NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.2

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Chapter 7 Ex.7.2 Question 1

Integrate:

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

Solution

Put $${\rm{1 + }}{x^2} = t$$

Therefore, $$2xdx = dt$$

\begin{align}\int {\frac{{2x}}{{1 + {x^2}}}dx = \int {\frac{1}{t}dt} } &= \log \left| t \right| + C\\ &= \log \left| {1 + {x^2}} \right| + C\\ &= \log \left( {1 + {x^2}} \right) + C\end{align}

Chapter 7 Ex.7.2 Question 2

Integrate:

$$\frac{{{{\left( {\log x} \right)}^2}}}{x}$$

Solution

Put $$\log \left| x \right| = t$$

Therefore, $$\frac{1}{x}dx = dt$$

\begin{align}\int {\frac{{{{\left( {\log \left| x \right|} \right)}^2}}}{x}} dx &= \int {{t^2}} dt\\& = \frac{{{t^3}}}{3} + C\\& = \frac{{{{\left( {\log \left| x \right|} \right)}^3}}}{3} + C\end{align}

Chapter 7 Ex.7.2 Question 3

Integrate:

$$\frac{1}{{x + x\log x}}$$

Solution

$$\frac{1}{{x + x\log x}} = \frac{1}{{x\left( {1 + \log x} \right)}}$$

Put $$1 + \log x = t$$

Therefore, $$\frac{1}{x}dx = dt$$

\begin{align}\int {\frac{1}{{x\left( {1 + \log x} \right)}}dx = } \int {\frac{1}{t}dt} &= \log \left| t \right| + C\\ &= \log \left| {1 + \log x} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 4

Integrate:

$$\sin x\sin \left( {\cos x} \right)$$

Solution

Put $$\cos x = t$$

Therefore, $$- \sin xdx = dt$$

\begin{align}\int {\sin x\sin \left( {\cos x} \right)dx = - \int {\sin tdt} }& = - \left[ { - \cos t} \right] + C\\ &= \cos t + C\\ &= \cos \left( {\cos x} \right) + C\end{align}

Chapter 7 Ex.7.2 Question 5

Integrate:

$${\mathop{\rm sin}\nolimits} \left( {ax + b} \right)\cos \left( {ax + b} \right)$$

Solution

\begin{align} \sin (a x+b) \cos (a x+b) &=\frac{2 \sin (a x+b) \cos (a x+b)}{2} \\ &=\frac{\sin 2(a x+b)}{2} \end{align}

Put $$2\left( {ax + b} \right) = t$$

Therefore, $$2adx = dt$$

\begin{align}\int {\frac{{\sin 2\left( {ax + b} \right)}}{2}} dx &= \frac{1}{2}\int {\frac{{\sin tdt}}{{2a}}} \\& = \frac{1}{{4a}}\left[ { - \cos t} \right] + C\\& = \frac{{ - 1}}{{4a}}\cos 2\left( {ax + b} \right) + C\end{align}

Chapter 7 Ex.7.2 Question 6

Integrate:

$$\sqrt {ax + b}$$

Solution

Put $$ax + b = t$$

Therefore,

\begin{align}& \Rightarrow \;adx = dt\\ &\Rightarrow \;dx = \frac{1}{a}dt\end{align}

\begin{align}\int {{{\left( {ax + b} \right)}^{\frac{1}{2}}}dx = \frac{1}{a}\int {{t^{\frac{1}{2}}}dt} } &= \frac{1}{a}\left( {\frac{{{t^{\frac{1}{2}}}}}{{\frac{3}{2}}}} \right) + C\\& = \frac{2}{{3a}}{\left( {ax + b} \right)^{\frac{3}{2}}} + C\end{align}

