NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6

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Chapter 7 Ex.7.6 Question 1

Integrate \(x\sin x\)

Solution

Let \(I = \int {x\sin x} dx\)

Taking \(u = x\)and \(v = \sin x\) and integrating by parts,

\[\begin{align}I &= x\int {\sin xdx - \int {\left\{ {\left( {\frac{d}{{dx}}\left( x \right)} \right)\int {\sin xdx} } \right\}dx} } \\ &= x\left( { - \cos x} \right) - \int {1 \cdot \left( { - \cos x} \right)dx} \\& = - x\cos x + \sin x + C\end{align}\]

Chapter 7 Ex.7.6 Question 2

Integrate \(x\sin 3x\)

Solution

Let \(I = \int {x\sin 3x} dx\)

Taking \(u = x\)and \(v = \sin 3x\) and integrating by parts,

\[\begin{align}I &= x\int {\sin 3xdx - \int {\left\{ {\left( {\frac{d}{{dx}}x} \right)\int {\sin 3xdx} } \right\}dx} } \\ &= x\left( {\frac{{ - \cos 3x}}{3}} \right) - \int {1 \cdot \left( {\frac{{ - \cos 3x}}{3}} \right)dx} \\ &= \frac{{ - x\cos 3x}}{3} + \frac{1}{3}\int {\cos 3xdx} \\& = \frac{{ - x\cos 3x}}{3} + \frac{1}{9}\sin 3x + C\end{align}\]

Chapter 7 Ex.7.6 Question 3

Integrate \({x^2}{e^x}\)

Solution

Let \(I = \int {{x^2}{e^x}} dx\)

Taking \(u = {x^2}\)and \(v = {e^x}\) and integrating by parts, we get

\[\begin{align}I &= {x^2}\int {{e^x}dx - \int {\left\{ {\left( {\frac{d}{{dx}}{x^2}} \right)\int {{e^x}dx} } \right\}dx} } \\ &= {x^2}{e^x} - \int {2x.{e^x}dx} \\& = {x^2}{e^x} - 2\int {x.{e^x}dx} \end{align}\]

Again using integration by parts, we get

\[\begin{align}& = {x^2}{e^x} - 2\left[ {x\int {{e^x}dx - \int {\left\{ {\left( {\frac{d}{{dx}}x} \right)\int {{e^x}dx} } \right\}dx} } } \right]\\ &= {x^2}{e^x} - 2\left[ {x{e^x} - \int {{e^x}dx} } \right]\\ &= {x^2}{e^x} - 2\left[ {x{e^x} - {e^x}} \right]\\& = {x^2}{e^x} - 2x{e^x} + 2{e^x} + C\\ &= {e^x}\left( {{x^2} - 2x + 2} \right) + C\end{align}\]

Chapter 7 Ex.7.6 Question 4

Integrate \(x\log x\)

Solution

Let \(I = \int {x\log xdx} \)

Taking \(u = \log x\)and \(v = x\) and integrating by parts, we get

\[\begin{align}I &= \log x\int {xdx - \int {\left\{ {\left( {\frac{d}{{dx}}\log x} \right)\int {xdx} } \right\}dx} } \\ &= \log x.\frac{{{x^2}}}{2} - \int {\frac{1}{x}.\frac{{{x^2}}}{2}dx} \\ &= \frac{{{x^2}\log x}}{2} - \int {\frac{x}{2}dx} \\ &= \frac{{{x^2}\log x}}{2} - \frac{{{x^2}}}{4} + C\end{align}\]

Chapter 7 Ex.7.6 Question 5

Integrate \(x\log 2x\)

Solution

Let \(I = \int {x\log 2xdx} \)

Taking \(u = \log 2x\)and \(v = x\) and integrating by parts, we get

\[\begin{align}I &= \log 2x\int {xdx - \int {\left\{ {\left( {\frac{d}{{dx}}\log 2x} \right)\int {xdx} } \right\}dx} } \\& = \log 2x.\frac{{{x^2}}}{2} - \int {\frac{2}{{2x}}.\frac{{{x^2}}}{2}dx} \\& = \frac{{{x^2}\log 2x}}{2} - \int {\frac{x}{2}dx} \\& = \frac{{{x^2}\log 2x}}{2} - \frac{{{x^2}}}{4} + C\end{align}\]

Chapter 7 Ex.7.6 Question 6

Integrate \({x^2}\log x\)

