Exercise 7.9 Practice
Definite Integrals by Substitution – NCERT Solutions
Q1
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Evaluate the integral: $\int_0^1 \frac{x}{x^2+1} dx$
Let $x^2+1 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}$.
Limits: When $x=0, t=1$. When $x=1, t=2$.
$\int_1^2 \frac{1}{t} \frac{dt}{2} = \frac{1}{2} [\log t]_1^2$.
$= \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} \log 2$.
$\boxed{\frac{1}{2}\log 2}$
Q2
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Evaluate the integral: $\int_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi d\phi$
Let $\sin \phi = t \Rightarrow \cos \phi d\phi = dt$. Limits: $0 \to 1$.
$\cos^4 \phi = (1-\sin^2 \phi)^2 = (1-t^2)^2$.
$\int_0^1 \sqrt{t} (1-t^2)^2 dt = \int_0^1 (t^{1/2} - 2t^{5/2} + t^{9/2}) dt$.
$[\frac{2}{3}t^{3/2} - \frac{4}{7}t^{7/2} + \frac{2}{11}t^{11/2}]_0^1 = \frac{2}{3} - \frac{4}{7} + \frac{2}{11}$.
$\boxed{\frac{64}{231}}$
Q3
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Evaluate the integral: $\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2}\right) dx$
Let $x = \tan \theta \Rightarrow dx = \sec^2 \theta d\theta$. Limits: $0 \to \pi/4$.
$\sin^{-1}(\sin 2\theta) = 2\theta$. Integral: $\int_0^{\pi/4} 2\theta \sec^2 \theta d\theta$.
By parts: $2[\theta \tan \theta - \log|\sec \theta|]_0^{\pi/4}$.
$2[\frac{\pi}{4}(1) - \log \sqrt{2}] = \frac{\pi}{2} - 2(\frac{1}{2}\log 2)$.
$\boxed{\frac{\pi}{2} - \log 2}$
Q4
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Evaluate the integral: $\int_0^2 x\sqrt{x+2} dx$
Let $x+2 = t^2 \Rightarrow dx = 2t dt$. Limits: $x=0 \to t=\sqrt{2}, x=2 \to t=2$.
$\int_{\sqrt{2}}^2 (t^2-2) t (2t) dt = 2 \int_{\sqrt{2}}^2 (t^4 - 2t^2) dt$.
$2 [\frac{t^5}{5} - \frac{2t^3}{3}]_{\sqrt{2}}^2 = 2 [(\frac{32}{5} - \frac{16}{3}) - (\frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3})]$.
Simplify to get:
$\boxed{\frac{16\sqrt{2}(\sqrt{2}+1)}{15}}$
Q5
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Evaluate the integral: $\int_0^{\pi/2} \frac{\sin x}{1+\cos^2 x} dx$
Let $\cos x = t \Rightarrow -\sin x dx = dt$. Limits: $1 \to 0$.
$\int_1^0 \frac{-dt}{1+t^2} = \int_0^1 \frac{dt}{1+t^2}$.
$[\tan^{-1} t]_0^1 = \frac{\pi}{4} - 0$.
$\boxed{\frac{\pi}{4}}$
Q6
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Evaluate the integral: $\int_0^2 \frac{dx}{x+4-x^2}$
$x+4-x^2 = \frac{17}{4} - (x-\frac{1}{2})^2 = (\frac{\sqrt{17}}{2})^2 - (x-\frac{1}{2})^2$.
$\frac{1}{\sqrt{17}} \log|\frac{\sqrt{17} + 2x - 1}{\sqrt{17} - 2x + 1}|_0^2$.
Upper: $\frac{1}{\sqrt{17}} \log \frac{\sqrt{17}+3}{\sqrt{17}-3}$. Lower: $\frac{1}{\sqrt{17}} \log \frac{\sqrt{17}-1}{\sqrt{17}+1}$.
$\boxed{\frac{1}{\sqrt{17}}\log\left(\frac{21+5\sqrt{17}}{4}\right)}$
Q7
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Evaluate the integral: $\int_{-1}^1 \frac{dx}{x^2+2x+5}$
$x^2+2x+5 = (x+1)^2 + 2^2$.
$\int_{-1}^1 \frac{dx}{(x+1)^2+2^2} = [\frac{1}{2}\tan^{-1}(\frac{x+1}{2})]_{-1}^1$.
$\frac{1}{2}(\tan^{-1} 1 - \tan^{-1} 0) = \frac{1}{2}(\frac{\pi}{4})$.
$\boxed{\frac{\pi}{8}}$
Q8
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Evaluate the integral: $\int_1^2 (\frac{1}{x} - \frac{1}{2x^2}) e^{2x} dx$
Let $2x = t \Rightarrow dx = dt/2$. Limits: $2 \to 4$.
$\int_2^4 (\frac{2}{t} - \frac{2}{t^2}) e^t \frac{dt}{2} = \int_2^4 e^t (\frac{1}{t} - \frac{1}{t^2}) dt$.
$[e^t \frac{1}{t}]_2^4 = \frac{e^4}{4} - \frac{e^2}{2}$.
$\boxed{\frac{e^2(e^2-2)}{4}}$
Q9
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The value of the integral $\int_{1/3}^1 \frac{(x-x^3)^{1/3}}{x^4} dx$ is:
(A) 6
(B) 0
(C) 3
(D) 4
(A) 6
(B) 0
(C) 3
(D) 4
Rewrite integrand: $\frac{x(1-x^2)^{1/3}}{x^4} = (x^{-2}-1)^{1/3} x^{-3}$.
Let $x^{-2}-1 = t \Rightarrow -2x^{-3} dx = dt$. Limits: $8 \to 0$.
$\int_8^0 t^{1/3} (-\frac{dt}{2}) = \frac{1}{2} \int_0^8 t^{1/3} dt$.
$\frac{1}{2} [\frac{3}{4} t^{4/3}]_0^8 = \frac{3}{8} (16) = 6$.
$\boxed{\text{(A) } 6}$
Q10
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If $f(x) = \int_0^x t \sin t dt$, then $f'(x)$ is:
(A) $\cos x + x \sin x$
(B) $x \sin x$
(C) $x \cos x$
(D) $\sin x + x \cos x$
(A) $\cos x + x \sin x$
(B) $x \sin x$
(C) $x \cos x$
(D) $\sin x + x \cos x$
By Fundamental Theorem of Calculus, $\frac{d}{dx} \int_0^x g(t) dt = g(x)$.
Here $g(t) = t \sin t$.
So $f'(x) = x \sin x$.
$\boxed{\text{(B) } x \sin x}$