Chapter 7 Ex.7.2 Question 7

Integrate:

$$x\sqrt {x+2}$$

Solution

\begin{align}{\rm{Put,\; }}x + 2 &= t\\\therefore dx &= dt\\ &\Rightarrow\; \int {x\sqrt {x + 2} = \int {\left( {t - 2} \right)} \sqrt t dt} \\& = \int {\left( {{t^{\frac{3}{2}}} - 2{t^{\frac{1}{2}}}} \right)} dt\\ &= \int {{t^{\frac{3}{2}}}dt - 2\int {{t^{\frac{1}{2}}}dt} } \\& = \frac{{{t^{\frac{5}{2}}}}}{{\frac{5}{2}}} - 2\left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right) + C\\& = \frac{2}{5}{t^{\frac{5}{2}}} - \frac{4}{3}{t^{\frac{3}{2}}} + C\\& = \frac{2}{5}{\left( {x + 2} \right)^{\frac{5}{2}}} - \frac{4}{3}{\left( {x + 2} \right)^{\frac{3}{2}}} + C\end{align}

Chapter 7 Ex.7.2 Question 8

Integrate:

$$x\sqrt {1 + 2{x^2}}$$

Solution

\begin{align}{\rm{Put, }}1 + 2{x^2} &= t\\\therefore 4xdx &= dt\\& \Rightarrow\; \int {x\sqrt {1 + 2{x^2}} dx = \int {\frac{{\sqrt t }}{4}} } dt\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{1}{4}\int {{t^{\frac{1}{2}}}} dt\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{1}{4}\left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right) + C\\& = \frac{1}{6}{\left( {1 + 2{x^2}} \right)^{\frac{3}{2}}} + C\end{align}

Chapter 7 Ex.7.2 Question 9

Integrate:

$$\left( {4x + 2} \right)\sqrt {{x^2} + x + 1}$$

Solution

\begin{align}{\rm{Put\; }}{x^2} + x + 1 &= t\\ \therefore \left( {2x + 1} \right)dx &= dt\\&\int {\left( {4x + 2} \right)\sqrt {{x^2} + x + 1} dx} \\ &= \int {2\sqrt t dt} \\& = 2\int {\sqrt t dt} \\ &= 2\left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right) + C = \frac{4}{3}{\left( {{x^2} + x + 1} \right)^{\frac{3}{2}}} + C\end{align}

Chapter 7 Ex.7.2 Question 10

Integrate:

$$\frac{1}{{x - \sqrt x }}$$

Solution

\begin{align}\frac{1}{{x - \sqrt x }} &= \frac{1}{{\sqrt {x\left( {\sqrt x - 1} \right)} }}\\{\rm{Put, }}\left( {\sqrt x - 1} \right) &= t\\\therefore \frac{1}{{2\sqrt x }}dx &= dt\\& \Rightarrow\; \int {\frac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}dx = \int {\frac{2}{t}} } dt\\& = 2\log \left| t \right| + C\\& = 2\log \left| {\sqrt x - 1} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 11

Integrate:

$$\frac{x}{{\sqrt {x + 4} }},x > 0$$

Solution

\begin{align}{\rm{Put, }}x + 4 &= t\\\therefore dx &= dt\\\int {\frac{x}{{\sqrt {x + 4} }}dx}&= {\int {\frac{{\left( {t - 4} \right)}}{{\sqrt t }}dt = \int {\left( {\sqrt t - \frac{4}{{\sqrt t }}} \right)} dt} } \\& = \left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right) - 4\left( {\frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}}} \right) + C = \frac{2}{3}{\left( t \right)^{\frac{3}{2}}} - 8{\left( t \right)^{\frac{1}{2}}} + C\\& = \frac{2}{3}t.{t^{\frac{1}{2}}} - 8{t^{\frac{1}{2}}} + C\\ &= \frac{2}{3}{t^{\frac{1}{2}}}\left( {t - 12} \right) + C\\ &= \frac{2}{3}{\left( {x + 4} \right)^{\frac{1}{2}}}\left( {x + 4 - 12} \right) + C\\& = \frac{2}{3}\sqrt {x + 4} \left( {x - 8} \right) + C\end{align}

