \]

The integral is thus converted into a standard inverse tangent form, which can be evaluated directly.

Important observation: Completing the square is particularly useful when the denominator is a quadratic expression with no convenient real linear factorisation. It helps identify standard inverse trigonometric integral forms.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given quadratic denominator.

  2. Complete the square to express it as \((x+1)^2+4\).

  3. Put \(t=x+1\) and obtain \(dt=dx\).

  4. Transform the original limits \(x=-1\) and \(x=1\) into the corresponding limits of \(t\).

  5. Rewrite the integral in the standard inverse tangent form.

  6. Integrate using the formula for \(\int\frac{dt}{t^2+a^2}\).

  7. Apply the transformed limits and simplify the result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  13 steps
  1. Given —
    \[I=\int\limits_{-1}^{1}\frac{dx}{x^2+2x+5}\]
  2. Complete the square in the denominator
    \[ \begin{aligned} x^2+2x+5 &=x^2+2x+1+4\\ &=(x+1)^2+4\\ &=(x+1)^2+2^2 \end{aligned} \]
  3. Therefore,
    \[I=\int\limits_{-1}^{1}\frac{dx}{(x+1)^2+2^2}\]
  4. Apply substitution
  5. Let
    \[t=x+1\]
  6. Differentiating both sides,
    \[dt=dx\]
  7. Also,
    \[x^2+2x+5=t^2+4\]
  8. Change the limits of integration
  9. When \(x=-1\),
    \[ \begin{aligned} t&=-1+1\\ &=0 \end{aligned} \]
  10. When \(x=1\),
    \[ \begin{aligned} t&=1+1\\ &=2 \end{aligned} \]
  11. Thus, the new limits are \(t=0\) and \(t=2\)
  12. Transform the given integral
  13. Substituting \(t=x+1\) and \(dx=dt\), we obtain
    \[ \begin{aligned} I&=\int\limits_{-1}^{1} \frac{dx}{x^2+2x+5}\\[6pt] &=\int\limits_0^2\frac{dt}{t^2+4}\\[6pt] &=\int\limits_0^2\frac{dt}{t^2+2^2} \end{aligned} \]
  14. Apply the standard inverse tangent formula
  15. We know that
    \[\int\frac{dt}{t^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)+C\]
  16. Here, \(a=2\). Therefore,
    \[ \begin{aligned} I&=\int\limits_0^2\frac{dt}{t^2+2^2}\\[6pt] &=\frac12\left[ \tan^{-1}\left(\frac{t}{2}\right) \right]_0^2 \end{aligned} \]
  17. Apply the limits
    \[\begin{aligned}I&=\frac12\left[\tan^{-1}\left(\frac{2}{2}\right)-\tan^{-1}\left(\frac{0}{2}\right)\right]\\[6pt] &=\frac12\left[\tan^{-1}1-\tan^{-1}0\right]\\[6pt] &=\frac12\left[\frac{\pi}{4}-0\right]\\[6pt] &=\frac{\pi}{8}\end{aligned}\]
  18. Hence, the required value is
    \[\boxed{\displaystyle\int\limits_{-1}^{1}\frac{dx}{x^2+2x+5}=\frac{\pi}{8}}\]
💡 Answer
Final Answer
›
\[\boxed{\displaystyle\frac{\pi}{8}}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the application of substitution after completing the square in a quadratic expression.
  • It reinforces the standard inverse tangent integration formula.
  • It develops the ability to identify quadratic expressions that can be converted into standard integration forms.
  • Showing the completion of the square, substitution, transformed limits and integration formula provides a complete step-by-step solution.

JEE Main and Other Competitive Entrance Examinations:

  • Recognising the standard form \(\frac{1}{t^2+a^2}\) is an important integration skill.
  • The problem develops proficiency in completing the square and selecting an appropriate substitution.
  • It reinforces the relationship between rational algebraic expressions and inverse trigonometric functions.
  • The technique is useful in more advanced problems involving quadratic denominators and definite integrals.

CUET and Other Entrance Examinations:

  • Tests direct application of the standard inverse tangent integration formula.
  • Provides practice in evaluating definite integrals involving quadratic expressions.
  • Strengthens accuracy in handling substitutions and transformed limits in objective-type questions.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Completing the square: The quadratic expression \(x^2+2x+5\) can be written as \((x+1)^2+4\).

