# NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.10

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## Chapter 7 Ex.7.10 Question 1

Evaluate the integral $$\int_0^1 {\frac{x}{{{x^2} + 1}}} dx$$

### Solution

$$\int_0^1 {\frac{x}{{{x^2} + 1}}} dx$$

Put, $${x^2} + 1 = t \Rightarrow \; \; 2xdx = dt$$

When, $$x = 0,\;t = 1$$and when $$x = 1,\;t = 2$$

\begin{align}\therefore \int_0^1 {\frac{x}{{{x^2} + 1}}} dx &= \frac{1}{2}\int_1^2 {\frac{{dt}}{t}} \\& = \frac{1}{2}\left[ {\log \left| t \right|} \right]_1^2\\ &= \frac{1}{2}\left[ {\log 2 - \log 1} \right]\\& = \frac{1}{2}\log 2\end{align}

## Chapter 7 Ex.7.10 Question 2

Evaluate the integral $$\int_0^{\frac{\pi }{2}} {\sqrt {\sin \phi } {{\cos }^5}\phi d\phi }$$

### Solution

Consider, $$I = \int_0^{\frac{\pi }{2}} {\sqrt {\sin \phi } {{\cos }^5}\phi d\phi } = \int_0^{\frac{\pi }{2}} {\sqrt {\sin \phi } {{\cos }^4}\phi \cos \phi d\phi }$$

Let $$\sin \phi = t \Rightarrow \; \; \cos \phi d\phi = dt$$

When $$\phi = 0,\;t = 0\,\,{\rm{and when }}\phi = \frac{\pi }{2},\;t = 1$$

\begin{align}\therefore I& = \int_0^1 {\sqrt t {{\left( {1 - {t^2}} \right)}^2}dt} \\ &= \int_0^1 {{t^{\frac{1}{2}}}\left( {1 + {t^4} - 2{t^2}} \right)dt} \\ &= \int_0^1 {\left[ {{t^{\frac{1}{2}}} + {t^{\frac{9}{2}}} - 2{t^{\frac{5}{2}}}} \right]} dt\\ &= \left[ {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}} + \frac{{{t^{\frac{{11}}{2}}}}}{{\frac{{11}}{2}}} - \frac{{2{t^{\frac{7}{2}}}}}{{\frac{7}{2}}}} \right]_0^1\\ &= \frac{2}{3} + \frac{2}{{11}} - \frac{4}{7}\\& = \frac{{154 + 42 - 132}}{{231}} = \frac{{64}}{{231}}\end{align}

## Chapter 7 Ex.7.10 Question 3

Evaluate the integral $$\int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)} dx$$

### Solution

Consider, $$I = \int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)} dx$$

Let $$x = \tan \theta \Rightarrow \; \; dx = {\sec ^2}\theta d\theta$$

When $$x = 0,\theta = 0{\rm{ and when }}x = 1,\theta = \frac{\pi }{4}$$

\begin{align}I &= \int_0^{\frac{\pi }{4}} {{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right){{\sec }^2}\theta d\theta } \\ &= \int_0^{\frac{\pi }{4}} {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right){{\sec }^2}\theta d\theta } \\ &= \int_0^{\frac{\pi }{4}} {2\theta {{\sec }^2}\theta d\theta } \\& = 2\int_0^{\frac{\pi }{4}} {\theta {{\sec }^2}\theta d\theta }\end{align}