Chapter 7 Ex.7.2 Question 12

Integrate:

$${\left( {{x^3} - 1} \right)^{\frac{1}{3}}}{x^5}$$

Solution

\begin{align}{\rm{Put, }}{x^3} - 1 &= t\\\therefore 3{x^2}dx &= dt\\ \Rightarrow\; \int {{{\left( {{x^3} - 1} \right)}^{\frac{1}{3}}}{x^5}dx }&={ \int {{{\left( {{x^3} - 1} \right)}^{\frac{1}{3}}}{x^3}{x^2}dx} } \\ &\Rightarrow\; \int {{t^{\frac{1}{3}}}\left( {t + 1} \right)\frac{{dt}}{3} = \frac{1}{3}\int {\left( {{t^{\frac{4}{3}}} + {t^{\frac{1}{3}}}} \right)dt} } \\& = \frac{1}{3}\left[ {\frac{{{t^{\frac{7}{3}}}}}{{\frac{7}{3}}} + \frac{{{t^{\frac{4}{3}}}}}{{\frac{4}{3}}}} \right] + C\\ &= \frac{1}{3}\left[ {\frac{3}{7}{t^{\frac{7}{3}}} + \frac{3}{4}{t^{\frac{4}{3}}}} \right] + C\\& = \frac{1}{7}{\left( {{x^3} - 1} \right)^{\frac{7}{3}}} + \frac{1}{4}{\left( {{x^3} - 1} \right)^{\frac{4}{3}}} + C\end{align}

Chapter 7 Ex.7.2 Question 13

Integrate:

$$\frac{{{x^2}}}{{{{\left( {2 + 3{x^3}} \right)}^3}}}$$

Solution

\begin{align}{\rm{Put, }}2 + 3{x^3} &= t\\\therefore 9{x^2}dx &= dt\\ \Rightarrow\; \int {\frac{{{x^2}}}{{{{\left( {2 + 3{x^3}} \right)}^3}}}dx }&={ \frac{1}{9}\int {\frac{{dt}}{{{{\left( t \right)}^3}}}} } \\ &= \frac{1}{9}\left[ {\frac{{{t^{ - 2}}}}{{ - 2}}} \right] + C\\& = - \frac{1}{{18}}\left( {\frac{1}{{{t^2}}}} \right) + C\\& = \frac{{ - 1}}{{18{{\left( {2 + 3{x^3}} \right)}^2}}} + C\end{align}

Chapter 7 Ex.7.2 Question 14

Integrate:

$$\frac{1}{{x{{\left( {\log x} \right)}^m}}},x > 0$$

Solution

\begin{align}{\rm{Put, }}\log x &= t\\\therefore \frac{1}{x}dx &= dt\\ \Rightarrow \int {\frac{1}{{x{{\left( {\log x} \right)}^m}}}dx }&={ \int {\frac{{dt}}{{{{\left( t \right)}^m}}} = \left( {\frac{{{t^{ - m - 1}}}}{{1 - m}}} \right) + C} } \\& = \frac{{{{\left( {\log x} \right)}^{1 - m}}}}{{\left( {1 - m} \right)}} + C\end{align}

Chapter 7 Ex.7.2 Question 15

Integrate:

$$\frac{x}{{9 - 4{x^2}}}$$

Solution

Put,$$9 - 4{x^2} = t$$

\begin{align}&\therefore - 8xdx = dt\\ &\Rightarrow\; \int {\frac{x}{{9 - 4{x^2}}}dx = \frac{{ - 1}}{8}\int {\frac{1}{t}dt} } \\ &= \frac{{ - 1}}{8}\log \left| t \right| + C\\& = \frac{{ - 1}}{8}\log \left| {9 - 4{x^2}} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 16

Integrate:

$${e^{2x + 3}}$$

Solution

Put, $$2x + 3 = t$$

\begin{align}&\therefore 2dx = dt\\ &\Rightarrow\; \int {{e^{2x + 3}}dx = \frac{1}{2}\int {{e^t}dt} } \\ &= \frac{1}{2}\left( {{e^t}} \right) + C\\ &= \frac{1}{2}{e^{\left( {2x + 3} \right)}} + C\end{align}

Chapter 7 Ex.7.2 Question 17

Integrate:

$$\frac{x}{{{e^{{x^2}}}}}$$

Solution

Put, $${x^2} = t$$

\begin{align}&\therefore 2xdx = dt\\& \Rightarrow\; \int {\frac{x}{{{e^{{x^2}}}}}dx = \frac{1}{2}\int {\frac{1}{{{e^t}}}dt} } = \frac{1}{2}\int {{e^{ - t}}dt} \\& = \frac{1}{2}\left( {\frac{{{e^{ - t}}}}{{ - 1}}} \right) + C\\& = - \frac{1}{2}{e^{ - {x^2}}} + C\\&= \frac{{ - 1}}{{2{e^{{x^2}}}}} + C\end{align}

