Chapter 7: Integrals

Ans: Given expression ${{\cos }^{4}}2x$.

Given expression can be written as  

${{\cos }^{4}}2x={{\left( {{\cos }^{2}}2x \right)}^{2}}$

$\Rightarrow {{\cos }^{4}}2x={{\left( \frac{1+\cos 4x}{2} \right)}^{2}}$

$\Rightarrow {{\cos }^{4}}2x=\frac{1}{4}\left( 1+{{\cos }^{2}}4x+2\cos 4x \right)$

$\Rightarrow {{\cos }^{4}}2x=\frac{1}{4}\left( 1+\frac{1+\cos 8x}{2}+2\cos 4x \right)$

\[\Rightarrow {{\cos }^{4}}2x=\frac{1}{4}\left( 1+\frac{1}{2}+\frac{\cos 8x}{2}+2\cos 4x \right)\]

\[\Rightarrow {{\cos }^{4}}2x=\frac{1}{4}\left( \frac{3}{2}+\frac{\cos 8x}{2}+2\cos 4x \right)\]

Integration of given expression is 

\[\Rightarrow \int{{{\cos }^{4}}2xdx}=\int{\left( \frac{3}{8}+\frac{\cos 8x}{8}+\frac{\cos 4x}{2} \right)dx}\]

\[\therefore \int{{{\cos }^{4}}2xdx}=\frac{3}{8}x+\frac{\sin 8x}{64}+\frac{\sin 4x}{8}+C\]

Ans: Given expression $\frac{{{\sin }^{2}}x}{1+\cos x}$.

By applying the identity $\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$ and $\cos x=2{{\cos }^{2}}\frac{x}{2}-1$, given expression can be written as 

  $\Rightarrow \frac{{{\sin }^{2}}x}{1+\cos x}=\frac{{{\left( 2\sin \frac{x}{2}\cos \frac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\frac{x}{2}}$

$\Rightarrow \frac{{{\sin }^{2}}x}{1+\cos x}=\frac{4{{\sin }^{2}}\frac{x}{2}{{\cos }^{2}}\frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}}$

$\Rightarrow \frac{{{\sin }^{2}}x}{1+\cos x}=2{{\sin }^{2}}\frac{x}{2}$

$\Rightarrow \frac{{{\sin }^{2}}x}{1+\cos x}=1-\cos x$

Integration of given expression is

$\Rightarrow \int{\frac{{{\sin }^{2}}x}{1+\cos x}dx}=\int{1dx-\int{\cos xdx}}$

$\therefore \int{\frac{{{\sin }^{2}}x}{1+\cos x}dx}=x-\sin x+C$

Ans: Given expression $\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$.

We can apply the identity $\cos C-\cos D=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}$ , we get

 $\Rightarrow \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\frac{-2\sin \frac{2x+2\alpha }{2}\sin \frac{2x-2\alpha }{2}}{-2\sin \frac{x+\alpha }{2}\sin \frac{x-\alpha }{2}}$

$\Rightarrow \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\frac{\sin \frac{2\left( x+\alpha  \right)}{2}\sin \frac{2\left( x-\alpha  \right)}{2}}{\sin \frac{x+\alpha }{2}\sin \frac{x-\alpha }{2}}$

$\Rightarrow \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\frac{\sin \left( x+\alpha  \right)\sin \left( x-\alpha  \right)}{\sin \frac{x+\alpha }{2}\sin \frac{x-\alpha }{2}}$

We can apply the identity $\sin 2x=2\sin x\cos x$, we get

$\Rightarrow \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\frac{\left[ 2\sin \frac{x+\alpha }{2}\cos \frac{x+\alpha }{2} \right]\left[ 2\sin \frac{x-\alpha }{2}\cos \frac{x-\alpha }{2} \right]}{\sin \frac{x+\alpha }{2}\sin \frac{x-\alpha }{2}}$

$\Rightarrow \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=4\cos \frac{x+\alpha }{2}\cos \frac{x-\alpha }{2}$

$\Rightarrow \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=2\left[ \cos \frac{x+\alpha }{2}+\frac{x-\alpha }{2}+\cos \frac{x+\alpha }{2}-\frac{x-\alpha }{2} \right]$

$\Rightarrow \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=2\left[ \cos x+\cos \alpha  \right]$

Integration of given expression is

$\Rightarrow \int{\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx}=2\int{\left[ \cos x+\cos \alpha  \right]}dx$

$\therefore \int{\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx}=2\left[ \sin x+x\cos \alpha  \right]+C$

Ans: Given expression $\frac{\cos x-\sin x}{1+\sin 2x}$.