  2. Choice of substitution: Put \(t=x+1\) to convert the denominator into the standard form \(t^2+4\).

  3. Differential transformation: Since \(t=x+1\), we obtain \(dt=dx\).

  4. Change of limits: The original limits \(x=-1\) and \(x=1\) become \(t=0\) and \(t=2\), respectively.

  5. Standard formula: Remember that \(\displaystyle\int\frac{dt}{t^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)+C\).

  6. Important coefficient: Do not forget the factor \(\frac{1}{a}\) in the inverse tangent integration formula.

  7. Common error to avoid: The expression \(\int\frac{dt}{t^2+4}\) is not simply \(\tan^{-1}t\). Its correct antiderivative is \(\frac12\tan^{-1}\left(\frac{t}{2}\right)\).

  8. Exam strategy: Whenever a quadratic denominator has no convenient real factorisation, try completing the square to identify an inverse trigonometric standard form.
← Q6
7 / 10  ·  70%
Q8 →
Q8
LONG ANSWER4 marks
Evaluate the following definite integral using substitution. \[\int\limits_1^2\left(\frac{1}{x}-\frac{1}{2x^2}\right)e^{2x}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves identifying a suitable substitution by examining the product of an exponential function and an algebraic expression.

The key observation is that the derivative of \(\dfrac{e^{2x}}{2x}\) is closely related to the given integrand.

Using the product and quotient rules,

\[ \begin{aligned} \frac{d}{dx}\left(\frac{e^{2x}}{2x}\right) &=\frac{1}{2}\frac{d}{dx}\left(e^{2x}x^{-1}\right)\\ &=\frac12\left(2e^{2x}x^{-1}-e^{2x}x^{-2}\right)\\ &=e^{2x}\left(\frac1x-\frac{1}{2x^2}\right) \end{aligned} \]

Thus, the integrand is the derivative of \(\dfrac{e^{2x}}{2x}\). This suggests the substitution

\[ t=\frac{e^{2x}}{2x} \]

Consequently, the integral can be transformed into a simple integral of \(dt\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Recognise the integrand as the derivative of \(\dfrac{e^{2x}}{2x}\).

  2. Substitute \(t=\dfrac{e^{2x}}{2x}\).

  3. Differentiate the substitution to obtain \(dt\).

  4. Convert the original limits \(x=1\) and \(x=2\) into the corresponding limits of \(t\).

  5. Evaluate the resulting definite integral.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Given —
    \[I=\int\limits_1^2\left(\frac{1}{x}-\frac{1}{2x^2}\right)e^{2x}\,dx\]
  2. Choose a suitable substitution
  3. Let
    \[t=\frac{e^{2x}}{2x}\]
  4. Differentiating both sides with respect to \(x\),
    \[ \begin{aligned} \frac{dt}{dx} &=\frac{d}{dx}\left(\frac{e^{2x}}{2x}\right)\\[6pt] &=\frac12\frac{d}{dx}\left(e^{2x}x^{-1}\right)\\[6pt] &=\frac12\left(2e^{2x}x^{-1} -e^{2x}x^{-2}\right)\\[6pt] &=\frac{e^{2x}}{x}-\frac{e^{2x}}{2x^2}\\[6pt] &=e^{2x}\left(\frac1x-\frac{1}{2x^2}\right) \end{aligned} \]
  5. Multiplying by \(dx\),
    \[dt=e^{2x}\left(\frac1x-\frac{1}{2x^2}\right)dx\]
  6. Change the limits of integration
  7. When \(x=1\),
    \[ \begin{aligned} t&=\frac{e^{2(1)}}{2(1)}\\ &=\frac{e^2}{2} \end{aligned} \]
  8. When \(x=2\),
    \[ \begin{aligned} t&=\frac{e^{2(2)}}{2(2)}\\ &=\frac{e^4}{4} \end{aligned} \]
  9. Therefore, the new limits are
    \[t=\frac{e^2}{2}\quad\text{and}\quad t=\frac{e^4}{4}\]
  10. Transform the integral
  11. Substituting the expression for \(dt\), we obtain
    \[ \begin{aligned} I&=\int\limits_1^2 e^{2x}\left(\frac1x-\frac{1}{2x^2}\right)dx\\[6pt] &=\int\limits_{\frac{e^2}{2}}^{\frac{e^4}{4}}dt \end{aligned} \]
  12. Integrate and apply the limits
    \[ \begin{aligned} I&=\left[t\right]_{\frac{e^2}{2}}^{\frac{e^4}{4}}\\[6pt] &=\frac{e^4}{4}-\frac{e^2}{2}\\[6pt] &=\frac{e^4-2e^2}{4}\\[6pt] &=\frac{e^2(e^2-2)}{4} \end{aligned} \]
  13. Hence, the required value is
    \[\boxed{\displaystyle I=\frac{e^2(e^2-2)}{4}}\]
💡 Answer
Final Answer
›
\[\boxed{\displaystyle\frac{e^2(e^2-2)}{4}}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the ability to identify a suitable substitution by recognising the derivative of a composite expression.
  • It reinforces the application of the product rule and quotient rule in differentiation.
  • It demonstrates how a complicated-looking integrand can be reduced to a simple integral through an appropriate substitution.
  • Writing the transformed limits correctly is essential for obtaining the final value of a definite integral.

