Exercise 7.6, Question 1
\(\displaystyle\int x\sin x\,dx\)
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- \(u = x\), \(v = \sin x\): \(x(-\cos x) - \int(-\cos x)\,dx\).
Answer: \(-x\cos x + \sin x + C\)
NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.6
Exercise 7.6: Integration by parts. \(\int u\,v\,dx = u\int v\,dx - \int\left(u'\int v\,dx\right)dx\); choose u by ILATE (inverse trig, log, algebraic, trig, exponential). Useful result: \(\int e^x[f(x) + f'(x)]\,dx = e^xf(x)\). For \(\int e^{ax}\sin bx\,dx\), apply parts twice and solve for the integral.
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Exercise 7.6 questions and solutions
Exercise 7.6, Question 2
\(\displaystyle\int x\sin 3x\,dx\)
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- \[x\left(-\dfrac{\cos 3x}{3}\right) + \int\dfrac{\cos 3x}{3}\,dx\]
Answer: \[-\dfrac{x\cos 3x}{3} + \dfrac{\sin 3x}{9} + C\]
Exercise 7.6, Question 3
\(\displaystyle\int x^2e^x\,dx\)
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- \(x^2e^x - \int 2xe^x\,dx\), and \(\int xe^x\,dx = xe^x - e^x\).
Answer: \(e^x(x^2 - 2x + 2) + C\)
Exercise 7.6, Question 4
\(\displaystyle\int x\log x\,dx\)
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- \(u = \log x\): \[\dfrac{x^2}{2}\log x - \int\dfrac{x^2}{2} \cdot \dfrac1x\,dx\]
Answer: \[\dfrac{x^2}{2}\log x - \dfrac{x^2}{4} + C\]
Exercise 7.6, Question 5
\(\displaystyle\int x\log 2x\,dx\)
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- \(u = \log 2x\), \(u' = \dfrac1x\).
Answer: \[\dfrac{x^2}{2}\log 2x - \dfrac{x^2}{4} + C\]
Exercise 7.6, Question 6
\(\displaystyle\int x^2\log x\,dx\)
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- \[\dfrac{x^3}{3}\log x - \int\dfrac{x^2}{3}\,dx\]
Answer: \[\dfrac{x^3}{3}\log x - \dfrac{x^3}{9} + C\]
Exercise 7.6, Question 7
\(\displaystyle\int x\sin^{-1}x\,dx\)
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- \[\dfrac{x^2}{2}\sin^{-1}x - \tfrac12\int\dfrac{x^2}{\sqrt{1 - x^2}}\,dx\]
- \[\dfrac{x^2}{\sqrt{1 - x^2}} = \dfrac{1}{\sqrt{1 - x^2}} - \sqrt{1 - x^2}\], so \[\int = \sin^{-1}x - \left[\dfrac{x}{2}\sqrt{1 - x^2} + \dfrac12\sin^{-1}x\right]\]
Answer: \[\dfrac{2x^2 - 1}{4}\sin^{-1}x + \dfrac{x}{4}\sqrt{1 - x^2} + C\]
Exercise 7.6, Question 8
\(\displaystyle\int x\tan^{-1}x\,dx\)
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- \[\dfrac{x^2}{2}\tan^{-1}x - \tfrac12\int\dfrac{x^2}{1 + x^2}\,dx = \dfrac{x^2}{2}\tan^{-1}x - \tfrac12(x - \tan^{-1}x)\]
Answer: \[\dfrac{x^2 + 1}{2}\tan^{-1}x - \dfrac{x}{2} + C\]
Exercise 7.6, Question 9
\(\displaystyle\int x\cos^{-1}x\,dx\)
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- \[\dfrac{x^2}{2}\cos^{-1}x + \tfrac12\int\dfrac{x^2}{\sqrt{1 - x^2}}\,dx\], and \[\int\dfrac{x^2}{\sqrt{1 - x^2}}\,dx = \tfrac12\sin^{-1}x - \tfrac{x}{2}\sqrt{1 - x^2}\]
- Using \(\sin^{-1}x = \tfrac{\pi}{2} - \cos^{-1}x\) (the \(\tfrac{\pi}{8}\) goes into C).
Answer: \[\dfrac{2x^2 - 1}{4}\cos^{-1}x - \dfrac{x}{4}\sqrt{1 - x^2} + C\]
Exercise 7.6, Question 10
\(\displaystyle\int (\sin^{-1}x)^2\,dx\)
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- Parts with \(v = 1\): \[x(\sin^{-1}x)^2 - \int\dfrac{2x\sin^{-1}x}{\sqrt{1 - x^2}}\,dx\]
- \[\int\dfrac{x\sin^{-1}x}{\sqrt{1 - x^2}}\,dx = -\sqrt{1 - x^2}\sin^{-1}x + x\] (parts again).
Answer: \[x(\sin^{-1}x)^2 + 2\sqrt{1 - x^2}\sin^{-1}x - 2x + C\]
Exercise 7.6, Question 11
\(\displaystyle\int \dfrac{x\cos^{-1}x}{\sqrt{1 - x^2}}\,dx\)
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- \(u = \cos^{-1}x\), \[\int\dfrac{x}{\sqrt{1 - x^2}}\,dx = -\sqrt{1 - x^2}\]
- \[-\sqrt{1 - x^2}\cos^{-1}x - \int(-\sqrt{1 - x^2})\left(-\dfrac{1}{\sqrt{1 - x^2}}\right)dx\]
Answer: \(-\sqrt{1 - x^2}\cos^{-1}x - x + C\)
Exercise 7.6, Question 12
\(\displaystyle\int x\sec^2 x\,dx\)
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- \(x\tan x - \int\tan x\,dx\).