Solution

Let \(I = \int {{x^2}\log xdx} \)

Taking \(u = \log x\)and \(v = {x^2}\) and integrating by parts, we get

\[\begin{align}I &= \log x\int {{x^2}dx - \int {\left\{ {\left( {\frac{d}{{dx}}\log x} \right)\int {{x^2}dx} } \right\}dx} } \\& = \log x.\left( {\frac{{{x^3}}}{3}} \right) - \int {\frac{1}{x}.\frac{{{x^3}}}{3}dx} \\ &= \frac{{{x^3}\log x}}{3} - \int {\frac{{{x^2}}}{3}dx} \\ &= \frac{{{x^3}\log x}}{3} - \frac{{{x^3}}}{9} + C\end{align}\]

Chapter 7 Ex.7.6 Question 7

Integrate \(x{\sin ^{ - 1}}x\)

Solution

Let \(I = \int {x{{\sin }^{ - 1}}xdx} \)

Taking \(u = {\sin ^{ - 1}}x\)and \(v = x\) and integrating by parts, we get

\[\begin{align}I &= {\sin ^{ - 1}}x\int {xdx - \int {\left\{ {\left( {\frac{d}{{dx}}{{\sin }^{ - 1}}x} \right)\int {xdx} } \right\}dx} } \\& = {\sin ^{ - 1}}x\left( {\frac{{{x^2}}}{2}} \right) - \int {\frac{1}{{\sqrt {1 - {x^2}} }}.\frac{{{x^2}}}{2}dx} \\ &= \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{1}{2}\int {\frac{{ - {x^2}}}{{\sqrt {1 - {x^2}} }}dx} \\ &= \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{1}{2}\int {\left\{ {\frac{{1 - {x^2}}}{{\sqrt {1 - {x^2}} }} - \frac{1}{{\sqrt {1 - {x^2}} }}} \right\}dx} \\ &= \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{1}{2}\int {\left\{ {\sqrt {1 - {x^2}} - \frac{1}{{\sqrt {1 - {x^2}} }}} \right\}dx} \\ &= \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{1}{2}\left\{ {\int {\sqrt {1 - {x^2}} } dx - \int {\frac{1}{{\sqrt {1 - {x^2}} }}dx} } \right\}\\ &= \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{1}{2}\left\{ {\frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}x - {{\sin }^{ - 1}}x} \right\} + C\\ &= \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{x}{4}\sqrt {1 - {x^2}} + \frac{1}{4}{\sin ^{ - 1}}x - \frac{1}{2}{\sin ^{ - 1}}x + C\\& = \frac{1}{4}\left( {2{x^2} - 1} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + C\end{align}\]

Chapter 7 Ex.7.6 Question 8

Integrate \(x{\tan ^{ - 1}}x\)

Solution

Let \(I = \int {x{{\tan }^{ - 1}}xdx} \)

Taking \(u = {\tan ^{ - 1}}x\) and \(v = x\) and integrating by parts, we get

\[\begin{align}I &= {\tan ^{ - 1}}x\int {xdx - \int {\left\{ {\left( {\frac{d}{{dx}}{{\tan }^{ - 1}}x} \right)\int {xdx} } \right\}dx} } \\ &= {\tan ^{ - 1}}x\left( {\frac{{{x^2}}}{2}} \right) - \int {\frac{1}{{1 + {x^2}}}.\frac{{{x^2}}}{2}dx} \\ &= \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}dx} \\ &= \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left\{ {\frac{{{x^2} + 1}}{{1 + {x^2}}} - \frac{1}{{1 + {x^2}}}} \right\}dx} \\ &= \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {1 - \frac{1}{{1 + {x^2}}}} \right)dx} \\ &= \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\left( {x - {{\tan }^{ - 1}}x} \right) + C\\ &= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{x}{2} + \frac{1}{2}{\tan ^{ - 1}}x + C\end{align}\]

Chapter 7 Ex.7.6 Question 9

Integrate \(x{\cos ^{ - 1}}x\)

Solution

Let \(I = \int {x{{\cos }^{ - 1}}xdx} \)