Chapter 7 Ex.7.2 Question 18

Integrate:

$$\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}$$

Solution

Put, $${\tan ^{ - 1}}x = t$$

\begin{align}&\therefore \frac{1}{{1 + {x^2}}}dx = dt\\ &\Rightarrow\; \int {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx = \int {{e^t}dt} } \\& = {e^t} + C\\ &= {e^{{{\tan }^{ - 1}}x}} + C\end{align}

Chapter 7 Ex.7.2 Question 19

Integrate:

$$\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$$

Solution

$$\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$$

Dividing Nr and Dr by $${e^x}$$, we get

$$\frac{{\frac{{{e^{2x}} - 1}}{{{e^x}}}}}{{\frac{{{e^{2x}} + 1}}{{{e^x}}}}} = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$$

Let $${e^x} + {e^{ - x}} = t$$

\begin{align}&\left( {{e^x} - {e^{ - x}}} \right)dx = dt\\& \Rightarrow\; \int {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}dx = \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx} } \\ &= \int {\frac{{dt}}{t}} \\ &= \log \left| t \right| + C\\&= \log \left| {{e^x} + {e^{ - x}}} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 20

Integrate:

$$\frac{{{e^{2x}} - {e^{ - 2x}}}}{{{e^{2x}} + {e^{ - 2x}}}}$$

Solution

Put, $${e^{2x}} + {e^{ - 2x}} = t$$

\begin{align}&\left( {2{e^{2x}} - 2{e^{ - 2x}}} \right)dx = dt\\&\Rightarrow\; 2\left( {{e^{2x}} - {e^{ - 2x}}} \right)dx = dt\\ &\Rightarrow\; \int {\left( {\frac{{{e^{2x}} - {e^{ - 2x}}}}{{{e^{2x}} + {e^{ - 2x}}}}} \right)dx = \int {\frac{{dt}}{{2t}}} } \\ &= \frac{1}{2}\int {\frac{1}{t}dt} \\& = \frac{1}{2}\log \left| t \right| + C\\& = \frac{1}{2}\log \left| {{e^{2x}} + {e^{ - 2x}}} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 21

Integrate:

$${\tan ^2}\left( {2x - 3} \right)$$

Solution

$${\tan ^2}\left( {2x - 3} \right) = {\sec ^2}\left( {2x - 3} \right) - 1$$

Put, $$2x - 3 = t$$

\begin{align}&\therefore 2dx = dt\\& \Rightarrow\; \int {{{\tan }^2}\left( {2x - 3} \right)dx = \int {\left[ {{{\sec }^2}\left( {2x - 3} \right) - 1} \right]} dx} \\ &= \frac{1}{2}\int {\left( {{{\sec }^2}t} \right)dt - \int {1dx = \frac{1}{2}\int {{{\sec }^2}tdt - \int {1dx} } } } \\& = \frac{1}{2}\tan t - x + C\\ &= \frac{1}{2}\tan \left( {2x - 3} \right) - x + C\end{align}

Chapter 7 Ex.7.2 Question 22

Integrate:

$${\sec ^2}\left( {7 - 4x} \right)$$

Solution

Put, $$7 - 4x = t$$

\begin{align}&\therefore - 4dx = dt\\&\therefore \int {{{\sec }^2}\left( {7 - 4x} \right)dx = \frac{{ - 1}}{4}\int {{{\sec }^2}tdt} } \\ &= \frac{{ - 1}}{4}\left( {\tan t} \right) + C\\& = \frac{{ - 1}}{4}\tan \left( {7 - 4x} \right) + C\end{align}

Chapter 7 Ex.7.2 Question 23

Integrate:

$$\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$$

Solution

Put, $${\sin ^{ - 1}}x = t$$

\begin{align}&\frac{1}{{\sqrt {1 - {x^2}} }}dx = dt\\& \Rightarrow\; \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx = \int {tdt} } \\& = \frac{{{t^2}}}{2} + C = \frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}}}{2} + C\end{align}

