Exercise 7.9, Question 1
\(\displaystyle\int_{0}^{1} \dfrac{x}{x^2 + 1}\,dx\)
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- \(t = x^2 + 1\) runs from 1 to 2: \(\tfrac12\int_1^2\dfrac{dt}{t}\).
Answer: \(\tfrac12\log 2\)
NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.9
Exercise 7.9: Definite integrals by substitution. After substituting \(t = g(x)\), change the limits too (\(x = a \to t = g(a)\)) so you never need to go back to x.
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Exercise 7.9 questions and solutions
Exercise 7.9, Question 2
\(\displaystyle\int_0^{\pi/2}\sqrt{\sin\phi}\cos^5\phi\,d\phi\)
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- \(t = \sin\phi\), \(dt = \cos\phi\,d\phi\), \(\cos^4\phi = (1 - t^2)^2\); limits 0 to 1.
- \[\begin{aligned}\int_0^1(t^{1/2} - 2t^{5/2} + t^{9/2})\,dt &= \tfrac23 - \tfrac47 + \tfrac{2}{11} \\ &= \dfrac{154 - 132 + 42}{231}\end{aligned}\]
Answer: \(\tfrac{64}{231}\)
Exercise 7.9, Question 3
\(\displaystyle\int_{0}^{1} \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)\,dx\)
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- For \(0 \le x \le 1\) the integrand is \(2\tan^{-1}x\) (\(x = \tan\theta\)).
- \[2\left[x\tan^{-1}x - \tfrac12\log(1 + x^2)\right]_0^1 = 2\left(\tfrac{\pi}{4} - \tfrac12\log 2\right)\]
Answer: \(\dfrac{\pi}{2} - \log 2\)
Exercise 7.9, Question 4
\(\displaystyle\int_{0}^{2} x\sqrt{x + 2}\,dx\)
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- \(x + 2 = t^2\), \(dx = 2t\,dt\), t from \(\sqrt2\) to 2: \[\int(t^2 - 2)t \cdot 2t\,dt = 2\left[\dfrac{t^5}{5} - \dfrac{2t^3}{3}\right]_{\sqrt2}^{2}\]
- \[\begin{aligned}2\left[\left(\tfrac{32}{5} - \tfrac{16}{3}\right) - \left(\tfrac{4\sqrt2}{5} - \tfrac{4\sqrt2}{3}\right)\right] &= 2\left[\tfrac{16}{15} + \tfrac{8\sqrt2}{15}\right] \\ &= \dfrac{32 + 16\sqrt2}{15}\end{aligned}\]
Answer: \(\dfrac{16\sqrt2(\sqrt2 + 1)}{15}\)
Exercise 7.9, Question 5
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{\sin x}{1 + \cos^2 x}\,dx\)
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- \(t = \cos x\) from 1 to 0: \(\int_0^1\dfrac{dt}{1 + t^2}\).
Answer: \(\tfrac{\pi}{4}\)
Exercise 7.9, Question 6
\(\displaystyle\int_{0}^{2} \dfrac{1}{x + 4 - x^2}\,dx\)
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- \[x + 4 - x^2 = \tfrac{17}{4} - \left(x - \tfrac12\right)^2\]; with \(a = \tfrac{\sqrt{17}}{2}\): \[\dfrac{1}{\sqrt{17}}\left[\log\left|\dfrac{\frac{\sqrt{17}}{2} + x - \frac12}{\frac{\sqrt{17}}{2} - x + \frac12}\right|\right]_0^2\]
- \[= \dfrac{1}{\sqrt{17}}\left[\log\dfrac{\sqrt{17} + 3}{\sqrt{17} - 3} - \log\dfrac{\sqrt{17} - 1}{\sqrt{17} + 1}\right] = \dfrac{1}{\sqrt{17}}\log\dfrac{5 + \sqrt{17}}{5 - \sqrt{17}}\], and \[\begin{aligned}\dfrac{5 + \sqrt{17}}{5 - \sqrt{17}} &= \dfrac{(5 + \sqrt{17})^2}{8} \\ &= \dfrac{21 + 5\sqrt{17}}{4}\end{aligned}\]
Answer: \[\dfrac{1}{\sqrt{17}}\log\dfrac{21 + 5\sqrt{17}}{4}\]
Exercise 7.9, Question 7
\(\displaystyle\int_{-1}^{1} \dfrac{1}{x^2 + 2x + 5}\,dx\)
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- \((x + 1)^2 + 4\): \[\left[\tfrac12\tan^{-1}\dfrac{x + 1}{2}\right]_{-1}^{1} = \tfrac12 \cdot \tfrac{\pi}{4}\]
Answer: \(\tfrac{\pi}{8}\)
Exercise 7.9, Question 8
\(\displaystyle\int_{1}^{2} \left(\dfrac1x - \dfrac{1}{2x^2}\right)e^{2x}\,dx\)
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- \(t = 2x\) from 2 to 4: \[\int_2^4\left(\dfrac1t - \dfrac{1}{t^2}\right)e^t\,dt = \left[\dfrac{e^t}{t}\right]_2^4\] (form \(e^t[f + f']\)).
- \(\dfrac{e^4}{4} - \dfrac{e^2}{2}\).
Answer: \(\dfrac{e^2(e^2 - 2)}{4}\)
Exercise 7.9, Question 9
\(\displaystyle\int_{1/3}^{1}\dfrac{(x - x^3)^{1/3}}{x^4}\,dx\) is: (A) 6 (B) 0 (C) 3 (D) 4
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- \[\dfrac{(x - x^3)^{1/3}}{x^4} = \dfrac{1}{x^3}\left(\dfrac{1}{x^2} - 1\right)^{1/3}\]
- \(t = \dfrac{1}{x^2} - 1\), \(dt = -\dfrac{2}{x^3}\,dx\), t from 8 to 0: \[\begin{aligned}\tfrac12\int_0^8 t^{1/3}\,dt &= \tfrac12 \cdot \tfrac34 \cdot 8^{4/3} \\ &= 6\end{aligned}\]
Answer: (A) 6
Exercise 7.9, Question 10
If \(f(x) = \displaystyle\int_0^x t\sin t\,dt\), then \(f'(x)\) is: (A) \(\cos x + x\sin x\) (B) \(x\sin x\) (C) \(x\cos x\) (D) \(\sin x + x\cos x\)
Show solution
- First fundamental theorem: \[\dfrac{d}{dx}\displaystyle\int_0^x g(t)\,dt = g(x)\]
Answer: (B)
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