NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 - Integrals

Exercise 7.2

1. Integrate  $ \dfrac{2x}{1+{{x}^{2}}} $ 

Ans:

Let  $ 1+{{x}^{2}}=\text{t} $ 

$ \therefore 2\text{xdx}=\text{dt} $ 

$ \Rightarrow \int{\dfrac{2x}{1+{{x}^{2}}}}dx=\int{\dfrac{1}{t}}dt $ 

$ =\log |t|+C $ 

$ =\log \left| 1+{{x}^{2}} \right|+C $ 

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

where C is an arbitrary constant. 

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

Ans:

Let  $ \log x=t $ 

$ \therefore \dfrac{1}{x}dx=dt $ 

$ \Rightarrow \int{\dfrac{{{(\log |x|)}^{2}}}{x}}dx=\int{{{t}^{2}}}dt $ 

$ =\dfrac{{{t}^{3}}}{3}+C $ 

$ =\dfrac{{{(\log |x|)}^{3}}}{3}+C $ 

where C is an arbitrary constant. 

3. Integrate  $ \dfrac{1}{x+x\log x} $ 

Ans:

The given function can be written as 

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

Let  $ 1+\log x=t $ 

$ \therefore \dfrac{1}{x}dx=dt $ 

$ \Rightarrow \int{\dfrac{1}{x(1+\log x)}}dx=\int_{t}^{1}{d}t $ 

$ =\log |t|+C $ 

$ =\log |1+\log x|+C $ 

where C is an arbitrary constant.

4. Integrate  $ \sin x.\sin \left( \cos x \right) $ 

Ans:

Let  $ \cos x=t $ 

$ \therefore -\sin xdx=t $ 

$ \Rightarrow \int{\sin }x\cdot \sin (\cos x)dx=-\int{\sin }tdt $ 

$ =-[-\cos t]+C $ 

$ =\cos t+C $ 

$ =\cos (\cos x)+C $ 

where C is an arbitrary constant.

5. Integrate  $ \sin \left( ax+b \right)\cos \left( ax+b \right) $ 

Ans:

The given function can be rewritten as

 $ \sin (ax+b)\cos (ax+b)=\dfrac{2\sin (ax+b)\cos (ax+b)}{2}=\dfrac{\sin 2(ax+b)}{2} $ 

let  $ 2(ax+b)=t $ 

 $ \therefore 2adx=dt $ 

 $ \Rightarrow \int{\dfrac{\sin 2(ax+b)}{2}}dx=\dfrac{1}{2}\int{\dfrac{\sin tdt}{2a}} $ 

 $ =\dfrac{1}{4a}[-\cos t]+C $ 

 $ =\dfrac{-1}{4a}\cos 2(ax+b)+C $ 

where C is an arbitrary constant.

6. Integrate $ \sqrt{ax+b} $ 

Ans:

Let  $ ax+b=t $ 

\[\Rightarrow adx=dt\]

\[\therefore dx=\dfrac{1}{a}dt\]

\[\Rightarrow \int{{{(ax+b)}^{\dfrac{1}{2}}}}dx=\dfrac{1}{a}\int{{{t}^{\dfrac{1}{2}}}}dt\]

\[=\dfrac{1}{a}\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{3}{2}} \right)+C\]

\[=\dfrac{2}{3a}{{(ax+b)}^{\dfrac{3}{2}}}+C\]

where C is an arbitrary constant.

7. Integrate  $ x\sqrt{x+2} $ 

Ans:

Let  $ x+2=t $ 

 $ \therefore dx=dt $ 

 $ \Rightarrow \int{x}\sqrt{x+2}dx=\int{(t-2)}\sqrt{t}dt $ 

 $ =\int{\left( {{t}^{\dfrac{3}{2}}}-2{{t}^{\dfrac{1}{2}}} \right)}dt\ $ \ $ =\int{{{t}^{\dfrac{3}{2}}}}dt-2\int{{{t}^{\dfrac{1}{2}}}}dt $ 

 $ =\dfrac{{{t}^{\dfrac{5}{2}}}}{\dfrac{5}{2}}-2\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C $ 

 $ =\dfrac{2}{5}{{t}^{\dfrac{5}{2}}}-\dfrac{4}{3}{{t}^{\dfrac{3}{2}}}+C $ 

 $ =\dfrac{2}{5}{{(x+2)}^{\dfrac{5}{2}}}-\dfrac{4}{3}{{(x+2)}^{\dfrac{3}{2}}}+C $ 

where C is an arbitrary constant.