Chapter 7 Ex.7.2 Question 24

Integrate:

$$\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}$$

Solution

\begin{align}&\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}} = \frac{{2\cos x - 3\sin x}}{{2\left( {3\cos x + 2\sin x} \right)}}\\{\rm{Let}}\,\,3\cos x + 2\sin x &= t\\&\left( { - 3\sin x + 2\cos x} \right)dx = dt\\&\int {\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}dx} = \int {\frac{{dt}}{{2t}}} \\& = \frac{1}{2}\int {\frac{1}{t}dt} \\ &= \frac{1}{2}\log \left| t \right| + C\\& = \frac{1}{2}\log \left| {2\sin x + 3\cos x} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 25

Integrate:

$$\frac{1}{{{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}}$$

Solution

\begin{align}&\frac{1}{{{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}} = \frac{{{{\sec }^2}x}}{{{{\left( {1 - \tan x} \right)}^2}}}\\&{\rm{Let }}\left( {1 - \tan x} \right) = t\\ &- {\sec ^2}xdx = dt\\& \Rightarrow\; \int {\frac{{{{\sec }^2}x}}{{{{\left( {1 - \tan x} \right)}^2}}}dx} = \int {\frac{{ - dt}}{{{t^2}}}} \\&= - \int {{t^{ - 2}}dt} \\& = \frac{1}{t} + C\\& = \frac{1}{{\left( {1 - \tan x} \right)}} + C\end{align}

Chapter 7 Ex.7.2 Question 26

Integrate:

$$\frac{{\cos \sqrt x }}{{\sqrt x }}$$

Solution

Let $$\sqrt x = t$$

\begin{align}&\frac{1}{{2\sqrt x }}dx = dt\\& \Rightarrow\; \int {\frac{{\cos \sqrt x }}{{\sqrt x }}dx} = 2\int {\cos tdt} \\& = 2\sin t + C\\ &= 2\sin \sqrt x + C\end{align}

Chapter 7 Ex.7.2 Question 27

Integrate:

$$\sqrt {\sin 2x} \cos 2x$$

Solution

Put, $$\sin 2x = t$$

So, $$2\cos 2xdx = dt$$

\begin{align}& \Rightarrow\; \int {\sqrt {\sin 2x} \cos 2xdx = \frac{1}{2}\int {\sqrt t dt} } \\ &= \frac{1}{2}\left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right) + C\\ &= \frac{1}{3}{t^{\frac{3}{2}}} + C\\ &= \frac{1}{3}{\left( {\sin 2x} \right)^{\frac{3}{2}}} + C\end{align}

Chapter 7 Ex.7.2 Question 28

Integrate:

$$\frac{{\cos x}}{{\sqrt {1 + \sin x} }}$$

Solution

Put, $$1 + \sin x = t$$

\begin{align}&\therefore \cos xdx = dt\\& \Rightarrow\; \int {\frac{{\cos x}}{{\sqrt {1 + \sin x} }}dx = \int {\frac{{dt}}{{\sqrt t }}} } \\ &= \frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}} + C\\& = 2\sqrt t + C\\ &= 2\sqrt {1 + \sin x} + C\end{align}

Chapter 7 Ex.7.2 Question 29

Integrate:

$$\cot x\log \sin x$$

Solution

Let $$\log \sin x = t$$

\begin{align} &\Rightarrow\; \frac{1}{{\sin x}}\cos xdx = dt\\&\therefore \cot xdx = dt\\& \Rightarrow\; \int {\cot x\log \sin xdx = \int {tdt} } \\& = \frac{{{t^2}}}{2} + C\\& = \frac{1}{2}{\left( {\log \sin x} \right)^2} + C\end{align}

Chapter 7 Ex.7.2 Question 30

Integrate:

$$\frac{{\sin x}}{{1 + \cos x}}$$

Solution

Put, $$1 + \cos x = t$$

\begin{align}&\therefore - \sin xdx = dt\\ &\Rightarrow\; \int {\frac{{\sin x}}{{1 + \cos x}}dx = \int { - \frac{{dt}}{t}} } \\& = - \log \left| t \right| + C\\ &= - \log \left| {1 + \cos x} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 31

