Exercise 7.2, Question 1
\(\displaystyle\int \dfrac{2x}{1 + x^2}\,dx\)
Show solution
- \(t = 1 + x^2\), \(dt = 2x\,dx\): \(\int\dfrac{dt}{t}\).
Answer: \(\log(1 + x^2) + C\)
NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2
Exercise 7.2: Integration by substitution. Look for a function and its derivative in the integrand: put \(t = g(x)\), \(dt = g'(x)\,dx\), integrate in t, then put \(g(x)\) back. Useful results: \(\int\dfrac{f'(x)}{f(x)}\,dx = \log|f(x)|\), \(\int\tan x\,dx = \log|\sec x|\), \(\int\cot x\,dx = \log|\sin x|\).
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Exercise 7.2 questions and solutions
Exercise 7.2, Question 2
\(\displaystyle\int \dfrac{(\log x)^2}{x}\,dx\)
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- \(t = \log x\), \(dt = \dfrac{dx}{x}\): \(\int t^2\,dt\).
Answer: \(\dfrac{(\log x)^3}{3} + C\)
Exercise 7.2, Question 3
\(\displaystyle\int \dfrac{1}{x + x\log x}\,dx\)
Show solution
- \(\dfrac{1}{x(1 + \log x)}\); \(t = 1 + \log x\), \(dt = \dfrac{dx}{x}\).
Answer: \(\log|1 + \log x| + C\)
Exercise 7.2, Question 4
\(\displaystyle\int \sin x\sin(\cos x)\,dx\)
Show solution
- \(t = \cos x\), \(dt = -\sin x\,dx\): \(-\int\sin t\,dt = \cos t\).
Answer: \(\cos(\cos x) + C\)
Exercise 7.2, Question 5
\(\displaystyle\int \sin(ax + b)\cos(ax + b)\,dx\)
Show solution
- \[\sin\theta\cos\theta = \tfrac12\sin 2\theta\]: \(\tfrac12\int\sin(2ax + 2b)\,dx\).
Answer: \(-\dfrac{\cos 2(ax + b)}{4a} + C\)
Exercise 7.2, Question 6
\(\displaystyle\int \sqrt{ax + b}\,dx\)
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- \(t = ax + b\), \(dx = \dfrac{dt}{a}\): \(\dfrac1a\int t^{1/2}\,dt\).
Answer: \(\dfrac{2}{3a}(ax + b)^{3/2} + C\)
Exercise 7.2, Question 7
\(\displaystyle\int x\sqrt{x + 2}\,dx\)
Show solution
- \(t = x + 2\): \[\int(t - 2)t^{1/2}\,dt = \int(t^{3/2} - 2t^{1/2})\,dt\]
Answer: \[\tfrac25(x + 2)^{5/2} - \tfrac43(x + 2)^{3/2} + C\]
Exercise 7.2, Question 8
\(\displaystyle\int x\sqrt{1 + 2x^2}\,dx\)
Show solution
- \(t = 1 + 2x^2\), \(dt = 4x\,dx\): \(\tfrac14\int t^{1/2}\,dt\).
Answer: \(\tfrac16(1 + 2x^2)^{3/2} + C\)
Exercise 7.2, Question 9
\(\displaystyle\int (4x + 2)\sqrt{x^2 + x + 1}\,dx\)
Show solution
- \(t = x^2 + x + 1\), \(dt = (2x + 1)\,dx\): \(2\int t^{1/2}\,dt\).
Answer: \(\tfrac43(x^2 + x + 1)^{3/2} + C\)
Exercise 7.2, Question 10
\(\displaystyle\int \dfrac{1}{x - \sqrt x}\,dx\)
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- \(\dfrac{1}{\sqrt x(\sqrt x - 1)}\); \(t = \sqrt x - 1\), \(dt = \dfrac{dx}{2\sqrt x}\): \(2\int\dfrac{dt}{t}\).