Taking \(u = {\cos ^{ - 1}}x\)and \(v = x\) and integrating by parts, we get

\[\begin{align}I &= {\cos ^{ - 1}}x\int {xdx - \int {\left\{ {\left( {\frac{d}{{dx}}{{\cos }^{ - 1}}x} \right)\int {xdx} } \right\}dx} } \\ &= {\cos ^{ - 1}}x\left( {\frac{{{x^2}}}{2}} \right) - \int {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}.\frac{{{x^2}}}{2}dx} \\ &= \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}dx} \\ &= \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left\{ {\sqrt {1 - {x^2}} + \left( {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}} \right)} \right\}dx} \\& = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\sqrt {1 - {x^2}} dx - \frac{1}{2}\int {\left( {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx} \\& = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}{I_1} - \frac{1}{2}{\cos ^{ - 1}}x.....................\left( 1 \right)\end{align}\]

\[\begin{align}{\text{Where, }}{I_1} &= \int {\sqrt {1 - {x^2}} dx} \\& \Rightarrow \;\;\;\;{I_1} = \sqrt {1 - {x^2}} \int {1dx - \int {\frac{d}{{dx}}\sqrt {1 - {x^2}} \int {1dx} } } \\ &\Rightarrow \;\;\;\;{I_1} = x\sqrt {1 - {x^2}} - \int {\frac{{ - 2x}}{{2\sqrt {1 - {x^2}} }}xdx} \\ &\Rightarrow \;\;\;\;{I_1} = x\sqrt {1 - {x^2}} - \int {\frac{{ - {x^2}}}{{\sqrt {1 - {x^2}} }}dx} \\ &\Rightarrow \;\;\;\;{I_1} = x\sqrt {1 - {x^2}} - \int {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}dx} \\& \Rightarrow \;\;\;\;{I_1} = x\sqrt {1 - {x^2}} - \left\{ {\int {\sqrt {1 - {x^2}} } dx + \int {\frac{{ - dx}}{{\sqrt {1 - {x^2}} }}} } \right\}\\& \Rightarrow \;\;\;\;{I_1} = x\sqrt {1 - {x^2}} - \left\{ {{I_1} + {{\cos }^{ - 1}}x} \right\}\\& \Rightarrow \;\;\;\;2{I_1} = x\sqrt {1 - {x^2}} - {\cos ^{ - 1}}x\\&\therefore {I_1} = \frac{x}{2}\sqrt {1 - {x^2}} - \frac{1}{2}{\cos ^{ - 1}}x\\&{\text{Substituting in }}\left( 1 \right),\\I &= \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\left( {\frac{x}{2}\sqrt {1 - {x^2}} - \frac{1}{2}{{\cos }^{ - 1}}x} \right) - \frac{1}{2}{\cos ^{ - 1}}x\\ &= \frac{{\left( {2{x^2} - 1} \right)}}{4}{\cos ^{ - 1}}x - \frac{x}{4}\sqrt {1 - {x^2}} + C\end{align}\]

Chapter 7 Ex.7.6 Question 10

Integrate \({\left( {{{\sin }^{ - 1}}x} \right)^2}\)

Solution

Let \(I = \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}.1\;dx} \)

Taking \(u = {\left( {{{\sin }^{ - 1}}x} \right)^2}\)and \(v = 1\) and integrating by parts, we get

\[\begin{align}I &= \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}.\int {1dx - } } \int {\left\{ {\frac{d}{{dx}}{{\left( {{{\sin }^{ - 1}}x} \right)}^2}.\int {1dx} } \right\}dx} \\& = {\left( {{{\sin }^{ - 1}}x} \right)^2}.x - \int {\frac{{2{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}.xdx} \\& = x{\left( {{{\sin }^{ - 1}}x} \right)^2} + \int {{{\sin }^{ - 1}}x\left( {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}} \right)dx} \\ &= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + \left[ {{{\sin }^{ - 1}}x\int {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}dx - \int {\left\{ {\left( {\frac{d}{{dx}}{{\sin }^{ - 1}}x} \right)\int {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}dx} } \right\}dx} } } \right]\\ &= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + \left[ {{{\sin }^{ - 1}}x.2\sqrt {1 - {x^2}} - \int {\frac{1}{{\sqrt {1 - {x^2}} }}.2\sqrt {1 - {x^2}} dx} } \right]\\ &= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - \int {2dx} \\ &= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + C\end{align}\]

Chapter 7 Ex.7.6 Question 11

Integrate \(\frac{{x{{\cos }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\)

Solution

Let \(I = \int {\frac{{x{{\cos }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx} \)

\(I = \frac{{ - 1}}{2}\int {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}.{{\cos }^{ - 1}}xdx} \)