Integrate:

$$\frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}$$

Solution

Put, $$1 + \cos x = t$$

\begin{align}&\therefore - \sin xdx = dt\\& \Rightarrow\; \int {\frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}dx = \int { - \frac{{dt}}{{{t^2}}}} } \\ &= - \int {{t^{ - 2}}dt} \\& = \frac{1}{t} + C\\ &= \frac{1}{{\left( {1 + \cos x} \right)}} + C\end{align}

Chapter 7 Ex.7.2 Question 32

Integrate:

$$\frac{1}{{1 + \cos x}}$$

Solution

Let I$$= \int {\frac{1}{{1 + \cos x}}dx}$$

\begin{align} &= \int {\frac{1}{{1 + \frac{{\cos x}}{{\sin x}}}}dx} \\ &= \int {\frac{{\sin x}}{{\sin x + \cos x}}dx} \\& = \frac{1}{2}\int {\frac{{2\sin x}}{{\sin x + \cos x}}dx} \\& = \frac{1}{2}\int {\frac{{\left( {\sin x + \cos x} \right) + \left( {\sin x - \cos x} \right)}}{{\left( {\sin x + \cos x} \right)}}} dx\\ &= \frac{1}{2}\int {1dx + \frac{1}{2}\int {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} dx} \\& = \frac{1}{2}\left( x \right) + \frac{1}{2}\int {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}dx} \\{\rm{Let }}\sin x + \cos x = t \Rightarrow\; \left( {\cos x - \sin x} \right)dx = dt\\&\therefore {\rm{I}} = \frac{x}{2} + \frac{1}{2}\int {\frac{{ - \left( {dt} \right)}}{t}} \\& = \frac{x}{2} - \frac{1}{2}\log \left| t \right| + C = \frac{x}{2} - \frac{1}{2}\log \left| {\sin x + \cos x} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 33

Integrate:

$$\frac{1}{{1 - \tan x}}$$

Solution

Put, I$$= \int {\frac{1}{{1 - \tan x}}dx}$$

\begin{align} &= \int {\frac{1}{{1 - \frac{{\sin x}}{{\cos x}}}}dx = \int {\frac{{\cos x}}{{\cos x - \sin x}}dx} } \\& = \frac{1}{2}\int {\frac{{2\cos x}}{{\cos x - \sin x}}dx = \frac{1}{2}\int {\frac{{\left( {\cos x - \sin x} \right) + \left( {\cos x + \sin x} \right)}}{{\left( {\cos x - \sin x} \right)}}} } dx\\& = \frac{1}{2}\int {1dx + \frac{1}{2}\int {\frac{{\cos x + \sin x}}{{\cos x - \sin x}}dx = \frac{x}{2} + \frac{1}{2}\int {\frac{{\cos x + \sin x}}{{\cos x - \sin x}}dx} } } \\&{\rm{Put, }}\cos x - \sin x = t \Rightarrow\; \left( { - \sin x - \cos x} \right)dx = dt\\&\therefore {\rm{I}} = \frac{x}{2} + \frac{1}{2}\int {\frac{{ - \left( {dt} \right)}}{t} = \frac{x}{2} - \frac{1}{2}\log \left| t \right| + C} \\ &= \frac{x}{2} - \frac{1}{2}\log \left| {\cos x - \sin x} \right| + C\end{align}

Chapter 7 Ex.7.2 Question 34

Integrate:

$$\frac{{\sqrt {\tan x} }}{{\sin x\cos x}}$$

Solution

\begin{align}{\rm{Let I}}& = \int {\frac{{\sqrt {\tan x} }}{{\sin x\cos x}}dx = \int {\frac{{\sqrt {\tan x} \times \cos x}}{{\sin x\cos x \times \cos x}}dx} } \\& = \int {\frac{{\sqrt {\tan x} }}{{\tan x{{\cos }^2}x}}dx = \int {\frac{{{{\sec }^2}xdx}}{{\sqrt {\tan x} }}} } \\&{\rm{Let}}\,\tan x = t \Rightarrow\; {\sec ^2}xdx = dt\\&\therefore {\rm{I}} = \int {\frac{{dt}}{{\sqrt t }}} \\& = 2\sqrt t + C\\ &= 2\sqrt {\tan x} + C\end{align}

