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

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.

NCERT Solutions for Class 12 Maths Chapter 7 – Exercise 7.2 Questions

Ex 7.2 Class 12 Maths NCERT Solutions: Introduction

Integration is known to be one of the main operations of calculus and the other important one is differentiation. The operation of the integration is one of the additive constants to differentiate between characters. It is the fundamental theorem of calculus that connects the differentiation with the definite integral. 

The principles of integration were formulated independently by the famous Isaac Newton and another famous scholar Gottfried Wilhelm Leibniz in the late years of 17th century, who thought of the importance of integral as an infinite sum of rectangles of infinitesimal width. Bernhard Riemann, later on, gave a very rigorous mathematical definition of integrals, which is based on a limiting procedure that approximates the whole area of a curvilinear region by independently breaking the region into thin vertical slabs. 

Beginning of the 19th century, class 12 Maths NCERT solutions chapter 7 exercise 7.2 will tell you, was the time when more sophisticated concepts of integrals began to appear. During this period, when, where, the type of the function, as well as the domain over which the integration is performed, used to be generalised. A line integral is something that covers the function of variables (2 or more in number). Here a curve connects the two endpoints and replaces the interval of integration. When surface integral is used in case of a 3D space, a piece of surface replaces the curve.

NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2: Formalization

Though Newton and Leibniz came up with a systematic approach of integration, there were some drawbacks of it regarding the degree of rigour. The concept proposed by Newton was opposed by Bishop Berkeley and was termed as “ghosts of departed quantities”.  Concerning the development of limits, Calculus was more accepted. In 1854 Riemann was the first to formalise integration including limits.

All bounded piecewise continuous functions are Riemann-integrable on a bounded interval, but along with that more general functions were also considered—particularly in the context of Fourier analysis—to which Riemann's definition does not apply. Lebesgue came up with a definition of integral which was different from the existing ones and took in the concept of measure theory which concerned real analysis or subfield analysis.

Gradually, integral was defined in other ways also bypassing the concepts proposed by Riemann and Lebesgue. Some of them were based on the real number system and those are the ones that are popularly used today. Alternative concepts like integral based on hyperreal number system are also extensively used.

Exercise 7.2 Class 12 Maths NCERT Solutions 

The topics and sub-topics included in the Integrals chapter are the following:

Class 12 Maths Ch Ex 7.2

Q1. Give the Solution of \[\int \frac{2x}{1+x^{2}}\] dx

Solution:

Let 1 + x2 = t

2x dx = dt

=>  \[\int \frac{2x}{1+x^{2}}\] dx = \[\int \frac{1}{t}\] dt

= log|t| + C

= log|1 + x2| + C

=log(1 + x2) + C

Question 2

\[\int \frac{dx}{sin^{2}x \; \; cos^{2}x}\]

(a) tanx + cotx + c

(b) tanx – cotx + c

(c) tanx cotx + c

(d) tanx – cot2x + c

Solution:

Let x10 + 10x = t

(10x9 + 10x loge 10)dx = dt

=> \[\int \frac{10x^{9} + 10^{x}log_{e}10}{x^{10} + 10^{x}}\] dx = \[\int \frac{dx}{t}\]

= log t +C

= log (10x + x10) + C

Hence, the correct answer is D

Question 3

\[\int \frac{10x^{9} + 10^{x}log_{e}^{10}}{x^{10} + 10^{x}}\] dx

(a) 10x – x10 + C

(b) 10x + x10 + C

(c) (10x – x10) + C

(d) log (10x + x10) + C

Solution:

Let x10 + 10x = t

(10x9 + 10x loge 10)dx = dt

=> \[\int \frac{10x^{9} + 10^{x}log_{e}10}{x^{10} + 10^{x}}\] dx = \[\int \frac{dx}{t}\]

= log t +C

= log (10x + x10) + C

Hence, the correct answer is D

Question 4

\[\frac{x^{3}sin(tan^{-1}x^{4})}{1 + x^{8}}\] dx

Solution:

Let x4 = t

4x3dx = dt

=> \[\int \frac{x^{3}sin(tan^{-1}x^{4})}{1 + x^{8}}\] dx = \[\frac{1}{4}\int \frac{sin(tan^{-1}t)}{1 + t^{2}}\] dt ……..(1)

let tan-1t = u

\[\frac{1}{1+t^{2}}\] dt = du

from (1) we obtain, 

\[\int \frac{x^{3}sin(tan^{-1}x^{4})}{1 + x^{8}}\] dx = ¼ ∫sin u du

= ¼ (- cos u) + C

-¼ cos (tan-1 t) + C

-¼ cos (tan-1 x4) + C

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solution Class 12 Maths of Chapter 7 All Exercises

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