Taking $$u = \theta$$and $$v = {\sec ^2}\theta$$and integrating by parts, we get

\begin{align}I &= 2\left[ {\theta \int {{{\sec }^2}\theta } d\theta - \int {\left\{ {\left( {\frac{d}{{d\theta }}\theta } \right)\int {{{\sec }^2}\theta } d\theta } \right\}d\theta } } \right]_0^{\frac{\pi }{4}}\\& = 2\left[ {\theta \tan \theta - \int {\tan \theta d\theta } } \right]_0^{\frac{\pi }{4}}\\ &= 2\left[ {\theta \tan \theta + \log \left| {\cos \theta } \right|} \right]_0^{\frac{\pi }{4}}\\ &= 2\left[ {\frac{\pi }{4}\tan \frac{\pi }{4} + \log \left| {\cos \frac{\pi }{4}} \right| - \log \left| {\cos 0} \right|} \right] = 2\left[ {\frac{\pi }{4} + \log \left( {\frac{1}{{\sqrt 2 }}} \right) - \log 1} \right]\\& = 2\left[ {\frac{\pi }{4} - \frac{1}{2}\log 2} \right]\\& = \frac{\pi }{2} - \log 2\end{align}

## Chapter 7 Ex.7.10 Question 4

Evaluate the integral $$\int_0^2 {x\sqrt {x + 2} } \;\;\;\;\left( {{\rm{Put}}\;x + 2 = {t^2}} \right)$$

### Solution

$$\int_0^2 {x\sqrt {x + 2} } dx$$

Put, $$x + 2 = {t^2} \Rightarrow \; \; dx = 2tdt$$

\begin{align}{\text{When}}\;x &= 0,t = \sqrt 2 \,\,{\text{and}}\,\,{\text{when}}\,\,x = 2,t = 2\\\therefore &\int_0^2 {x\sqrt {x + 2} } dx = \int_{\sqrt 2 }^2 {\left( {{t^2} - 2} \right)\sqrt {{t^2}} 2tdt} \\ &= 2\int_{\sqrt 2 }^2 {\left( {{t^2} - 2} \right){t^2}dt} \\ &= 2\int_{\sqrt 2 }^2 {\left( {{t^4} - 2{t^2}} \right)dt} \\ &= 2\left[ {\frac{{{t^5}}}{5} - \frac{{2{t^3}}}{3}} \right]_{\sqrt 2 }^2\\ &= 2\left[ {\frac{{32}}{5} - \frac{{16}}{3} - \frac{{4\sqrt 2 }}{5} + \frac{{4\sqrt 2 }}{3}} \right] = 2\left[ {\frac{{96 - 80 - 12\sqrt 2 + 20\sqrt 2 }}{{15}}} \right] = 2\left[ {\frac{{16 + 8\sqrt 2 }}{{15}}} \right]\\ &= \frac{{16\left( {2 + \sqrt 2 } \right)}}{{15}}\\ &= \frac{{16\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{15}}\end{align}

## Chapter 7 Ex.7.10 Question 5

Evaluate the integral $$\int_0^{\frac{\pi }{2}} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} dx$$

### Solution

$$\int_0^{\frac{\pi }{2}} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} dx$$

Put, $$\cos x = t \Rightarrow \; \; - \sin xdx = dt$$

When $$x = 0,t = 1\,\,{\text{and when }}x = \frac{\pi }{2},t = 0$$

\begin{align} \Rightarrow \; \; \int_0^{\frac{\pi }{2}} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} dx& = - \int_1^0 {\frac{{dt}}{{1 + {t^2}}}} \\ &= - \left[ {{{\tan }^{ - 1}}t} \right]_1^0\\ &= - \left[ {{{\tan }^{ - 1}}0 - {{\tan }^{ - 1}}1} \right]\\ &= - \left[ { - \frac{\pi }{4}} \right]\\& = \frac{\pi }{4}\end{align}

## Chapter 7 Ex.7.10 Question 6

Evaluate the integral $$\int_0^2 {\frac{{dx}}{{x + 4 - {x^2}}}}$$

### Solution

\begin{align}\int_0^2 {\frac{{dx}}{{x + 4 - {x^2}}}} &= \int_0^2 {\frac{{dx}}{{ - \left( {{x^2} - x - 4} \right)}}} \\ &= \int_{}^{} {\frac{{dx}}{{ - \left( {{x^2} - x + \frac{1}{4} - \frac{1}{4} - 4} \right)}}} = \int_0^2 {\frac{{dx}}{{ - \left[ {{{\left( {x - \frac{1}{2}} \right)}^2} - \frac{{17}}{4}} \right]}}} \\ &= \int_0^2 {\frac{{dx}}{{{{\left( {\frac{{\sqrt {17} }}{2}} \right)}^2} - {{\left( {x - \frac{1}{2}} \right)}^2}}}}\end{align}