8. Integrate  $ x\sqrt{1+2{{x}^{2}}} $ 

Ans:

Let  $ 1+2{{x}^{2}}=t $ 

\[\therefore 4\text{xdx}=\text{dt}\]

\[\Rightarrow \int{x}\sqrt{1+2{{x}^{2}}}dx=\int{\dfrac{\sqrt{t}}{4}}dt\]

\[=\dfrac{1}{4}\int{{{t}^{\dfrac{1}{2}}}}dt\]

\[=\dfrac{1}{4}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C\]

\[=\dfrac{1}{6}{{\left( 1+2{{x}^{2}} \right)}^{\dfrac{3}{2}}}+C\]

where C is an arbitrary constant.

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

Ans:

Let  $ {{\text{x}}^{2}}+\text{x}+1=\text{t} $ 

 $ \therefore (2\text{x}+1)\text{dx}=\text{dt} $ 

 $ \int{(4x+2)}\sqrt{{{x}^{2}}+x+1}dx $ 

 $ =\int{2}\sqrt{t}dt $ 

 $ =2\int{\sqrt{t}}dt $ 

 $ =2\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C $ 

 $ =\dfrac{4}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{3}{2}}}+C $ 

where C is an arbitrary constant.

10. Integrate  $ \dfrac{1}{x-\sqrt{x}} $ 

Ans:

The given function can be rewritten as

 $ \dfrac{1}{x-\sqrt{x}}=\dfrac{1}{\sqrt{x}\left( \sqrt{x}-1 \right)} $ 

Let $ \left( \sqrt{x}-1 \right)=t $ 

 $ \therefore \dfrac{1}{2\sqrt{x}}dx=dt $ 

 $ \Rightarrow \int{\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}}dx=\int{\dfrac{2}{t}}dt $ 

 $ =2\log |t|+C $ 

 $ =2\log |\sqrt{x}-1|+C $ 

where C is an arbitrary constant.

11. Integrate  $ \dfrac{x}{\sqrt{x+4}},x>0 $ 

Ans:

Let  $ x+4=t $ 

 $ \therefore dx=dt $ 

 $ \int{\dfrac{x}{\sqrt{x+4}}}dx=\int{\dfrac{(t-4)}{\sqrt{t}}}dt $ 

 $ =\int{\left( \sqrt{t}-\dfrac{4}{\sqrt{t}} \right)}dt $ 

 $ =\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-4\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right)+C $ 

 $ =\dfrac{2}{3}{{(t)}^{\dfrac{3}{2}}}-8{{(t)}^{\dfrac{1}{2}}}+C $ 

 $ =\dfrac{2}{3}t\cdot {{t}^{\dfrac{1}{2}}}-8{{t}^{\dfrac{1}{2}}}+C $ 

 $ =\dfrac{2}{3}{{t}^{\dfrac{1}{2}}}(t-12)+C $ 

 $ =\dfrac{2}{3}{{(x+4)}^{\dfrac{1}{2}}}(x+4-12)+C $ 

 $ =\dfrac{2}{3}\sqrt{x+4}(x-8)+C $ 

where C is an arbitrary constant.