Taking \(u = {\cos ^{ - 1}}x\)and \(v = \left( {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}} \right)\) and integrating by parts, we get

\[\begin{align}I &= \frac{{ - 1}}{2}\left[ {{{\cos }^{ - 1}}x\int {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}dx - \int {\left\{ {\left( {\frac{d}{{dx}}{{\cos }^{ - 1}}x} \right)\int {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}dx} } \right\}dx} } } \right]\\ &= \frac{{ - 1}}{2}\left[ {{{\cos }^{ - 1}}x.2\sqrt {1 - {x^2}} - \int {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}.2\sqrt {1 - {x^2}} dx} } \right]\\& = \frac{{ - 1}}{2}\left[ {2\sqrt {1 - {x^2}} {{\cos }^{ - 1}}x + \int {2dx} } \right]\\ &= \frac{{ - 1}}{2}\left[ {2\sqrt {1 - {x^2}} {{\cos }^{ - 1}}x + 2x} \right] + C\\ &= - \left[ {\sqrt {1 - {x^2}} {{\cos }^{ - 1}}x + x} \right] + C\end{align}\]

Chapter 7 Ex.7.6 Question 12

Integrate \(x{\sec ^2}x\)

Solution

Let \(I = \int {x{{\sec }^2}x} dx\)

Taking \(u = x\)and \(v = {\sec ^2}x\) and integrating by parts, we get

\[\begin{align}I &= x\int {{{\sec }^2}xdx - } \int {\left\{ {\left( {\frac{d}{{dx}}x} \right)\int {{{\sec }^2}xdx} } \right\}dx} \\ &= x\tan x - \int {1.\tan xdx} \\ & = x\tan x + \log \left| {\cos x} \right| + C\end{align}\]

Chapter 7 Ex.7.6 Question 13

Integrate \({\tan ^{ - 1}}x\)

Solution

Let \(I = \int {1.{{\tan }^{ - 1}}x} dx\)

Taking \(u = {\tan ^{ - 1}}x\)and \(v = 1\) and integrating by parts, we get

\[\begin{align}I &= {\tan ^{ - 1}}x\int {1dx - } \int {\left\{ {\left( {\frac{d}{{dx}}{{\tan }^{ - 1}}x} \right)\int {1.dx} } \right\}dx} \\& = {\tan ^{ - 1}}x.x - \int {\frac{1}{{1 + {x^2}}}} xdx\\ &= x{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{2}{{1 + {x^2}}}dx} \\ &= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left| {1 + {x^2}} \right| + C\\ &= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right) + C\end{align}\]

Chapter 7 Ex.7.6 Question 14

Integrate \(x{\left( {\log x} \right)^2}\)

Solution

Let \(I = \int {x{{\left( {\log x} \right)}^2}} dx\)

Taking \(u = {\left( {\log x} \right)^2}\)and \(v = x\) and integrating by parts, we get

\[\begin{align}I &= {\left( {\log x} \right)^2}\int {xdx - } \int {\left[ {\left\{ {\frac{d}{{dx}}{{\left( {\log x} \right)}^2}} \right\}\int {xdx} } \right]dx} \\& = \frac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \left[ {\int {2\log x.\frac{1}{x}.\frac{{{x^2}}}{2}dx} } \right]\\ &= \frac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \int {x\log xdx} \end{align}\]

Again, using integration by parts, we get

\[\begin{align}I &= \frac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \left[ {\log x\int {xdx - \int {\left\{ {\left( {\frac{d}{{dx}}\log x} \right)\int {xdx} } \right\}dx} } } \right]\\ &= \frac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \left[ {\frac{{{x^2}}}{2}\log x - \int {\frac{1}{x}.\frac{{{x^2}}}{2}dx} } \right]\\& = \frac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \frac{{{x^2}}}{2}\log x + \frac{1}{2}\int {xdx} \\ &= \frac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \frac{{{x^2}}}{2}\log x + \frac{{{x^2}}}{4} + C\end{align}\]

Chapter 7 Ex.7.6 Question 15

Integrate \(\left( {{x^2} + 1} \right)\log x\)

Solution

Let \(I = \int {\left( {{x^2} + 1} \right)\log x} dx = \int {{x^2}\log xdx} + \int {\log xdx} \)

Let \(I = {I_1} + {I_2}...........\left( 1 \right)\)

