Chapter 7: Integrals

Ans: Given expression ${{\sin }^{2}}\left( 2x+5 \right)$.

Given expression can be written as  

${{\sin }^{2}}\left( 2x+5 \right)=\frac{1-\cos 2\left( 2x+5 \right)}{2}$

$\Rightarrow {{\sin }^{2}}\left( 2x+5 \right)=\frac{1-\cos \left( 4x+10 \right)}{2}$ 

Integration of given expression is

\[\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\int{\frac{1-\cos \left( 4x+10 \right)}{2}dx}\]

\[\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\frac{1}{2}\int{1dx-\frac{1}{2}\int{\cos \left( 4x+10 \right)dx}}\]

\[\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\frac{1}{2}x-\frac{1}{2}\left( \frac{\sin \left( 4x+10 \right)}{4} \right)+C\]

\[\therefore \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\frac{1}{2}x-\frac{1}{8}\sin \left( 4x+10 \right)+C\]

Ans: Given expression $\sin 3x\cos 4x$.

Using the identity $\sin A\cos B=\frac{1}{2}\left\{ \sin \left( A+B \right)+\sin \left( A-B \right) \right\}$ given expression can be written as

$\sin 3x\cos 4x=\frac{1}{2}\left\{ \sin \left( 3x+4x \right)+\sin \left( 3x-4x \right) \right\}$

Integration of above expression is

\[\Rightarrow \int{\sin 3x\cos 4x}dx=\int{\frac{1}{2}\left\{ \sin 7x+\sin \left( -x \right) \right\}dx}\]

\[\Rightarrow \int{\sin 3x\cos 4x}dx=\frac{1}{2}\int{\left\{ \sin 7x+\sin x \right\}dx}\]

\[\Rightarrow \int{\sin 3x\cos 4x}dx=\frac{1}{2}\int{\sin 7xdx+\frac{1}{2}\int{\sin x}dx}\]

\[\Rightarrow \int{\sin 3x\cos 4x}dx=\frac{1}{2}\left( \frac{-\cos 7x}{7} \right)-\frac{1}{2}\left( -\cos x \right)+C\]

\[\therefore \int{\sin 3x\cos 4x}dx=\frac{-\cos 7x}{14}+\frac{\cos x}{2}+C\]

Ans: Given expression $\cos 2x\cos 4x\cos 6x$.

Using the identity $\cos A\cos B=\frac{1}{2}\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$ given expression can be written as

$\cos 2x\left( \cos 4x\cos 6x \right)=\cos 2x\frac{1}{2}\left\{ \cos \left( 4x+6x \right)+\cos \left( 4x-6x \right) \right\}$

Integration of the above expression is

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\int{\cos 2x}\left[ \frac{1}{2}\left\{ \cos 10x+\cos \left( -2x \right) \right\} \right]dx$

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\int{\left[ \frac{1}{2}\left\{ \cos 2x\cos 10x+\cos 2x\cos \left( -2x \right) \right\} \right]}dx$

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\frac{1}{2}\int{\left[ \left\{ \cos 2x\cos 10x+{{\cos }^{2}}2x \right\} \right]}dx$

Again applying the identity $\cos A\cos B=\frac{1}{2}\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$, we get

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\frac{1}{2}\int{\left[ \left\{ \frac{1}{2}\cos \left( 2x+10x \right)+\cos \left( 2x-10x \right)+\left( \frac{1+\cos 4x}{2} \right) \right\} \right]}dx$$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\frac{1}{4}\int{\left[ \cos 12x+\cos 8x+\cos 4x \right]}dx$

$\therefore \int{\cos 2x\cos 4x\cos 6x}=\frac{1}{4}\left[ \frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4} \right]+C$

Ans: Given expression ${{\sin }^{3}}\left( 2x+1 \right)$. 

Let $I=\int{{{\sin }^{3}}\left( 2x+1 \right)dx}$

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

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

Let \[\cos \left( 2x+1 \right)=t\] 

$\therefore -2\sin \left( 2x+1 \right)dx=dt$ 

Integration becomes

$\Rightarrow I=-\frac{1}{2}\int{\left( 1-{{t}^{2}} \right)dt}$

$\Rightarrow I=-\frac{1}{2}\left( t-\frac{{{t}^{3}}}{3} \right)+C$

Substitute \[\cos \left( 2x+1 \right)=t\],

$\Rightarrow I=-\frac{1}{2}\left( \cos \left( 2x+1 \right)-\frac{{{\cos }^{3}}\left( 2x+1 \right)}{3} \right)+C$

$\therefore \int{{{\sin }^{3}}\left( 2x+1 \right)dx}=\frac{-\cos \left( 2x+1 \right)}{2}+\frac{{{\cos }^{3}}\left( 2x+1 \right)}{3}+C$

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

Let $I=\int{{{\sin }^{3}}x{{\cos }^{3}}x}dx$ 

$\Rightarrow I=\int{{{\sin }^{2}}x\sin x{{\cos }^{3}}x}dx$

$\Rightarrow I=\int{{{\cos }^{3}}x\left( 1-{{\cos }^{2}}x \right)\sin x}dx$

Let $\cos x=t$ 

$\therefore -\sin dx=dt$ 

Integration becomes

$\Rightarrow I=-\int{{{t}^{3}}\left( 1-{{t}^{2}} \right)}dt$

$\Rightarrow I=-\int{\left( {{t}^{3}}-{{t}^{5}} \right)}dt$

$\Rightarrow I=-\left[ \frac{{{t}^{4}}}{4}-\frac{{{t}^{6}}}{6} \right]+C$

Substitute $\cos x=t$,

$\Rightarrow I=-\left[ \frac{{{\cos }^{4}}x}{4}-\frac{{{\cos }^{6}}x}{6} \right]+C$

$\therefore \int{{{\sin }^{3}}x{{\cos }^{3}}x}dx=\frac{{{\cos }^{6}}x}{6}-\frac{{{\cos }^{4}}x}{4}+C$

Ans: Given expression $\sin x\sin 2x\sin 3x$.