JEE Main and Other Competitive Entrance Examinations:

  • Recognising a hidden derivative is an important technique for solving integration problems efficiently.
  • The question connects differentiation and integration through the Fundamental Theorem of Calculus.
  • It develops pattern recognition involving exponential and rational functions.
  • Similar techniques are useful in problems involving products of exponential functions and algebraic expressions.

CUET and Other Entrance Examinations:

  • Tests the identification of an appropriate substitution.
  • Reinforces differentiation-based integration techniques.
  • Provides practice in evaluating definite integrals by directly transforming their limits.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Recognise the derivative: The given integrand is exactly the derivative of \(\dfrac{e^{2x}}{2x}\).

  2. Choose the substitution: Put \(t=\dfrac{e^{2x}}{2x}\).

  3. Differentiate carefully: Applying the product or quotient rule gives \(dt=e^{2x}\left(\dfrac1x-\dfrac{1}{2x^2}\right)dx\).

  4. Transform the limits: The limits \(x=1\) and \(x=2\) become \(t=\dfrac{e^2}{2}\) and \(t=\dfrac{e^4}{4}\), respectively.

  5. Use substitution efficiently: Once the integrand is recognised as \(dt\), the integral reduces to \(\int dt\).

  6. Exam strategy: In integrals involving exponential and algebraic factors, check whether the integrand resembles the derivative of a product or quotient before attempting more complicated integration methods.
← Q7
8 / 10  ·  80%
Q9 →
Q9
LONG ANSWER4 marks
Choose the correct answer
The value of the integral \[I=\int\limits_{\frac13}^{1}\frac{(x-x^3)^{\frac13}}{x^4}\,dx\] is
(A) 6
(B) 0
(C) 3
(D) 4
📘 Concept & Theory
Theory and Concept
›

This problem involves simplifying an algebraic expression containing a fractional power and selecting a substitution that reduces the integral to a standard power integral.

First, factorise the expression inside the cube root:

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

Therefore,

\[ \begin{aligned} \frac{(x-x^3)^{\frac13}}{x^4} &=\frac{x^{\frac13}(1-x^2)^{\frac13}}{x^4}\\[6pt] &=\frac{(1-x^2)^{\frac13}}{x^{\frac{11}{3}}} \end{aligned} \]

The expression suggests using the substitution

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

This substitution is useful because it transforms the fractional-power expression into a simple power of \(t\), while the differential \(dx\) combines with the remaining powers of \(x\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Factorise \(x-x^3\) and simplify the integrand.