Where, \({I_1} = \int {{x^2}\log xdx\,\,{\text{and}}\,} {I_2} = \int {\log xdx} \)

\({I_1} = \int {{x^2}\log xdx} \)

Taking \(u = \log x\)and \(v = {x^2}\) and integrating by parts, we get

\[\begin{align}{I_1} &= \log x\int {{x^2}dx} - \int {\left[ {\left( {\frac{d}{{dx}}\log x} \right)\int {{x^2}dx} } \right]dx} \\ &= \log x.\frac{{{x^3}}}{3} - \int {\frac{1}{x}.\frac{{{x^3}}}{3}dx} \\ &= \frac{{{x^3}}}{3}\log x - \frac{1}{3}\left( {\int {{x^2}dx} } \right)\\ &= \frac{{{x^3}}}{3}\log x - \frac{{{x^3}}}{9} + {C_1}...........\left( 2 \right)\end{align}\]

\({I_2} = \int {\log xdx} \)

Taking \(u = \log x\) and \(v = 1\)and integrating by parts,

\[\begin{align}{I_2} &= \log x\int {1.dx} - \int {\left[ {\left( {\frac{d}{{dx}}\log x} \right)\int {1.dx} } \right]} \\ &= \log x.x - \int {\frac{1}{x}.xdx} \\& = x\log x - \int {1.dx} \\ &= x\log x - x + {C_2}...........\left( 3 \right)\end{align}\]

Using equations \(\left( 2 \right)\)and \(\left( 3 \right)\)in \(\left( 1 \right)\),

\[\begin{align}I&= \frac{{{x^3}}}{3}\log x - \frac{{{x^3}}}{9} + {C_1} + x\log x - x + {C_2}\\& = \frac{{{x^3}}}{3}\log x - \frac{{{x^3}}}{9} + x\log x - x + \left( {{C_1} + {C_2}} \right)\\& = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{{{x^3}}}{9} - x + C\end{align}\]

Chapter 7 Ex.7.6 Question 16

Integrate \({e^x}\left( {\sin x + \cos x} \right)\)

Solution

Let \(I = \int {{e^x}\left( {\sin x + \cos x} \right)dx} \)

Let \(f\left( x \right) = \sin x\)

\[\begin{align}f'\left( x \right) &= \cos x\\I& = \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx\\{\text{Since, }}\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx}&= {e^x}f\left( x \right) + C\\\therefore I &= {e^x}\sin x + C\end{align}\]

Chapter 7 Ex.7.6 Question 17

Integrate \(\frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}\)

Solution

\[\begin{align}{\text{Let}},\;I& = \int {\frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}} dx = \int {{e^x}\left\{ {\frac{x}{{{{\left( {1 + x} \right)}^2}}}} \right\}dx} \\& = \int {{e^x}\left\{ {\frac{{1 + x - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right\}dx} = \int {{e^x}\left\{ {\frac{1}{{1 + x}} - \frac{1}{{{{\left( {1 + x} \right)}^2}}}} \right\}dx} \end{align}\]

\({\text{Here, }}f\left( x \right) = \frac{1}{{1 + x}}{\text{ }}f'\left( x \right){\text{ = }}\frac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}\)

\[\begin{align} &\Rightarrow \;\;\;\;\int {\frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx = \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx} \\&{\text{Since, }}\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx = {e^x}f\left( x \right) + C\\&\therefore \int {\frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx = {e^x}\frac{1}{{1 + x}} + C} \end{align}\]

Chapter 7 Ex.7.6 Question 18

Integrate \({e^x}\left( {\frac{{1 + \sin x}}{{1 + \cos x}}} \right)\)

Solution

\[\begin{align}{e^x}\left( {\frac{{1 + \sin x}}{{1 + \cos x}}} \right) &= {e^x}\left( {\frac{{{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}} \right)\\& = \frac{{{e^x}{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}}}{{2{{\cos }^2}\frac{x}{2}}} = \frac{1}{2}{e^x}{\left( {\frac{{\sin \frac{x}{2} + \cos \frac{x}{2}}}{{\cos \frac{x}{2}}}} \right)^2}\\ &= \frac{1}{2}{e^x}{\left[ {\tan \frac{x}{2} + 1} \right]^2}\\& = \frac{1}{2}{e^x}{\left[ {1 + \tan \frac{x}{2}} \right]^2}\\ &= \frac{1}{2}{e^x}\left[ {1 + {{\tan }^2}\frac{x}{2} + 2\tan \frac{x}{2}} \right]\\ &= \frac{1}{2}{e^x}\left[ {{{\sec }^2}\frac{x}{2} + 2\tan \frac{x}{2}} \right]\\&\frac{{{e^x}\left( {1 + \sin x} \right)dx}}{{\left( {1 + \cos x} \right)}} = {e^x}\left[ {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right]............\left( 1 \right)\end{align}\]