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.

Given expression can be written as  

$\Rightarrow \frac{\cos x-\sin x}{1+\sin 2x}=\frac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+\sin 2x}$ .

We can apply the identity $\sin 2x=2\sin x\cos x$, we get

 $\Rightarrow \frac{\cos x-\sin x}{1+\sin 2x}=\frac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}$

$\Rightarrow \frac{\cos x-\sin x}{1+\sin 2x}=\frac{\cos x-\sin x}{{{\left( \sin x+\cos x \right)}^{2}}}$

Let $\sin x+\cos x=t$ 

$\therefore \left( \cos x-\sin x \right)dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\frac{\cos x-\sin x}{1+\sin 2x}dx=\int{\frac{\cos x-\sin x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}}$ 

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

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

\[\Rightarrow \int{\frac{\cos x-\sin x}{1+\sin 2x}dx=-{{t}^{-1}}+C}\]

\[\Rightarrow \int{\frac{\cos x-\sin x}{1+\sin 2x}dx=-\frac{1}{t}+C}\]

Substitute $\sin x+\cos x=t$,

\[\therefore \int{\frac{\cos x-\sin x}{1+\sin 2x}dx=-\frac{1}{\sin x+\cos x}+C}\]

Ans: Given expression ${{\tan }^{3}}2x\sec 2x$.

Given expression can be written as

${{\tan }^{3}}2x\sec 2x={{\tan }^{2}}2x\tan 2x\sec 2x$

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

$\Rightarrow {{\tan }^{3}}2x\sec 2x={{\sec }^{2}}2x\tan 2x\sec 2x-\tan 2x\sec 2x$

Integration of given expression is

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\int{{{\sec }^{2}}2x\tan 2x\sec 2xdx}-\int{\tan 2x\sec 2xdx}$

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\int{{{\sec }^{2}}2x\tan 2x\sec 2xdx}-\frac{\sec 2x}{2}+C$

Let $\sec 2x=t$ 

$\therefore 2\sec 2x\tan 2xdx=dt$ 

Above integral becomes

\[\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\frac{1}{2}\int{{{t}^{2}}dt}-\frac{\sec 2x}{2}+C\]

\[\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\frac{{{t}^{3}}}{6}-\frac{\sec 2x}{2}+C\]

Substitute $\sec 2x=t$,

\[\therefore \int{{{\tan }^{3}}2x\sec 2xdx}=\frac{{{\left( \sec 2x \right)}^{3}}}{6}-\frac{\sec 2x}{2}+C\]

Ans: Given expression ${{\tan }^{4}}x$.

Given expression can be written as 

$\Rightarrow {{\tan }^{4}}x={{\tan }^{2}}x{{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x=\left( {{\sec }^{2}}x-1 \right){{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-{{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-\left( {{\sec }^{2}}x-1 \right)$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1$

Integration of given expression is

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1 \right)}dx$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\tan }^{2}}x \right)}dx-\int{{{\sec }^{2}}xdx+\int{1dx}}$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{{{\sec }^{2}}x{{\tan }^{2}}x}dx-\tan x+x+C$

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{{{t}^{2}}}dt-\tan x+x+C$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\frac{{{t}^{3}}}{3}-\tan x+x+C$

Substitute $\tan x=t$,

$\therefore \int{{{\tan }^{4}}xdx}=\frac{1}{3}{{\tan }^{3}}x-\tan x+x+C$

Ans: Given expression $\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$.