  2. Use the substitution \(t=\dfrac{1}{x^2}-1\).

  3. Differentiate to express \(dx\) in terms of \(dt\).

  4. Change the limits of integration.

  5. Convert the integral into a simple power integral.

  6. Evaluate and match the result with the given options.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Given —
    \[I=\int\limits_{\frac13}^{1}\frac{(x-x^3)^{\frac13}}{x^4}\,dx\]
  2. Factorising the expression inside the cube root,
    \[ \begin{aligned} I&=\int\limits_{\frac13}^{1} \frac{\left[x(1-x^2)\right]^{\frac13}}{x^4}\,dx\\[6pt] &=\int\limits_{\frac13}^{1} \frac{x^{\frac13}(1-x^2)^{\frac13}}{x^4}\,dx\\[6pt] &=\int\limits_{\frac13}^{1} \frac{(1-x^2)^{\frac13}}{x^{\frac{11}{3}}}\,dx \end{aligned} \]
  3. Apply substitution
  4. Let
    \[t=\frac{1}{x^2}-1\]
  5. Rearranging,
    \[1-x^2=tx^2\]
  6. Also, differentiating with respect to \(x\),
    \[ \begin{aligned} \frac{dt}{dx}&=-2x^{-3}\\[4pt] dt&=-\frac{2}{x^3}\,dx \end{aligned} \]
  7. Thus,
    \[dx=-\frac{x^3}{2}\,dt\]
  8. Change the limits of integration.
  9. When \(x=\dfrac13\),
    \[ \begin{aligned} t&=\frac{1}{\left(\frac13\right)^2}-1\\[4pt] &=9-1\\ &=8 \end{aligned} \]
  10. When \(x=1\),
    \[ \begin{aligned} t&=\frac{1}{1^2}-1\\ &=0 \end{aligned} \]
  11. Therefore, the new limits are \(t=8\) and \(t=0\).
  12. Transform the integrand
  13. Since \(1-x^2=tx^2\),
    \[ \begin{aligned} \frac{(1-x^2)^{\frac13}}{x^{\frac{11}{3}}}\,dx &=\frac{(tx^2)^{\frac13}}{x^{\frac{11}{3}}} \left(-\frac{x^3}{2}\,dt\right)\\[6pt] &=-\frac12 \frac{t^{\frac13}x^{\frac23}x^3} {x^{\frac{11}{3}}}\,dt\\[6pt] &=-\frac12t^{\frac13} x^{\frac{11}{3}-\frac{11}{3}}\,dt\\[6pt] &=-\frac12t^{\frac13}\,dt \end{aligned} \]
  14. Hence,
    \[ \begin{aligned} I&=-\frac12\int\limits_8^0t^{\frac13}\,dt\\[6pt] &=\frac12\int\limits_0^8t^{\frac13}\,dt \end{aligned} \]
  15. Integrate and evaluate the limits
  16. Using the power rule,
    \[ \begin{aligned} I&=\frac12\left[ \frac{t^{\frac43}}{\frac43} \right]_0^8\\[6pt] &=\frac38\left[t^{\frac43}\right]_0^8\\[6pt] &=\frac38\left(8^{\frac43}-0\right)\\[6pt] &=\frac38\left(8\cdot 8^{\frac13}\right)\\[6pt] &=\frac38(8\cdot2)\\[6pt] &=\frac38\cdot16\\[6pt] &=6 \end{aligned} \]
💡 Answer
Final Answer
›
Correct Option: Option (A) – 6
\[\boxed{\displaystyle I=6}\]
🎯 Exam Significance
Exam Significance
›
  • Tests simplification of algebraic expressions involving fractional powers.
  • Reinforces substitution and the transformation of definite integral limits.
  • Develops the ability to convert complicated algebraic integrands into standard power integrals.

JEE Main and Other Competitive Entrance Examinations:

  • Tests recognition of a suitable substitution involving reciprocal powers.
  • Requires careful handling of fractional indices and algebraic simplification.
  • Highlights the importance of choosing a substitution that eliminates the original variable completely.
  • Builds speed and accuracy in definite-integral multiple-choice questions.

CUET and Other Entrance Examinations:

  • Tests direct application of substitution in definite integration.
  • Reinforces the power rule for integration.
  • Emphasises correct handling of reversed limits and negative signs.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Factorisation: Use \(x-x^3=x(1-x^2)\) to simplify the fractional-power expression.

  2. Substitution: The choice \(t=\dfrac{1}{x^2}-1\) eliminates the variable \(x\) from the transformed integrand.

  3. Differentiation: Remember that \(dt=-\dfrac{2}{x^3}dx\).

  4. Limits: The limits change from \(\dfrac13,1\) to \(8,0\).

  5. Sign management: Reversing the limits removes the negative sign introduced by differentiation.

  6. Exam strategy: In integrals involving expressions such as \(1-x^2\) and negative powers of \(x\), consider substitutions involving \(\dfrac{1}{x^2}\).
← Q8
9 / 10  ·  90%
Q10 →
Q10
LONG ANSWER4 marks
If \(f(x)=\int\limits_0^x t\sin t\,dt\;\) then \(f'(x)\) is
(A) \(\cos x+x\sin x\)
(B) \(x\sin x\)
(C) \(x\cos x\)
(D) \(\sin x+x\cos x\)
📘 Concept & Theory
Theory and Concept
›

This question is based on the Fundamental Theorem of Calculus (Part I), which establishes the relationship between definite integration and differentiation.

If a function is defined as

\[ F(x)=\int\limits_a^x g(t)\,dt \]

where \(g(t)\) is continuous, then

\[ \boxed{F'(x)=g(x)} \]

In other words, differentiating a definite integral with respect to its upper limit gives the integrand evaluated at that upper limit.

