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​ 

Class 12 Maths Chapter 7: Exercises Breakdown

CBSE Class 12 Maths Chapter 7 Other Study Materials

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.