12. Integrate  $ {{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{5}} $ 

Ans:

Let  $ {{x}^{3}}-1=t $ 

 $ \therefore 3{{x}^{2}}dx=dt $ 

 $ \Rightarrow \int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}}{{x}^{5}}dx=\int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}}{{x}^{3}}{{x}^{2}}dx $ 

 $ =\int{{{t}^{\dfrac{1}{3}}}}(t+1)\dfrac{dt}{3}\ $ \ $ =\dfrac{1}{3}\int{\left({{t}^{\dfrac{4}{3}}}+{{t}^{\dfrac{1}{3}}}\right)}dt $ 

 $ =\dfrac{1}{3}\left[ \dfrac{{{t}^{\dfrac{7}{3}}}}{\dfrac{7}{3}}+\dfrac{{{t}^{\dfrac{4}{3}}}}{\dfrac{4}{3}} \right]+C $ 

 $ =\dfrac{1}{3}\left[ \dfrac{3}{7}{{t}^{\dfrac{7}{3}}}+\dfrac{3}{4}{{t}^{\dfrac{4}{3}}} \right]+C $ 

 $ =\dfrac{1}{7}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{7}{3}}}+\dfrac{1}{4}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{4}{3}}}+C $ 

where C is an arbitrary constant.

13. Integrate  $ \dfrac{{{x}^{2}}}{{{\left( 2+3{{x}^{3}} \right)}^{3}}} $ 

Ans:

Let  $ 2+3{{x}^{3}}=t $ 

$ \therefore 9{{x}^{2}}dx=dt $ 

$ \Rightarrow \int{\dfrac{{{x}^{2}}}{{{\left( 2+3{{x}^{3}}           \right)}^{3}}}}dx=\dfrac{1}{9}\int{\dfrac{dt}{{{(t)}^{3}}}} $ 

$ =\dfrac{1}{9}\left[ \dfrac{{{t}^{-2}}}{-2} \right]+C $ 

$ =\dfrac{-1}{18}\left( \dfrac{1}{{{t}^{2}}} \right)+C $ 

$ =\dfrac{-1}{18{{\left( 2+3{{x}^{3}} \right)}^{2}}}+C $ 

where C is an arbitrary constant.

14. Integrate \[\dfrac{1}{x{{\left( \log x \right)}^{m}}},x>0\]

Ans:

Let  $ \log \text{x}=\text{t} $ 

 $ \dfrac{1}{x}dx=dt $ 

 $ \Rightarrow \int{\dfrac{1}{x{{(\log x)}^{m}}}}dx=\int{\dfrac{dt}{{{(t)}^{m}}}} $ 

 $ =\dfrac{{{t}^{-m+1}}}{-m+1}+C $ 

 $ =\dfrac{{{(\log x)}^{1-m}}}{(1-m)}+C $ 

where C is an arbitrary constant.

15. Integrate  $ \dfrac{x}{9-4{{x}^{2}}} $ 

Ans:

Let  $ 9-4{{x}^{2}}=t $ 

 $ \therefore -8xdx=dt $ 

 $ \Rightarrow \int{\dfrac{x}{9-4{{x}^{2}}}}dx=\dfrac{-1}{8}\int{\dfrac{1}{t}}dt $ 

 $ =\dfrac{-1}{8}\log |t|+C $ 

 $ =\dfrac{-1}{8}\log \left| 9-4{{x}^{2}} \right|+C $  

where C is an arbitrary constant.

16. Integrate  $ {{e}^{2x+3}} $ 

Ans:

Let  $ 2x+3=t $ 

 $ \therefore 2dx=dt $ 

 $ \Rightarrow \int{{{e}^{2x+3}}}dx=\dfrac{1}{2}\int{{{e}^{t}}}dt $ 

 $ =\dfrac{1}{2}\left( {{e}^{t}} \right)+C $ 

 $ =\dfrac{1}{2}\left( {{e}^{2x+3}} \right)+C $ 

where C is an arbitrary constant.

17. Integrate  $ \dfrac{x}{{{e}^{{{x}^{2}}}}} $ 

Ans:

Let  $ {{\text{x}}^{2}}=\text{t} $ 

 $ \therefore 2\text{xdx}=\text{dt} $ 

 $ \Rightarrow \int{\dfrac{x}{{{e}^{{{x}^{2}}}}}}dx=\dfrac{1}{2}\int{\dfrac{1}{{{e}^{t}}}}dt $ 

 $ =\dfrac{1}{2}\int{{{e}^{-t}}}dt $ 

 $ =\dfrac{1}{2}\left( \dfrac{{{e}^{-t}}}{-1} \right)+C $ 

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

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

 where C is an arbitrary constant.