For the given function, the integrand is \(t\sin t\), and the upper limit is \(x\). Therefore, we can directly apply the Fundamental Theorem of Calculus.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the integrand and the variable upper limit.

  2. Apply the Fundamental Theorem of Calculus.

  3. Replace the dummy variable \(t\) with the upper limit \(x\).

  4. Compare the resulting expression with the given options.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  3 steps
  1. Given —
    \[f(x)=\int\limits_0^x t\sin t\,dt\]
  2. Apply the Fundamental Theorem of Calculus
  3. According to the theorem,
    \[\frac{d}{dx}\left[\int\limits_a^x g(t)\,dt\right]=g(x)\]
  4. Therefore, differentiating the given function,
    \[ \begin{aligned} f'(x) &=\frac{d}{dx}\left[\int\limits_0^x t\sin t\,dt\right]\[6pt] &=x\sin x \end{aligned} \]
  5. Identify the correct option
  6. Comparing the result with the given choices, we obtain
    \[\boxed{f'(x)=x\sin x}\]
💡 Answer
Final Answer
›
Correct Option Option (B) – \(x\sin x\)
🎯 Exam Significance
Exam Significance
›
  • Tests the application of the Fundamental Theorem of Calculus.
  • Establishes the direct connection between differentiation and definite integration.
  • Reinforces the concept of a variable upper limit in a definite integral.

JEE Main and Other Competitive Entrance Examinations:

  • This is a standard application of the Fundamental Theorem of Calculus and can be solved directly without evaluating the integral first.
  • It tests the distinction between a dummy variable of integration and the variable upper limit.
  • It develops speed in solving objective questions involving functions defined through definite integrals.

CUET and Other Entrance Examinations:

  • Tests direct differentiation of a function defined by a definite integral.
  • Requires recognition of the integrand at the variable upper limit.
  • Is particularly useful for quick multiple-choice problem solving.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Fundamental Theorem of Calculus: \(\displaystyle\frac{d}{dx}\left(\int\limits_a^x g(t)\,dt\right)=g(x)\).

  2. Dummy variable: The variable \(t\) is used only for integration and is replaced by \(x\) when applying the theorem.

  3. Variable upper limit: The derivative is obtained by evaluating the integrand at the upper limit.

  4. Time-saving technique: There is no need to calculate the original definite integral before differentiating it.

  5. Common mistake to avoid: Do not apply the product rule or integration by parts here. The Fundamental Theorem of Calculus gives the derivative directly.
← Q9
10 / 10  ·  100%
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🎓

Chapter Complete!

All 10 solutions for Integrals covered.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 12 Maths Ex 7.9 Integrals Solutions
NCERT Class 12 Maths Ex 7.9 Integrals Solutions — Complete Notes & Solutions · academia-aeternum.com
Explore the step-by-step solution to NCERT Class 12 Mathematics Chapter 7, Integrals, Exercise 7.7, Question 11. This multiple-choice problem focuses on evaluating the integral of the square root of a quadratic expression using the method of completing the square and standard integration formulae. The solution explains how to transform the given quadratic into the form \((x-a)^2-b^2\), apply the standard formula for integrating \(\sqrt{u^2-a^2}\), and obtain the correct antiderivative. Special…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    Exercise 7.9 covers important problems on definite integrals, substitution methods, standard integration formulas and the Fundamental Theorem of Calculus.

    You can find detailed NCERT Class 12 Mathematics Exercise 7.9 solutions on Academia Aeternum, with step-by-step calculations and explanations.

    Yes, the solutions include relevant concepts, substitution steps, changes in integration limits, mathematical calculations and final answers.

    Important techniques include algebraic substitution, completing the square, inverse trigonometric integration, fractional-power simplification and differentiation of functions defined by definite integrals.

    Substitution simplifies complicated integrands by transforming them into standard forms and requires corresponding changes in the limits of integration.

    Substitute the original lower and upper limits into the new variable expression to obtain the corresponding transformed limits.

    The Fundamental Theorem of Calculus connects differentiation and integration. In particular, if f(x) = ??? g(t) dt, then f'(x) = g(x), provided g is continuous.

    Yes, these solutions help students understand definite integration techniques, practise NCERT-based questions and develop accuracy in mathematical calculations.

    Yes, the substitution techniques, standard integral forms and Fundamental Theorem of Calculus covered in this exercise are useful for JEE Main, CUET and other entrance examinations.

    Students should revise standard integration formulas, practise substitution and limit transformations, understand each solution step and solve the exercise independently for better accuracy and confidence.

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