Chapter 7: Integrals

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.