18. Integrate  $ \dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}} $ 

Ans:

Let  $ {{\tan }^{-1}}x=t $ 

 $ \therefore \dfrac{1}{1+{{x}^{2}}}dx=dt $ 

 $ \Rightarrow \int{\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}}dx=\int{{{e}^{t}}}dt $ 

 $ ={{e}^{t}}+C $ 

 $ ={{e}^{{{\tan }^{-1}}x}}+C $ 

where C is an arbitrary constant.

19. Integrate \[\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1}\]

Ans:

Dividing the given function’s numerator and denominator by  $ {{e}^{x}} $ , we obtain,

\[\dfrac{\dfrac{\left( {{e}^{2x}}-1 \right)}{{{e}^{x}}}}{\dfrac{\left( {{e}^{2x}}+1 \right)}{{{e}^{x}}}}=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}\]

Let \[{{e}^{x}}+{{e}^{-x}}=t\]

\[\left( {{e}^{x}}-{{e}^{-x}} \right)dx=dt\]

\[\Rightarrow \int{\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1}}dx=\int{\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}}dx\]

\[=\int{\dfrac{dt}{t}}\]

\[=\log |t|+C\]

\[=\log |{{e}^{x}}-{{e}^{-x}}|+C\]

where C is an arbitrary constant.

20. Integrate \[\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}\]

Ans:

Let  $ {{e}^{2x}}+{{e}^{-2x}}=t $ 

 $ \Rightarrow 2{{e}^{2x}}-2{{e}^{-2x}}dx=dt $ 

 $ \Rightarrow 2\left( {{e}^{2x}}-{{e}^{-2x}} \right)dx=dt $ 

 $ \Rightarrow \int{\left( \dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}} \right)}dx=\int{\dfrac{dt}{2t}} $ 

 $ =\dfrac{1}{2}\int{\dfrac{1}{t}}d $ 

 $ =\dfrac{1}{2}\log |t|+C $ 

 $ =\dfrac{1}{2}\log \left| {{e}^{2x}}+{{e}^{-2x}} \right|+C $ 

where C is an arbitrary constant.

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

Ans:

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

Let  $ 2\text{x}-3=\text{t} $ 

 $ \therefore 2\text{dx}=\text{dt} $ 

 $ \Rightarrow \int{{{\tan }^{2}}}(2x-3)dx=\int{\left[ {{\sec }^{2}}(2x-3)-1 \right]}dx $ 

 $ =\dfrac{1}{2}\int{\left( {{\sec }^{2}}t \right)}dt-\int{1}dx $ 

 $ =\dfrac{1}{2}\int{{{\sec }^{2}}}tdt-\int{1}dx $ 

 $ =\dfrac{1}{2}\tan t-x+C $ 

 $ =\dfrac{1}{2}\tan (2x-3)-x+C $ 

where C is an arbitrary constant.

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

Ans:

Let  $ 7-4x=t $ 

 $ \therefore -4~\text{d}x=dt $ 

 $ \therefore \int{{{\sec }^{2}}}(7-4x)dx=\dfrac{-1}{4}\int{{{\sec }^{2}}}tdt $ 

 $ =\dfrac{-1}{4}(\tan t)+C $ 

 $ =\dfrac{-1}{4}\tan (7-4x)+C $ 

where C is an arbitrary constant.

23. Integrate  $ \dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} $ 

Ans:

Let  $ {{\sin }^{-1}}x=t $ 

 $ \dfrac{1}{\sqrt{1-{{x}^{2}}}}dx=dt $ 

 $ \Rightarrow \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx=\int{t}dt $ 

 $ =\dfrac{{{t}^{2}}}{2}+C $ 

 $ =\dfrac{{{\left( {{\sin }^{-1}}x \right)}^{2}}}{2}+C $ 

where C is an arbitrary constant.