Chapter 7 Ex.7.2 Question 35

Integrate:

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

Solution

Put, $$1 + \log x = t$$

\begin{align}&\therefore \frac{1}{x}dx = dt\\ &\Rightarrow\; \int {\frac{{{{\left( {1 + \log x} \right)}^2}}}{x}dx} = \int {{t^2}dt} \\& = \frac{{{t^3}}}{3} + C\\& = \frac{{{{\left( {1 + \log x} \right)}^3}}}{3} + C\end{align}

Chapter 7 Ex.7.2 Question 36

Integrate:

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

Solution

\begin{align}&\frac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x} = \left( {\frac{{x + 1}}{x}} \right){\left( {x + \log } \right)^2} = \left( {1 + \frac{1}{x}} \right){\left( {x + \log x} \right)^2}\\&{\rm{Put, }}\left( {x + \log x} \right) = t\\&\therefore \left( {1 + \frac{1}{x}} \right)dx = dt\\& \Rightarrow\; \int {\left( {1 + \frac{1}{x}} \right)} {\left( {x + \log x} \right)^2}dx = \int {{t^2}dt} \\& = \frac{{{t^3}}}{3} + C\\&= \frac{1}{3}{\left( {x + \log x} \right)^3} + C\end{align}

Chapter 7 Ex.7.2 Question 37

Integrate:

$$\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}$$

Solution

Put, $${x^4} = t$$

\begin{align}&\therefore 4{x^3}dx = dt\\& \Rightarrow\; \int {\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}} dx = \frac{1}{4}\int {\frac{{\sin \left( {{{\tan }^{ - 1}}t} \right)}}{{1 + {t^2}}}} dt\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\end{align}

Let $${\tan ^{ - 1}}t = u$$

$$\therefore \frac{1}{{1 + {t^2}}}dt = du$$

From (1), we get

\begin{align}\int {\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)dx}}{{1 + {x^8}}}} &= \frac{1}{4}\int {\sin udu} \\ &= \frac{1}{4}\left( { - \cos u} \right) + C\\& = - \frac{1}{4}\cos \left( {{{\tan }^{ - 1}}t} \right) + C\\ &= \frac{{ - 1}}{4}\cos \left( {{{\tan }^{ - 1}}{x^4}} \right) + C\end{align}

Chapter 7 Ex.7.2 Question 38

$$\int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{x^{10}} + {{10}^x}}}dx}$$ equals

(A) $${10^x} - {x^{10}} + C$$

(B) $${10^x} + {x^{10}} + C$$

(C)$${\left( {{{10}^x} - {x^{10}}} \right)^{ - 1}} + C$$

(D) $$\log \left( {{{10}^x} + {x^{10}}} \right) + C$$

Solution

Put, $${x^{10}} + {10^x} = t$$

\begin{align}&\therefore \left( {10{x^9} + {{10}^x}{{\log }_e}10} \right)dx = \int {\frac{{dt}}{t}} \\& \Rightarrow\; \int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{x^{10}} + 10x}}} dx = \int {\frac{{dt}}{t}} \\& = \log t + C\\ &=\log \left( {{{10}^x} + {x^{10}}} \right) + C\end{align}

Thus, the correct option is D.

Chapter 7 Ex.7.2 Question 39

$$\int {\frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}}}$$ equals

(A)$$\tan x + \cot x + C$$

(B) $$\tan x - \cot x + C$$

(C) $$\tan x\cot x + C$$

(D) $$\tan x - \cot 2x + C$$

Solution

Put, $${\rm{I}} = \int {\frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}} = \int {\frac{1}{{{{\sin }^2}x{{\cos }^2}x}}dx} }$$

\begin{align}&= \int {\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx\\ &= \int {\frac{{{{\sin }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx} + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx\\&= \int {{{\sec }^2}xdx + \int {\cos e{c^2}} dx} \\&= \tan x - \cot x + C\end{align}

Thus, the correct option is B.

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