Let \(\tan \frac{x}{2} = f\left( x \right){\text{ so }}f'\left( x \right) = \frac{1}{2}{\sec ^2}\frac{x}{2}\)

It is known that, \(\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} = {e^x}f\left( x \right) + C\)

From equation \(\left( 1 \right)\), we get

\(\int {\frac{{{e^x}\left( {1 + \sin x} \right)}}{{\left( {1 + \cos x} \right)}}dx} = {e^x}\tan \frac{x}{2} + C\)

Chapter 7 Ex.7.6 Question 19

Integrate \({e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\)

Solution

Let \(I = \int {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)dx} \)

Here, \(\frac{1}{x} = f\left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( x \right) = \frac{{ - 1}}{{{x^2}}}\)

It is known that,

\[\begin{align}&\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} \\ &= {e^x}f\left( x \right) + C\\\therefore I &= \frac{{{e^x}}}{x} + C\end{align}\]

Chapter 7 Ex.7.6 Question 20

Integrate \(\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}\)

Solution

\[\begin{align}&\int {{e^x}\left\{ {\frac{{x - 3}}{{{{\left( {x - 1} \right)}^3}}}} \right\}dx} = \int {{e^x}\left\{ {\frac{{x - 1 - 2}}{{{{\left( {x - 1} \right)}^3}}}} \right\}} dx\\& = \int {{e^x}\left\{ {\frac{1}{{{{\left( {x - 1} \right)}^2}}} - \frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right\}dx} \end{align}\]

Let \({\text{ }}f\left( x \right) = \frac{1}{{{{\left( {x - 1} \right)}^2}}}{\text{ }}f'\left( x \right) = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}\)

It is known that,

\[\begin{align}&\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} = {e^x}f\left( x \right) + C\\&\therefore \int {{e^x}\left\{ {\frac{{\left( {x - 3} \right)}}{{{{\left( {x - 1} \right)}^2}}}} \right\}dx} = \frac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + C\end{align}\]

Chapter 7 Ex.7.6 Question 21

Integrate \({e^{2x}}\sin x\)

Solution

Let \(I = {e^{2x}}\sin xdx................\left( 1 \right)\)

Taking \(u = \sin x\) and \(v = {e^{2x}}\)and integrating by parts, we get

\[\begin{align} &I= \sin x\int {{e^{2x}}dx - \int {\left\{ {\left( {\frac{d}{{dx}}\sin x} \right)\int {{e^{2x}}dx} } \right\}dx} } \\ &\Rightarrow \;\;\;\;I = \sin x.\frac{{{e^{2x}}}}{2} - \int {\cos x.\frac{{{e^{2x}}}}{2}dx} \\ &\Rightarrow \;\;\;\;I = \frac{{{e^{2x}}\sin x}}{2} - \frac{1}{2}\int {{e^{2x}}\cos xdx} \end{align}\]

Again, using integration by parts, we get

\[\begin{align}&I = \frac{{{e^{2x}}\sin x}}{2} - \frac{1}{2}\left[ {\cos x\int {{e^{2x}}dx - } \int {\left\{ {\left( {\frac{d}{{dx}}\cos x} \right)\int {{e^{2x}}dx} } \right\}dx} } \right]\\& \Rightarrow \;\;\;\;I = \frac{{{e^{2x}}\sin x}}{2} - \frac{1}{2}\left[ {\cos x.\frac{{{e^{2x}}}}{2} - \int {\left( { - \sin x} \right)\frac{{{e^{2x}}}}{2}dx} } \right]\\ &\Rightarrow \;\;\;\;I = \frac{{{e^{2x}}\sin x}}{2} - \frac{1}{2}\left[ {\frac{{{e^{2x}}\cos x}}{2} + \frac{1}{2}\int {{e^{2x}}\sin xdx} } \right]\\ &\Rightarrow \;\;\;\;I = \frac{{{e^{2x}}\sin x}}{2} - \frac{{{e^{2x}}\cos x}}{4} - \frac{1}{4}I{\text{ }}\left[ {{\text{From}}\left( 1 \right)} \right]\\& \Rightarrow \;\;\;\;I + \frac{1}{4}I = \frac{{{e^{2x}}\sin x}}{2} - \frac{{{e^{2x}}\cos x}}{4}\\ &\Rightarrow \;\;\;\;\frac{5}{4}I = \frac{{{e^{2x}}\sin x}}{2} - \frac{{{e^{2x}}\cos x}}{4}\\& \Rightarrow \;\;\;\;I = \frac{4}{5}\left[ {\frac{{{e^{2x}}\sin x}}{2} - \frac{{{e^{2x}}\cos x}}{4}} \right] + C\\ &\Rightarrow \;\;\;\;I = \frac{{{e^{2x}}}}{5}\left[ {2\sin x - \cos x} \right] + C\end{align}\]