Let $$x - \frac{1}{2} = t \Rightarrow \; \; dx = dt$$

\begin{align}{\text{when}}\,\,x = 0,t = - \frac{1}{2}\,\,{\text{and}}\,\,{\text{when}}\,\,x = 2,t = \frac{3}{2}\\\therefore \int_0^2 {\frac{{dx}}{{{{\left( {\frac{{\sqrt {17} }}{2}} \right)}^2} - {{\left( {x - \frac{1}{2}} \right)}^2}}}} = \int_{ - \frac{1}{2}}^{\frac{3}{2}} {\frac{{dt}}{{{{\left( {\frac{{\sqrt {17} }}{2}} \right)}^2} - {t^2}}}}\end{align}

\begin{align}{\text{when}}\,\,x &= 0,t = - \frac{1}{2}\,\,{\text{and}}\,\,{\text{when}}\,\,x = 2,t = \frac{3}{2}\\&\therefore \int_0^2 {\frac{{dx}}{{{{\left( {\frac{{\sqrt {17} }}{2}} \right)}^2} - {{\left( {x - \frac{1}{2}} \right)}^2}}}} = \int_{ - \frac{1}{2}}^{\frac{3}{2}} {\frac{{dt}}{{{{\left( {\frac{{\sqrt {17} }}{2}} \right)}^2} - {t^2}}}}\\&= \left[ {\frac{1}{{2\left( {\frac{{\sqrt {17} }}{2}} \right)}}\log \frac{{\frac{{\sqrt {17} }}{2} + t}}{{\frac{{\sqrt {17} }}{2} - t}}} \right]_{ - \frac{1}{2}}^{\frac{3}{2}} = \frac{1}{{\sqrt {17} }}\left[ {\log \frac{{\frac{{\sqrt {17} }}{2} + \frac{3}{2}}}{{\frac{{\sqrt {17} }}{2} - \frac{3}{2}}} - \frac{{\log \frac{{\sqrt {17} }}{2} - \frac{1}{2}}}{{\log \frac{{\sqrt {17} }}{2} + \frac{1}{2}}}} \right]\\ &= \frac{1}{{\sqrt {17} }}\left[ {\log \frac{{\sqrt {17} + 3}}{{\sqrt {17} - 3}} - \log \frac{{\sqrt {17} - 1}}{{\sqrt {17} + 1}}} \right] = \frac{1}{{\sqrt {17} }}\log \frac{{\sqrt {17} + 3}}{{\sqrt {17} - 3}} \times \frac{{\sqrt {17} + 1}}{{\sqrt {17} - 1}}\\& = \frac{1}{{\sqrt {17} }}\log \left[ {\frac{{17 + 3 + 4\sqrt {17} }}{{17 + 3 - 4\sqrt {17} }}} \right] = \frac{1}{{\sqrt {17} }}\log \left[ {\frac{{20 + 4\sqrt {17} }}{{20 - 4\sqrt {17} }}} \right]\\ &= \frac{1}{{\sqrt {17} }}\log \left( {\frac{{5 + \sqrt {17} }}{{5 - \sqrt {17} }}} \right) = \frac{1}{{\sqrt {17} }}\log \left[ {\frac{{\left( {5 + \sqrt {17} } \right)\left( {5 + \sqrt {17} } \right)}}{{25 - 17}}} \right]\\ &= \frac{1}{{\sqrt {17} }}\log \left[ {\frac{{25 + 17 + 10\sqrt {17} }}{8}} \right] = \frac{1}{{\sqrt {17} }}\log \left( {\frac{{42 + 10\sqrt {17} }}{8}} \right)\\& = \frac{1}{{\sqrt {17} }}\log \left( {\frac{{21 + 5\sqrt {17} }}{4}} \right)\end{align}