24. Integrate  $ \dfrac{2\cos x-3\sin x}{6\cos x+4\sin x} $ 

Ans:

The given function is 

 $ \dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}=\dfrac{2\cos x-3\sin x}{2(3\cos x+2\sin x)} $ 

Let  $ 3\cos x+2\sin x=t $ 

 $ (-3\sin x+2\cos x)dx=dt $ 

 $ \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\int{\dfrac{dt}{2t}} $ 

 $ =\dfrac{1}{2}\int{\dfrac{1}{t}}dt $ 

 $ =\dfrac{1}{2}\log |t|+C $ 

 $ =\dfrac{1}{2}\log |2\sin x+3\cos x|+C $ 

where C is an arbitrary constant.

25. Integrate  $ \dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}} $ 

Ans:

The given function is,

 $ \dfrac{1}{{{\cos }^{2}}x{{(1-\tan x)}^{2}}}=\dfrac{{{\sec }^{2}}}{{{(1-\tan x)}^{2}}} $ 

Let  $ (1-\tan x)=t $ 

 $ -{{\sec }^{2}}xdx=dt $ 

 $ \Rightarrow \int{\dfrac{{{\sec }^{2}}}{{{(1-\tan x)}^{2}}}}dx=\int{\dfrac{-dt}{{{t}^{2}}}} $ 

 $ =-\int{{{t}^{-2}}}dt $ 

 $ =+\dfrac{1}{t}+C $ 

 $ =\dfrac{1}{\left( 1-\tan x \right)}+C $ 

where C is an arbitrary constant.

26. Integrate  $ \dfrac{\cos \sqrt{x}}{\sqrt{x}} $ 

Ans:

let  $ \sqrt{x}=t $ 

 $ \dfrac{1}{2\sqrt{x}}dx=dt $ 

 $ \Rightarrow \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}}dx=2\int{\cos }tdt=2\sin t+C=2\sin \sqrt{x}+C $ 

where C is an arbitrary constant.

27. Integrate  $ \sqrt{\sin 2x}\cos 2x $ 

Ans:

Let  $ \sin 2\text{x}=\text{t} $ 

So,  $ 2\cos 2\text{xdx}=\text{dt} $ 

 $ \Rightarrow \int{\sqrt{\sin 2x}}\cos 2xdx $ 

 $ =\dfrac{1}{2}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C $ 

 $ =\dfrac{1}{2}\int{\sqrt{t}}dt $ 

 $ =\dfrac{1}{3}{{t}^{\dfrac{3}{2}}}+C $ 

 $ =\dfrac{1}{3}{{(\sin 2x)}^{\dfrac{3}{2}}}+C $ 

where C is an arbitrary constant.

28. Integrate  $ \dfrac{\cos x}{\sqrt{1+\sin x}} $ 

Ans:

Let  $ 1+\sin \text{x}=\text{t} $ 

 $ \therefore \cos \text{xdx}=\text{dt} $ 

 $ \Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}}dx=\int{\dfrac{dt}{\sqrt{t}}} $ 

 $ =\dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}+C $ 

 $ =2\sqrt{t}+\text{C} $ 

 $ =2\sqrt{1+\sin x}+C $ 

where C is an arbitrary constant.

29. Integrate  $ \cot x\log \sin x $ 

Ans:

Let \[\log \sin x=t\]

\[\Rightarrow \dfrac{1}{\sin x}\cdot \cos xdx=dt\]

\[\therefore \cot xdx=dt\]

\[\Rightarrow \int{\cot }x\log \sin xdx=\int{t}dt\]

\[=\dfrac{{{t}^{2}}}{2}+C\]

\[=\dfrac{1}{2}{{(\log \sin x)}^{2}}+C\]

where C is an arbitrary constant.

30. Integrate  $ \dfrac{\sin x}{1+\cos x} $ 

Ans:

$ \Rightarrow \int{\dfrac{\sin x}{1+\cos x}}dx=\int{-}\dfrac{dt}{t} $ 

$ =-\log |t|+C $ 

$ =-\log |1+\cos x|+C $ 

where C is an arbitrary constant.