Answer: \(x\tan x + \log|\cos x| + C\)
Exercise 7.6, Question 13
\(\displaystyle\int \tan^{-1}x\,dx\)
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- Parts with \(v = 1\): \(x\tan^{-1}x - \int\dfrac{x}{1 + x^2}\,dx\).
Answer: \(x\tan^{-1}x - \tfrac12\log(1 + x^2) + C\)
Exercise 7.6, Question 14
\(\displaystyle\int x(\log x)^2\,dx\)
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- \[\dfrac{x^2}{2}(\log x)^2 - \int x\log x\,dx\], then Q4.
Answer: \[\dfrac{x^2}{2}(\log x)^2 - \dfrac{x^2}{2}\log x + \dfrac{x^2}{4} + C\]
Exercise 7.6, Question 15
\(\displaystyle\int (x^2 + 1)\log x\,dx\)
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- \[\left(\dfrac{x^3}{3} + x\right)\log x - \int\left(\dfrac{x^2}{3} + 1\right)dx\]
Answer: \[\left(\dfrac{x^3}{3} + x\right)\log x - \dfrac{x^3}{9} - x + C\]
Exercise 7.6, Question 16
\(\displaystyle\int e^x(\sin x + \cos x)\,dx\)
Show solution
- Form \(e^x[f + f']\) with \(f = \sin x\).
Answer: \(e^x\sin x + C\)
Exercise 7.6, Question 17
\(\displaystyle\int \dfrac{xe^x}{(1 + x)^2}\,dx\)
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- \[\dfrac{x}{(1 + x)^2} = \dfrac{1}{1 + x} - \dfrac{1}{(1 + x)^2}\]: form \(e^x[f + f']\) with \(f = \dfrac{1}{1 + x}\).
Answer: \(\dfrac{e^x}{1 + x} + C\)
Exercise 7.6, Question 18
\(\displaystyle\int e^x\left(\dfrac{1 + \sin x}{1 + \cos x}\right)\,dx\)
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- \[\begin{aligned}\dfrac{1 + \sin x}{1 + \cos x} &= \dfrac{1}{2\cos^2\frac{x}{2}} + \dfrac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} \\ &= \tfrac12\sec^2\dfrac{x}{2} + \tan\dfrac{x}{2}\end{aligned}\]
- Form \(e^x[f + f']\) with \(f = \tan\tfrac{x}{2}\).
Answer: \(e^x\tan\dfrac{x}{2} + C\)
Exercise 7.6, Question 19
\(\displaystyle\int e^x\left(\dfrac1x - \dfrac{1}{x^2}\right)\,dx\)
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- \(f = \dfrac1x\), \(f' = -\dfrac{1}{x^2}\).
Answer: \(\dfrac{e^x}{x} + C\)
Exercise 7.6, Question 20
\(\displaystyle\int \dfrac{(x - 3)e^x}{(x - 1)^3}\,dx\)
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- \[\dfrac{x - 3}{(x - 1)^3} = \dfrac{1}{(x - 1)^2} - \dfrac{2}{(x - 1)^3}\]: form \(e^x[f + f']\) with \(f = (x - 1)^{-2}\).
Answer: \(\dfrac{e^x}{(x - 1)^2} + C\)
Exercise 7.6, Question 21
\(\displaystyle\int e^{2x}\sin x\,dx\)
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- Parts twice: \[\begin{aligned}I &= -e^{2x}\cos x + 2\int e^{2x}\cos x\,dx \\ &= -e^{2x}\cos x + 2\left[e^{2x}\sin x - 2I\right]\end{aligned}\]
- \(5I = e^{2x}(2\sin x - \cos x)\).
Answer: \(\dfrac{e^{2x}}{5}(2\sin x - \cos x) + C\)
Exercise 7.6, Question 22
\(\displaystyle\int \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)\,dx\)
Show solution
- For \(|x| \le 1\), \[\sin^{-1}\dfrac{2x}{1 + x^2} = 2\tan^{-1}x\] (put \(x = \tan\theta\)).
- Then use Q13.
Answer: \(2x\tan^{-1}x - \log(1 + x^2) + C\)
Exercise 7.6, Question 23
\(\displaystyle\int x^2e^{x^3}\,dx\) equals: (A) \(\tfrac13e^{x^3} + C\) (B) \(\tfrac13e^{x^2} + C\) (C) \(\tfrac12e^{x^3} + C\) (D) \(\tfrac12e^{x^2} + C\)
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- \(t = x^3\), \(dt = 3x^2\,dx\).
Answer: (A)
Exercise 7.6, Question 24
\(\displaystyle\int e^x\sec x(1 + \tan x)\,dx\) equals: (A) \(e^x\cos x + C\) (B) \(e^x\sec x + C\) (C) \(e^x\sin x + C\) (D) \(e^x\tan x + C\)
Show solution
- \(e^x[\sec x + \sec x\tan x]\): form \(e^x[f + f']\) with \(f = \sec x\).
Answer: (B)
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