NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.3 - 2025-26

Exercise 7.3

1. Solve the following: ${{\sin }^{2}}\left( 2x+5 \right)$.

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

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

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

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

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

2. Solve the following: $\sin 3x\cos 4x$.

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

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

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

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

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

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

3. Solve the following: $\cos 2x\cos 4x\cos 6x$.

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$

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

4. Solve the following: ${{\sin }^{3}}\left( 2x+1 \right)$.

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

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

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

5. Solve the following: ${{\sin }^{3}}x{{\cos }^{3}}x$.

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 $\cos x=t$,

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

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

6. Solve the following: $\sin x\sin 2x\sin 3x$.  

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

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

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

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

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

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

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

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

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

7. Solve the following: $\sin 4x\sin 8x$. 

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

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

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

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

8. Solve the following: $\frac{1-\cos x}{1+\cos x}$. 

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 

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

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

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

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

9. Solve the following: $\frac{\cos x}{1+\cos x}$. 

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 

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

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

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

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

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

10. Solve the following: ${{\sin }^{4}}x$. 

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

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

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

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

11. Solve the following: ${{\cos }^{4}}2x$.

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 

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

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

12. Solve the following: $\frac{{{\sin }^{2}}x}{1+\cos x}$.

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

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

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

13. Solve the following: $\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$. 

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

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

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

14. Solve the following: $\frac{\cos x-\sin x}{1+\sin 2x}$. 

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

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

15. Solve the following: ${{\tan }^{3}}2x\sec 2x$.

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 $\sec 2x=t$,

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

16. Solve the following: ${{\tan }^{4}}x$. 

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 $\tan x=t$,

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

17. Solve the following: $\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$.

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 

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

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

18. Solve the following: $\frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}$.

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 

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

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

19. Solve the following: $\frac{1}{\sin x{{\cos }^{3}}x}$.

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 $\tan x=t$,

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

20. Solve the following: $\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}$. 

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 $1+\sin 2x=t$

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

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

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

21. Solve the following: ${{\sin }^{-1}}\left( \cos x \right)$. 

Ans: Given expression ${{\sin }^{-1}}\left( \cos x \right)$.

Let $\cos x=t$ 

$\therefore \sin x=\sqrt{1-{{t}^{2}}}$ 

$\Rightarrow -\sin xdx=dt$

$\Rightarrow dx=-\frac{dt}{\sin x}$

$\Rightarrow dx=-\frac{dt}{\sqrt{1-{{t}^{2}}}}$

Integration of given expression is

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

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

Let ${{\sin }^{-1}}t=u$ 

$\Rightarrow \frac{1}{\sqrt{1-{{t}^{2}}}}dt=du$ 

Integration becomes

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\int{4du}$

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

Substitute ${{\sin }^{-1}}t=u$

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

Substitute $\cos x=t$

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

We know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2}$ 

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

Substitute in eq. (1), we get

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

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

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

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

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

22. Solve the following: $\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}$. 

Ans: Given expression $\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}$.

Given expression can be written as

$\Rightarrow \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\frac{1}{\sin \left( a-b \right)}\left[ \frac{\sin \left( a-b \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\frac{1}{\sin \left( a-b \right)}\left[ \frac{\sin \left[ \left( x-b \right)-\left( x-a \right) \right]}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\frac{1}{\sin \left( a-b \right)}\left[ \frac{\sin \left( x-b \right)\cos \left( x-a \right)-\cos \left( x-b \right)\sin \left( x-a \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\frac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]$

Integration of given expression is

$\Rightarrow \int{\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\frac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]dx}$

$\Rightarrow \int{\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\frac{1}{\sin \left( a-b \right)}\left[ -\log \left| \cos \left( x-b \right) \right|+\log \cos \left( x-a \right) \right]}$

$\Rightarrow \int{\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\frac{1}{\sin \left( a-b \right)}\left[ \log \left| \frac{\cos \left( x-a \right)}{\cos \left( x-b \right)} \right| \right]}+C$

23. Solve the following: $\int{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$ is equal to

1. $\tan x+\cot x+C$ 

2. $\tan x+cosecx+C$ 

3. $-\tan x+\cot x+C$ 

4. $\tan x+\sec x+C$ 

Ans: Given expression $\int{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.

Given expression can be written as

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

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

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

Therefore, option A is the correct answer.

24. Solve the following: $\int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}$ equals

1. $-\cot \left( e{{x}^{x}} \right)+C$ 

2. $\tan \left( x{{e}^{x}} \right)+C$ 

3. $\tan \left( {{e}^{x}} \right)+C$

4. $\cot \left( {{e}^{x}} \right)+C$

Ans: Given expression $\int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}$.

Let ${{e}^{x}}x=t$ 

$\therefore \left( {{e}^{x}}.x+{{e}^{x}}.1 \right)dx=dt$ 

$\Rightarrow {{e}^{x}}\left( x+1 \right)dx=dt$

Integration of given expression is

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

\[\Rightarrow \int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{{{\sec }^{2}}t}dt\]

\[\Rightarrow \int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan t+C\]

Substitute ${{e}^{x}}x=t$,

\[\therefore \int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan \left( {{e}^{x}}x \right)+C\]

Therefore, option B is the correct answer.

Conclusion

Chapter 7 of Class 12 Mathematics, focusing on Integrals, is crucial for understanding calculus. Exercise 7.3 Class 12 Maths delves into techniques of integration, which are essential for solving complex problems in higher mathematics. Important topics include integration by substitution, integration by parts, and partial fractions. Mastery of these methods is vital, as they are frequently tested in exams. Students should focus on practicing various types of problems, understanding the underlying principles, and applying the correct techniques in class 12 maths ex 7.3.

Class 12 Maths Chapter 7: Exercises Breakdown

CBSE Class 12 Maths Chapter 7 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

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