## Chapter 7 Ex.7.10 Question 7

Evaluate the integral $$\int_{ - 1}^1 {\frac{{dx}}{{{x^2} + 2x + 5}}}$$

### Solution

$$\int_{ - 1}^1 {\frac{{dx}}{{{x^2} + 2x + 5}}} = \int_{ - 1}^1 {\frac{{dx}}{{\left( {{x^2} + 2x + 1} \right) + 4}}} = \int_{ - 1}^1 {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + {{\left( 2 \right)}^2}}}}$$

Put, $$x + 1 = t \Rightarrow \; \; dx = dt$$

When $$x = - 1,t = 0\,\,{\text{and when }}x = 1,t = 2$$

\begin{align}\int_{ - 1}^1 {\frac{{dx}}{{{{\left( {x - 1} \right)}^2} + {{\left( 2 \right)}^2}}}} &= \int_0^2 {\frac{{dt}}{{{t^2} + {2^2}}}} \\ &= \left[ {\frac{1}{2}{{\tan }^{ - 1}}\frac{t}{2}} \right]_0^2 = \frac{1}{2}{\tan ^{ - 1}}1 - \frac{1}{2}{\tan ^{ - 1}}0\\ &= \frac{1}{2}\left( {\frac{\pi }{4}} \right) = \frac{\pi }{8}\end{align}

## Chapter 7 Ex.7.10 Question 8

Evaluate the integral $$\int_1^2 {\left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right)} {e^{2x}}dx$$

### Solution

$$\int_1^2 {\left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right)} {e^{2x}}dx$$

Put, $$2x = t \Rightarrow \; \; 2dx = dt$$

When $$x = 1,t = 2\,\,{\text{and}}\,{\text{when}}\,\,x = 2,t = 4$$

\begin{align}\therefore \int_1^2 {\left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right){e^{2x}}dx}& = \frac{1}{2}\int_2^4 {\left( {\frac{2}{t} - \frac{2}{{{t^2}}}} \right){e^t}dt} \\& = \int_2^4 {\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right)} {e^t}dt\end{align}

Let $$\frac{1}{t} = f\left( t \right)$$

Then, $$f'\left( t \right) = - \frac{1}{{{t^2}}}$$

\begin{align} \Rightarrow \; \; \int_2^4 {\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right)} {e^t}dt &= \int_2^4 {{e^t}\left[ {f\left( t \right) + f'\left( t \right)} \right]} dt\\& = \left[ {{e^t}f\left( t \right)} \right]_2^4\\ &= \left[ {{e^t}.\frac{1}{t}} \right]_2^4\\ &= \left[ {\frac{{{e^t}}}{t}} \right]_2^4\\& = \frac{{{e^4}}}{4} - \frac{{{e^2}}}{2}\\ &= \frac{{{e^2}\left( {{e^2} - 2} \right)}}{4}\end{align}

## Chapter 7 Ex.7.10 Question 9

The value of the integral $$\int_{\frac{1}{3}}^1 {\frac{{{{\left( {x - {x^3}} \right)}^{\frac{1}{3}}}}}{{{x^4}}}} dx$$ is

A. $$6$$

B. $$0$$

C. $$3$$

D. $$4$$

### Solution

Consider, $$I = \int_{\frac{1}{3}}^1 {\frac{{{{\left( {x - {x^3}} \right)}^{\frac{1}{3}}}}}{{{x^4}}}} dx$$