Given expression can be written as 

$\Rightarrow \frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\frac{{{\sin }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}+\frac{{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$

$\Rightarrow \frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\frac{\sin x}{{{\cos }^{2}}x}+\frac{\cos x}{{{\sin }^{2}}x}$

$\Rightarrow \frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\tan x\sec x+\cot xcosecx$

Integration of given expression is 

$\Rightarrow \int{\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\tan x\sec xdx}+\int{\cot x\operatorname{cosecx}dx}$

$\therefore \int{\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\sec x-cosecx+C$

Ans: Given expression $\frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}$.

By applying the identity $\cos 2x=1-2{{\sin }^{2}}x$, we get

 $\Rightarrow \frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}=\frac{\cos 2x+1-\cos 2x}{{{\cos }^{2}}x}$

$\Rightarrow \frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}=\frac{1}{{{\cos }^{2}}x}$

$\Rightarrow \frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\sec }^{2}}x$

Integration of given expression is 

$\Rightarrow \int{\frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx=\int{{{\sec }^{2}}xdx}$

$\therefore \int{\frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx=\tan x+C$

Ans: Given expression $\frac{1}{\sin x{{\cos }^{3}}x}$.

We can apply the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get

$\Rightarrow \frac{1}{\sin x{{\cos }^{3}}x}=\frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x}$

$\Rightarrow \frac{1}{\sin x{{\cos }^{3}}x}=\frac{{{\sin }^{2}}x}{\sin x{{\cos }^{3}}x}+\frac{{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x}$

$\Rightarrow \frac{1}{\sin x{{\cos }^{3}}x}=\frac{\sin x}{{{\cos }^{3}}x}+\frac{1}{\sin x\cos x}$

$\Rightarrow \frac{1}{\sin x{{\cos }^{3}}x}=\tan x{{\sec }^{2}}x+\frac{{{\cos }^{2}}x}{\frac{\sin x\cos x}{{{\cos }^{2}}x}}$

$\Rightarrow \frac{1}{\sin x{{\cos }^{3}}x}=\tan x{{\sec }^{2}}x+\frac{{{\sec }^{2}}x}{\tan x}$

Integration of given expression is

$\Rightarrow \int{\frac{1}{\sin x{{\cos }^{3}}x}}dx=\int{\tan x{{\sec }^{2}}x}dx+\int{\frac{{{\sec }^{2}}x}{\tan x}}dx$

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

\[\Rightarrow \int{\frac{1}{\sin x{{\cos }^{3}}x}}dx=\int{t}dt+\int{\frac{1}{\operatorname{t}}}dt\]

\[\Rightarrow \int{\frac{1}{\sin x{{\cos }^{3}}x}}dx=\frac{{{t}^{2}}}{2}+\log \left| t \right|+C\]

Substitute $\tan x=t$,

\[\therefore \int{\frac{1}{\sin x{{\cos }^{3}}x}}dx=\frac{1}{2}{{\tan }^{2}}x+\log \left| \tan x \right|+C\]

Ans: Given expression $\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}$.

Given expression can be written as 

$\Rightarrow \frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\frac{\cos 2x}{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}$

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $2\sin x\cos x=\sin 2x$, we get

 $\Rightarrow \frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\frac{\cos 2x}{1+\sin 2x}$

Integration of given expression is

$\Rightarrow \int{\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\int{\frac{\cos 2x}{1+\sin 2x}dx}$

Let $1+\sin 2x=t$ 

$\therefore 2\cos 2xdx=dt$ 

Integration becomes

$\Rightarrow \int{\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\frac{1}{2}\int{\frac{1}{t}dt}$

$\Rightarrow \int{\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\frac{1}{2}\log \left| t \right|+C$

Substitute $1+\sin 2x=t$

$\Rightarrow \int{\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\frac{1}{2}\log \left| 1+\sin 2x \right|+C$

$\Rightarrow \int{\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\frac{1}{2}\log \left| {{\left( \cos x+\sin x \right)}^{2}} \right|+C$

$\therefore \int{\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\log \left| \left( \cos x+\sin x \right) \right|+C$