Answer: \(2\log|\sqrt x - 1| + C\)
Exercise 7.2, Question 11
\(\displaystyle\int \dfrac{x}{\sqrt{x + 4}}\,dx\), \(x > 0\)
Show solution
- \(t = x + 4\): \[\int\dfrac{t - 4}{\sqrt t}\,dt = \tfrac23t^{3/2} - 8t^{1/2}\]
- \(= \tfrac23\sqrt{x + 4}\,[(x + 4) - 12]\).
Answer: \(\tfrac23\sqrt{x + 4}\,(x - 8) + C\)
Exercise 7.2, Question 12
\(\displaystyle\int (x^3 - 1)^{1/3}x^5\,dx\)
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- \(t = x^3 - 1\), \(dt = 3x^2\,dx\), \(x^3 = t + 1\): \[\tfrac13\int(t + 1)t^{1/3}\,dt = \tfrac13\left[\tfrac37t^{7/3} + \tfrac34t^{4/3}\right]\]
Answer: \[\tfrac17(x^3 - 1)^{7/3} + \tfrac14(x^3 - 1)^{4/3} + C\]
Exercise 7.2, Question 13
\(\displaystyle\int \dfrac{x^2}{(2 + 3x^3)^3}\,dx\)
Show solution
- \(t = 2 + 3x^3\), \(dt = 9x^2\,dx\): \[\tfrac19\int t^{-3}\,dt = -\tfrac{1}{18}t^{-2}\]
Answer: \(-\dfrac{1}{18(2 + 3x^3)^2} + C\)
Exercise 7.2, Question 14
\(\displaystyle\int\dfrac{1}{x(\log x)^m}\,dx\), \(x > 0\), \(m \ne 1\)
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- \(t = \log x\), \(dt = \dfrac{dx}{x}\): \(\int t^{-m}\,dt = \dfrac{t^{1-m}}{1 - m}\).
Answer: \(\dfrac{(\log x)^{1-m}}{1 - m} + C\)
Exercise 7.2, Question 15
\(\displaystyle\int \dfrac{x}{9 - 4x^2}\,dx\)
Show solution
- \(t = 9 - 4x^2\), \(dt = -8x\,dx\): \(-\tfrac18\int\dfrac{dt}{t}\).
Answer: \(-\tfrac18\log|9 - 4x^2| + C\)
Exercise 7.2, Question 16
\(\displaystyle\int e^{2x + 3}\,dx\)
Show solution
- \(t = 2x + 3\).
Answer: \(\tfrac12e^{2x + 3} + C\)
Exercise 7.2, Question 17
\(\displaystyle\int \dfrac{x}{e^{x^2}}\,dx\)
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- \(t = -x^2\), \(dt = -2x\,dx\): \(-\tfrac12\int e^t\,dt\).
Answer: \(-\dfrac{1}{2e^{x^2}} + C\)
Exercise 7.2, Question 18
\(\displaystyle\int \dfrac{e^{\tan^{-1}x}}{1 + x^2}\,dx\)
Show solution
- \(t = \tan^{-1}x\), \(dt = \dfrac{dx}{1 + x^2}\).
Answer: \(e^{\tan^{-1}x} + C\)
Exercise 7.2, Question 19
\(\displaystyle\int \dfrac{e^{2x} - 1}{e^{2x} + 1}\,dx\)
Show solution
- Divide top and bottom by \(e^x\): \(\dfrac{e^x - e^{-x}}{e^x + e^{-x}}\); the top is the derivative of the bottom.
Answer: \(\log(e^x + e^{-x}) + C\)
Exercise 7.2, Question 20
\(\displaystyle\int \dfrac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}\,dx\)
Show solution
- \(t = e^{2x} + e^{-2x}\), \(dt = 2(e^{2x} - e^{-2x})\,dx\).