31. Integrate  $ \dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}} $ 

Ans:

Let  $ 1+\cos \text{x}=\text{t} $ 

 $ \therefore -\sin \text{xdx}=\text{dt} $ 

 $ \Rightarrow \int{\dfrac{\sin x}{{{(1+\cos x)}^{2}}}}dx=\int{-}\dfrac{dt}{{{t}^{2}}} $ 

 $ =-\int{{{t}^{-2}}}dt $ 

 $ =\dfrac{1}{t}+C $ 

 $ =\dfrac{1}{1+\cos x}+C $ 

where C is an arbitrary constant.

32. Integrate $ \dfrac{1}{1+\cot x} $ 

Ans:

Let \[I=\int{\dfrac{1}{1+\cot x}}dx\]

 $ =\int{\dfrac{1}{1+\dfrac{\cos x}{\sin x}}}dx $ 

 $ =\int{\dfrac{\sin x}{\sin x+\cos x}}dx $ 

 $ =\dfrac{1}{2}\int{\dfrac{2\sin x}{\sin x+\cos x}}dx $ 

\[=\dfrac{1}{2}\int{\dfrac{(\sin x+\cos x)+(\sin x-\cos x)}{(\sin x+\cos x)}}dx\]

 $ =\dfrac{1}{2}\int{1}dx+\dfrac{1}{2}\int{\dfrac{\sin x-\cos x}{\sin x+\cos x}}dx $ 

 $ =\dfrac{1}{2}(x)+\dfrac{1}{2}\int{\dfrac{\sin x-\cos x}{\sin x+\cos x}}dx $ 

Let  $ \sin x+\cos x=t\Rightarrow (\cos x-\sin x)dx=dt $ 

 $ \therefore I=\dfrac{x}{2}+\dfrac{1}{2}\int{\dfrac{-(dt)}{t}} $ 

 $ =\dfrac{x}{2}-\dfrac{1}{2}\log |t|+C $ 

 $ =\dfrac{x}{2}-\dfrac{1}{2}\log |\sin x+\cos x|+C $ 

where C is an arbitrary constant.

33. Integrate  $ \dfrac{1}{1-\tan x} $ 

Ans:

Let  $ I=\int{\dfrac{1}{1-\tan x}}dx $ 

 $ =\int{\dfrac{1}{1-\dfrac{\sin x}{\cos x}}}dx $ 

 $ =\int{\dfrac{\cos x}{\cos x-\sin x}}dx $ 

 $ =\dfrac{1}{2}\int{\dfrac{2\cos x}{\cos x-\sin x}}dx $ 

 $ =\dfrac{1}{2}\int{\dfrac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)}}dx $ 

 $ =\dfrac{1}{2}\int{1}dx+\dfrac{1}{2}\int{\dfrac{\cos x+\sin x}{\cos x-\sin x}}dx $ 

 $ =\dfrac{x}{2}+\dfrac{1}{2}\int{\dfrac{\cos x+\sin x}{\cos x-\sin x}}dx $ 

Put  $ \cos x-\sin x=t\Rightarrow (-\sin x-\cos x)dx=dt $ 

 $ \therefore I=\dfrac{x}{2}+\dfrac{1}{2}\int{\dfrac{-(dt)}{t}} $ 

 $ =\dfrac{x}{2}-\dfrac{1}{2}\log |t|+C $ 

 $ =\dfrac{x}{2}-\dfrac{1}{2}\log |\cos x-\sin x|+C $ 

where C is an arbitrary constant.

34. Integrate  $ \dfrac{\sqrt{\tan x}}{\sin x\cos x} $ 

Ans:

Let  $ I=\int{\dfrac{\sqrt{\tan x}}{\sin x\cos x}}dx $ 

 $ =\int{\dfrac{\sqrt{\tan x}\times \cos x}{\sin x\cos x\times \cos x}}dx $ 

 $ =\int{\dfrac{\sqrt{\tan x}}{\tan x{{\cos }^{2}}x}}dx $ 

 $ =\int{\dfrac{{{\sec }^{2}}xdx}{\sqrt{\tan x}}} $ 

Let  $ \tan x=t\Rightarrow {{\sec }^{2}}xdx=dt $ 

 $ \therefore I=\int{\dfrac{dt}{\sqrt{t}}} $ 

 $ =2\sqrt{t}+C $ 

 $ =2\sqrt{\tan x+C} $ 

where C is an arbitrary constant.