Chapter 7 Ex.7.6 Question 22

Integrate \({\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\)

Solution

Let \(x = \tan \theta {\text{ }}dx = {\sec ^2}\theta d\theta \)

\[\begin{align}&\therefore {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) = {\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) = {\sin ^{ - 1}}\left( {\sin 2\theta } \right) = 2\theta \\&\int {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)} dx = \int {2\theta .{{\sec }^2}\theta d\theta = 2\int {\theta .{{\sec }^2}\theta d\theta } } \end{align}\]

Using integration by parts, we get

\[\begin{align}&2\left[ {\theta .\int {{{\sec }^2}\theta d\theta } - \int {\left\{ {\left( {\frac{d}{{d\theta }}\theta } \right)\int {{{\sec }^2}\theta d\theta } } \right\}d\theta } } \right]\\ &= 2\left[ {\theta .\tan \theta - \int {\tan \theta d\theta } } \right]\\& = 2\left[ {\theta .\tan \theta + \log \left| {\cos \theta } \right|} \right] + C\\& = 2\left[ {x{{\tan }^{ - 1}}x + \log \left| {\frac{1}{{\sqrt {1 + {x^2}} }}} \right|} \right] + C\\& = 2x{\tan ^{ - 1}}x + 2\log {\left( {1 + {x^2}} \right)^{\frac{{ - 1}}{2}}} + C\\ &= 2x{\tan ^{ - 1}}x + 2\left[ {\frac{{ - 1}}{2}\log \left( {1 + {x^2}} \right)} \right] + C\\ &= 2x{\tan ^{ - 1}}x - \log \left( {1 + {x^2}} \right) + C\end{align}\]

Chapter 7 Ex.7.6 Question 23

\(\int {{x^2}{e^{{x^3}}}dx} \)equals

A.\(\frac{1}{3}{e^{{x^3}}} + C\)

B. \(\frac{1}{3}{e^{{x^2}}} + C\)

C. \(\frac{1}{2}{e^{{x^3}}} + C\)

D. \(\frac{1}{2}{e^{{x^2}}} + C\)

Solution

Let \(I = \int {{x^2}{e^{{x^3}}}dx} \)

Also, let \({x^3} = t{\text{ }}\;{\text{so, }}3{x^2}dx = dt\)

\[\begin{align} &\Rightarrow \;\;\;\;I = \frac{1}{3}\int {{e^t}dt} \\ &= \frac{1}{3}\left( {{e^t}} \right) + C\\ &= \frac{1}{3}{e^{{x^3}}} + C\end{align}\]

Thus, the correct option is A.

Chapter 7 Ex.7.6 Question 24

\(\int {{e^x}\sec x\left( {1 + \tan x} \right)dx} \) equals

A. \({e^x}\cos x + C\)

B. \({e^x}\sec x + C\)

C. \({e^x}\sin x + C\)

D. \({e^x}\tan x + C\)

Solution

\(\int {{e^x}\sec x\left( {1 + \tan x} \right)dx} \)

Consider, \(I = \int {{e^x}\sec x\left( {1 + \tan x} \right)dx} = \int {{e^x}\left( {\sec x + \sec x\tan x} \right)dx} \)

Let \(\sec x = f\left( x \right){\text{ }}\sec x\tan x = f'\left( x \right)\)

It is known that, \(\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx = {e^x}f\left( x \right) + C\)

\(\therefore I = {e^x}\sec x + C\)

Thus, the correct option is B.

  
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