Let $$x = \sin \theta \Rightarrow \; \; dx = \cos \theta d\theta$$

When $$x = \frac{1}{3},\theta = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$$ and when $$x = 1,\theta = \frac{\pi }{2}$$

\begin{align} \Rightarrow \; \; I &= \int_{{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)}^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \theta - {{\sin }^3}\theta } \right)}^{\frac{1}{3}}}}}{{{{\sin }^4}\theta }}} \cos \theta d\theta \\ &= \int_{{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)}^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \theta } \right)}^{\frac{1}{3}}}{{\left( {1 - {{\sin }^2}\theta } \right)}^{\frac{1}{3}}}}}{{{{\sin }^4}\theta }}} \cos \theta d\theta = \int_{{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)}^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \theta } \right)}^{\frac{1}{3}}}{{\left( {\cos \theta } \right)}^{\frac{2}{3}}}}}{{{{\sin }^4}\theta }}} \cos \theta d\theta \\ &= \int_{{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)}^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \theta } \right)}^{\frac{1}{3}}}{{\left( {\cos \theta } \right)}^{\frac{2}{3}}}}}{{{{\sin }^2}\theta {{\sin }^2}\theta }}} \cos \theta d\theta = \int_{{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)}^{\frac{\pi }{2}} {\frac{{{{\left( {\cos \theta } \right)}^{\frac{5}{3}}}}}{{{{\left( {\sin \theta } \right)}^{\frac{5}{3}}}}}} \cos e{c^2}\theta d\theta \\& = \int_{{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)}^{\frac{\pi }{2}} {{{\left( {\cot \theta } \right)}^{\frac{5}{3}}}} \cos e{c^2}\theta d\theta\end{align}

Put $$\cot \theta = t \Rightarrow \; \; - \cos e{c^2}\theta d\theta = dt$$

When $$\theta = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right),t = 2\sqrt 2$$ and when $$\theta = \frac{\pi }{2},t = 0$$

\begin{align}\therefore I &= - \int_{2\sqrt 2 }^0 {{{\left( t \right)}^{\frac{5}{3}}}} dt\\& = - \left[ {\frac{3}{8}{{\left( t \right)}^{\frac{8}{3}}}} \right]_{2\sqrt 2 }^0\\& = - \frac{3}{8}\left[ { - {{\left( {2\sqrt 2 } \right)}^{\frac{8}{3}}}} \right]_{2\sqrt 2 }^0 = \frac{3}{8}\left[ {{{\left( {\sqrt 8 } \right)}^{\frac{8}{3}}}} \right]\\ &= \frac{3}{8}\left[ {{{\left( 8 \right)}^{\frac{4}{3}}}} \right]\\& = \frac{3}{8}\left[ {16} \right]\\& = 6\end{align}

Thus, the correct option is A.

## Chapter 7 Ex.7.10 Question 10

If $$f\left( x \right) = \int_0^x {t\sin tdt}$$, then $$f'\left( x \right)$$ is

\begin{align}&A.\,\,\cos x + x\sin x\\&B.\,\,x\sin x\\&C.\,\,x\cos x\\&D.\,\,\sin x + x\cos x\end{align}

### Solution

$$f\left( x \right) = \int_0^x {t\sin tdt}$$

Using integration by parts, we get

\begin{align}f\left( x \right) &= t\int_0^x {\sin tdt} - \int_0^x {\left\{ {\left( {\frac{d}{{dt}}t} \right)\int {\sin tdt} } \right\}} dt\\ &= \left[ {t\left( { - \cos t} \right)} \right]_0^x - \int_0^x {\left( { - \cos t} \right)} dt\\& = \left[ { - t\cos t + \sin t} \right]_0^x\\& = - x\cos x + \sin x\\& \Rightarrow \; \; f'\left( x \right) = - \left[ {\left\{ {x\left( { - \sin x} \right)} \right\} + \cos x} \right] + \cos x\\ &= x\sin x - \cos x + \cos x\\ &= x\sin x\end{align}

Thus, the correct option is B.

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