35. Integrate  $ \dfrac{{{\left( 1+\log x \right)}^{2}}}{x} $ 

Ans:

Let  $ 1+\log x=t $ 

 $ \therefore \dfrac{1}{x}dx=dt $ 

 $ \Rightarrow \int{\dfrac{{{(1+\log x)}^{2}}}{x}}dx\text{ }=\int{{{t}^{2}}}dt\text{ } $ 

 $ =\dfrac{{{t}^{3}}}{3}+C $

 $ =\dfrac{{{(1+\log x)}^{3}}}{3}+C $

where C is an arbitrary constant.

36. Integrate  $ \dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x} $ 

Ans:

The given function can be rewritten as

 $ \dfrac{(x+1){{(x+\log x)}^{2}}}{x} $ 

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

Let  $ (x+\log x)=t $ 

 $ \therefore \left( 1+\dfrac{1}{x} \right)dx=dt $ 

 $ \Rightarrow \int{\left( 1+\dfrac{1}{x} \right)}{{(x+\log x)}^{2}}dx\text{ } $ 

 $ =\int{{{t}^{2}}}dt $ 

 $ =\dfrac{{{t}^{3}}}{3}+C $ 

 $ =\dfrac{1}{3}{{(x+\log x)}^{3}}+C $ 

where C is an arbitrary constant. 

37. Integrate  $ \dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}} $ 

Ans:

$ \therefore 4{{\text{x}}^{3}}\text{dx}=\text{dt} $ 

 $ \Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}}dx=\dfrac{1}{4}\int{\dfrac{\sin \left( {{\tan }^{-1}}t \right)}{1+{{t}^{2}}}}dt $ 

Let  $ {{\tan }^{-1}}t=u $ 

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

From (1), we obtain

 $ \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)dx}{1+{{x}^{8}}}}=\dfrac{1}{4}\int{\sin }udu $ 

 $ =\dfrac{1}{4}(-\cos u)+C $ 

 $ =-\dfrac{1}{4}\cos \left( {{\tan }^{-1}}t \right)+C $ 

 $ =\dfrac{-1}{4}\cos \left( {{\tan }^{-1}}{{x}^{4}} \right)+C $ 

where C is an arbitrary constant. 

38. Here, $ \int{\dfrac{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 $ 

Ans:

Let  $ {{x}^{10}}+{{10}^{x}}=t $ 

 $ \therefore \left( 10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10 \right)dx=\int{\dfrac{dt}{t}} $ 

 $ \Rightarrow \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{\varepsilon }}10}{{{x}^{10}}+10x}}dx=\int{\dfrac{dt}{t}} $ 

 $ =\log t+C $ 

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

Hence, the correct Answer is D.

39. Here, \[\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}\]equals 

A.\[\tan x+\cot x+C\]

B.\[\tan x-\cot x+C\]

C.\[\tan x\cot x+C\]

D.\[\tan x-\cot 2x+C\]

Ans:

Let  $ I=\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}} $ 

 $ =\int{\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx $ 

 $ =\int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx+\int{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx $ 

 $ =\int{{{\sec }^{2}}}xdx+\int{{{\operatorname{cosec}}^{2}}}dx $ 

 $ =\tan x-\cot x+C $ 

Hence, the correct Answer is B.

Conclusion

The NCERT Solutions for Class 12 Maths Exercise 7.2 Chapter 7 - Integrals, Ex 7.2 class 12 Maths, provided by Vedantu, offers comprehensive guidance for mastering integration techniques. This exercise focuses on the application of various integration methods, including substitution and integration by parts, to solve complex integrals. Key topics include Integration using substitution. Integration by parts. Solving integrals involving trigonometric functions and logarithms. By solving these questions, students can solidify their understanding and improve their problem-solving skills in integration​.

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