Answer: \(\tfrac12\log(e^{2x} + e^{-2x}) + C\)
Exercise 7.2, Question 21
\(\displaystyle\int \tan^2(2x - 3)\,dx\)
Show solution
- \(\tan^2\theta = \sec^2\theta - 1\).
Answer: \(\tfrac12\tan(2x - 3) - x + C\)
Exercise 7.2, Question 22
\(\displaystyle\int \sec^2(7 - 4x)\,dx\)
Show solution
- \[\int\sec^2(ax + b)\,dx = \tfrac1a\tan(ax + b)\], with \(a = -4\).
Answer: \(-\tfrac14\tan(7 - 4x) + C\)
Exercise 7.2, Question 23
\(\displaystyle\int \dfrac{\sin^{-1}x}{\sqrt{1 - x^2}}\,dx\)
Show solution
- \(t = \sin^{-1}x\), \(dt = \dfrac{dx}{\sqrt{1 - x^2}}\).
Answer: \(\tfrac12(\sin^{-1}x)^2 + C\)
Exercise 7.2, Question 24
\(\displaystyle\int \dfrac{2\cos x - 3\sin x}{6\cos x + 4\sin x}\,dx\)
Show solution
- Take 2 out of the denominator: \[\dfrac12 \cdot \dfrac{2\cos x - 3\sin x}{3\cos x + 2\sin x}\]
- \((3\cos x + 2\sin x)' = 2\cos x - 3\sin x\), the numerator: \(t = 3\cos x + 2\sin x\) gives \(\tfrac12\int\dfrac{dt}{t}\).
Answer: \(\tfrac12\log|3\cos x + 2\sin x| + C\)
Exercise 7.2, Question 25
\(\displaystyle\int \dfrac{1}{\cos^2 x(1 - \tan x)^2}\,dx\)
Show solution
- \(t = 1 - \tan x\), \(dt = -\sec^2 x\,dx\): \(-\int t^{-2}\,dt = t^{-1}\).
Answer: \(\dfrac{1}{1 - \tan x} + C\)
Exercise 7.2, Question 26
\(\displaystyle\int \dfrac{\cos\sqrt x}{\sqrt x}\,dx\)
Show solution
- \(t = \sqrt x\), \(dt = \dfrac{dx}{2\sqrt x}\).
Answer: \(2\sin\sqrt x + C\)
Exercise 7.2, Question 27
\(\displaystyle\int \sqrt{\sin 2x}\cos 2x\,dx\)
Show solution
- \(t = \sin 2x\), \(dt = 2\cos 2x\,dx\): \(\tfrac12\int t^{1/2}\,dt\).
Answer: \(\tfrac13(\sin 2x)^{3/2} + C\)
Exercise 7.2, Question 28
\(\displaystyle\int \dfrac{\cos x}{\sqrt{1 + \sin x}}\,dx\)
Show solution
- \(t = 1 + \sin x\), \(dt = \cos x\,dx\): \(\int t^{-1/2}\,dt\).
Answer: \(2\sqrt{1 + \sin x} + C\)
Exercise 7.2, Question 29
\(\displaystyle\int \cot x\log\sin x\,dx\)
Show solution
- \(t = \log\sin x\), \(dt = \cot x\,dx\).
Answer: \(\tfrac12(\log\sin x)^2 + C\)
Exercise 7.2, Question 30
\(\displaystyle\int \dfrac{\sin x}{1 + \cos x}\,dx\)
Show solution
- \(t = 1 + \cos x\), \(dt = -\sin x\,dx\).
Answer: \(-\log|1 + \cos x| + C\)
Exercise 7.2, Question 31
\(\displaystyle\int \dfrac{\sin x}{(1 + \cos x)^2}\,dx\)
Show solution
- \(t = 1 + \cos x\): \(-\int t^{-2}\,dt = t^{-1}\).
