Chapter 7: Integrals

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.

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.

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.

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.

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.

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. 

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. 

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.

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.