Using the identity $\sin A\sin B=\frac{1}{2}\cos \left( A-B \right)-\cos \left( A+B \right)$, given expression can be written as

$\Rightarrow \sin x\sin 2x\sin 3x=\sin x.\frac{1}{2}\cos \left( 2x-3x \right)-\cos \left( 2x+3x \right)$

Integration of given expression is

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\frac{1}{2}\int{\left( \sin x\cos \left( -x \right)-\sin x\cos \left( 5x \right) \right)dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\frac{1}{2}\int{\left( \sin x\cos x-\sin x\cos 5x \right)dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\frac{1}{2}\int{\frac{\sin 2x}{2}dx-\frac{1}{2}\int{\left( \sin x\cos 5x \right)}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\frac{1}{4}\left( \frac{-\cos 2x}{2} \right)-\frac{1}{2}\int{\left\{ \frac{1}{2}\left( \sin \left( x+5x \right)+\sin \left( x-5x \right) \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\frac{-\cos 2x}{8}-\frac{1}{4}\int{\left\{ \left( \sin 6x+\sin \left( -4x \right) \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\frac{-\cos 2x}{8}-\frac{1}{4}\int{\left\{ \left( \sin 6x+\sin 4x \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\frac{-\cos 2x}{8}-\frac{1}{8}\left[ \frac{-\cos 6x}{3}+\frac{\cos 4x}{4} \right]+C$

$\therefore \int{\sin x\sin 2x\sin 3x}dx=\frac{1}{8}\left[ \frac{\cos 6x}{3}-\frac{\cos 4x}{4}-\cos 2x \right]+C$

Ans: Given expression $\sin 4x\sin 8x$.

Using the identity $\sin A\sin B=\frac{1}{2}\cos \left( A-B \right)-\cos \left( A+B \right)$, given expression can be written as

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

Integration of given expression is

$\Rightarrow \int{\sin 4x\sin 8x}dx=\frac{1}{2}\int{\left( \cos \left( -4x \right)-\cos \left( 12x \right) \right)dx}$

$\Rightarrow \int{\sin 4x\sin 8x}dx=\frac{1}{2}\int{\left( \cos 4x-\cos 12x \right)dx}$

$\therefore \int{\sin 4x\sin 8x}dx=\frac{1}{2}\left[ \frac{\sin 4x}{4}-\frac{\sin 12x}{12} \right]+C$

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

Using the identities $2{{\sin }^{2}}\frac{x}{2}=1-\cos x$ and $\cos x=2{{\cos }^{2}}\frac{x}{2}-1$ given expression can be written as 

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

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

Integration of given expression is 

$\Rightarrow \int{\frac{1-\cos x}{1+\cos x}dx}=\int{\left[ {{\tan }^{2}}\frac{x}{2} \right]dx}$

$\Rightarrow \int{\frac{1-\cos x}{1+\cos x}dx}=\int{\left[ {{\sec }^{2}}\frac{x}{2}-1 \right]dx}$

$\Rightarrow \int{\frac{1-\cos x}{1+\cos x}dx}=\left[ \frac{\tan \frac{x}{2}}{\frac{1}{2}}-x \right]+C$

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

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

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

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

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

$\Rightarrow \frac{\cos x}{1+\cos x}=\frac{1}{2}\left[ 1-{{\tan }^{2}}\frac{x}{2} \right]$

Integration of given expression is 

$\Rightarrow \int{\frac{\cos x}{1+\cos x}dx}=\frac{1}{2}\int{\left[ 1-{{\tan }^{2}}\frac{x}{2} \right]dx}$

$\Rightarrow \int{\frac{\cos x}{1+\cos x}dx}=\frac{1}{2}\int{\left[ 1-{{\sec }^{2}}\frac{x}{2}+1 \right]dx}$

$\Rightarrow \int{\frac{\cos x}{1+\cos x}dx}=\frac{1}{2}\int{\left[ 2-{{\sec }^{2}}\frac{x}{2} \right]dx}$

$\Rightarrow \int{\frac{\cos x}{1+\cos x}dx}=\frac{1}{2}\left[ 2x-\frac{\tan \frac{x}{2}}{\frac{1}{2}} \right]+C$

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

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

Given expression can be written as ${{\sin }^{4}}x={{\sin }^{2}}x{{\sin }^{2}}x$

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

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

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

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

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

$\Rightarrow {{\sin }^{4}}x=\frac{1}{4}\left( \frac{3}{2}+\frac{1}{2}\cos 4x-2\cos 2x \right)$

Integration of given expression is

$\Rightarrow \int{{{\sin }^{4}}x}dx=\frac{1}{4}\int{\left( \frac{3}{2}+\frac{1}{2}\cos 4x-2\cos 2x \right)}dx$

$\Rightarrow \int{{{\sin }^{4}}x}dx=\frac{1}{4}\left[ \frac{3}{2}x+\frac{\sin 4x}{8}-\sin 2x \right]+C$

$\therefore \int{{{\sin }^{4}}x}dx=\frac{3x}{8}+\frac{\sin 4x}{32}-\frac{1}{4}\sin 2x+C$