Answer: \(\dfrac{1}{1 + \cos x} + C\)
Exercise 7.2, Question 32
\(\displaystyle\int \dfrac{1}{1 + \cot x}\,dx\)
Show solution
- \[\dfrac{\sin x}{\sin x + \cos x} = \dfrac12 \cdot \dfrac{(\sin x + \cos x) - (\cos x - \sin x)}{\sin x + \cos x}\]
- The second part is \[\dfrac{(\sin x + \cos x)'}{\sin x + \cos x}\]
Answer: \[\dfrac{x}{2} - \tfrac12\log|\sin x + \cos x| + C\]
Exercise 7.2, Question 33
\(\displaystyle\int \dfrac{1}{1 - \tan x}\,dx\)
Show solution
- \[\dfrac{\cos x}{\cos x - \sin x} = \dfrac12 \cdot \dfrac{(\cos x - \sin x) + (\cos x + \sin x)}{\cos x - \sin x}\]
- \((\cos x - \sin x)' = -(\sin x + \cos x)\), so the second part integrates to \(-\log|\cos x - \sin x|\).
Answer: \[\dfrac{x}{2} - \tfrac12\log|\cos x - \sin x| + C\]
Exercise 7.2, Question 34
\(\displaystyle\int \dfrac{\sqrt{\tan x}}{\sin x\cos x}\,dx\)
Show solution
- \(\sin x\cos x = \tan x\cos^2 x\), so the integrand is \[\dfrac{\sqrt{\tan x}\sec^2 x}{\tan x} = \dfrac{\sec^2 x}{\sqrt{\tan x}}\]
- \(t = \tan x\): \(\int t^{-1/2}\,dt = 2\sqrt t\).
Answer: \(2\sqrt{\tan x} + C\)
Exercise 7.2, Question 35
\(\displaystyle\int \dfrac{(1 + \log x)^2}{x}\,dx\)
Show solution
- \(t = 1 + \log x\), \(dt = \dfrac{dx}{x}\).
Answer: \(\tfrac13(1 + \log x)^3 + C\)
Exercise 7.2, Question 36
\(\displaystyle\int \dfrac{(x + 1)(x + \log x)^2}{x}\,dx\)
Show solution
- \[\begin{aligned}\dfrac{x + 1}{x} &= 1 + \dfrac1x \\ &= (x + \log x)'\end{aligned}\]; \(t = x + \log x\).
Answer: \(\tfrac13(x + \log x)^3 + C\)
Exercise 7.2, Question 37
\(\displaystyle\int \dfrac{x^3\sin(\tan^{-1}x^4)}{1 + x^8}\,dx\)
Show solution
- \(t = \tan^{-1}x^4\), \(dt = \dfrac{4x^3}{1 + x^8}\,dx\): \(\tfrac14\int\sin t\,dt\).
Answer: \(-\tfrac14\cos(\tan^{-1}x^4) + C\)
Exercise 7.2, Question 38
\(\displaystyle\int\dfrac{10x^9 + 10^x\log_e 10}{x^{10} + 10^x}\,dx\) equals: (A) \(10^x - x^{10} + C\) (B) \(10^x + x^{10} + C\) (C) \((10^x - x^{10})^{-1} + C\) (D) \(\log(10^x + x^{10}) + C\)
Show solution
- The numerator is the derivative of \(x^{10} + 10^x\).
Answer: (D)
Exercise 7.2, Question 39
\(\displaystyle\int\dfrac{dx}{\sin^2 x\cos^2 x}\) equals: (A) \(\tan x + \cot x + C\) (B) \(\tan x - \cot x + C\) (C) \(\tan x\cot x + C\) (D) \(\tan x - \cot 2x + C\)
Show solution
- \[\begin{aligned}\dfrac{1}{\sin^2 x\cos^2 x} &= \dfrac{\sin^2 x + \cos^2 x}{\sin^2 x\cos^2 x} \\ &= \sec^2 x + \csc^2 x\end{aligned}\]
Answer: (B)
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