EXERCISE 7.3(Integration)

Find the integrals of the functions in Exercises 1 to 22:(Ex 7.3 integration ncert maths solution class 12)

Question 1: $\sin^2(2x+5)$

Solution: Let $I = \int \sin^2(2x+5)dx$

Using formula

$2\sin^2x=1+\cos 2x$

$\Rightarrow 2\sin^2(2x+5)=1+\cos2(2x+5)$

$I = \frac{1}{2}\int (1+\cos2(2x+5))dx$

$=\frac{1}{2}\int (1+\cos(4x+10))dx$

$=\frac{1}{2}(x+\frac{\sin(4x+10)}{4})+C$

$=\frac{x}{2}+\frac{\sin(4x+10)}{8}+C$

Question 2: $\sin 3x\cos 4x$

Question: Let $I =\int \sin 3x\cos 4x dx$

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

$I=\int{\frac{1}{2}\left\{ \sin 7x+\sin \left( -x \right) \right\}dx}$

$=\frac{1}{2}\int{\left\{ \sin 7x+\sin x \right\}dx}$

$=\frac{1}{2}\int{\sin 7xdx+\frac{1}{2}\int{\sin x}dx}$

$=\frac{1}{2}\left( \frac{-\cos 7x}{7} \right)-\frac{1}{2}\left( -\cos x \right)+C$

$=\frac{-\cos 7x}{14}+\frac{\cos x}{2}+C$

Question 3: $\cos2x\cos4x\cos6x.$

Solution: Let $I=\int \cos2x\cos4x\cos6x\ dx.$

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

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

$I= \int{\cos 2x\left( \cos 4x\cos 6x \right)}$

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

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}$

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

$=\frac{1}{2}\int{ \left\{ \cos 2x\cos 10x+{{\cos }^{2}}2x \right\} }dx$

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

$=\frac{1}{4}\int{\left[ \cos 12x+\cos 8x+1+\cos 4x \right]}dx$

$I=\frac{1}{4}\left[ \frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x \right]+C$

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

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

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

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

Let $\cos \left( 2x+1 \right)=t$

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

$\Rightarrow dx = \frac{1}{-2\sin(2x+1)}dt$

$\Rightarrow I=\int{[ 1-t^2 ]\sin \left( 2x+1 \right)\frac{1}{-2\sin(2x+1)}dt}$

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

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

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

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

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

Question 5: ${{\sin }^{3}}x{{\cos }^{3}}x.$

Solution: 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 $\cos x=t$

$\Rightarrow -\sin x dx=dt$

$\Rightarrow dx = -\frac{1}{\sin x}dt$

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

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

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

Substitute $\cos x=t$ ,

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

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

Question 6: $\sin x\sin 2x\sin 3x.$

Solution: Let $I=\int \sin x\sin 2x\sin 3x\ dx$

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

$I=\int \sin x.\frac{1}{2}[\cos \left( 2x-3x \right)-\cos \left( 2x+3x \right)]dx$

$I=\frac{1}{2}\int{\left[ \sin x\cos \left( -x \right)-\sin x\cos \left( 5x \right) \right]dx}$

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

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

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

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

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

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

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

Question 7: $\sin 4x\sin 8x.$

Solution: Let $I=\int \sin 4x\sin 8x\ dx$ .

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

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

Integration of given expression is

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

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

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

Question 8: $\frac{1-\cos x}{1+\cos x}.$

Solution: Let $I=\int \frac{1-\cos x}{1+\cos x}\ dx$ .

Using the identities

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

$\Rightarrow I= \int \frac{1-\cos x}{1+\cos x}\ dx$

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

$= \int \frac{1-\cos x}{1+\cos x}\ dx$

$=\int {{\tan }^{2}}\frac{x}{2}\ dx$

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

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

$=2\tan \frac{x}{2}-x+C$

Question 9: $\frac{\cos x}{1+\cos x}$ .

Solution: Let $I = \int \frac{\cos x}{1+\cos x}dx$ .

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

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

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

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

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

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

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

$=x-\tan \frac{x}{2}+C$

Question 10: ${{\sin }^{4}}x$ .

Solution: Let $I =\int {{\sin }^{4}}x\ dx$ .

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

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

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

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

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

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

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

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

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

Question 11: ${{\cos }^{4}}2x.$

Solution: Let $I = \int {{\cos }^{4}}2xdx$

$I=\int {{\left( {{\cos }^{2}}2x \right)}^{2}}\ dx$

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

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

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

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

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

$=\frac{3}{8}x+\frac{\sin 8x}{64}+\frac{\sin 4x}{8}+C$

Question 12: $\frac{{{\sin }^{2}}x}{1+\cos x}$

Solution: Let I= $\int \frac{{{\sin }^{2}}x}{1+\cos x}dx$ .

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

$=\int \frac{(1-\cos x)(1+\cos x)}{1+\cos x}dx$

$=\int (1-\cos x)dx$

$=x-\sin x+C$

Question 13: $\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$ .

Solution: Let $I = \int \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx$ .

$=\int\frac{(2\cos^2x-1)-(2\cos^2\alpha-1)}{\cos x-\cos \alpha}dx$

$=\int \frac{2\cos^2x-1-2\cos^\alpha+1}{\cos x-\cos\alpha}dx$

$=\int \frac{2(\cos^2x-\cos^2\alpha)}{\cos x-\cos\alpha}dx$

$=\int \frac{2(\cos x+\cos\alpha)(\cos x-\cos\alpha)}{\cos x-\cos \alpha}dx$

$=2\int(\cos x+\cos \alpha)dx$

$=2(\sin x+(\cos\alpha) x)+C$

$=2\sin x+2x\cos\alpha+C$

Question 14: $\frac{\cos x-\sin x}{1+\sin 2x}$

Solution: Let $I=\int \frac{\cos x-\sin x}{1+\sin 2x}dx$ .

We know that

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

$I=\int \frac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+\sin 2x}dx$ .

We can apply the identity

$\sin 2x=2\sin x\cos x$ ,

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

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

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

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

$\Rightarrow dx=\frac{1}{\cos x-\sin x}dx$

$I=\int{\frac{dt}{{{{t}^{2}}}}}$

$=\int{{{{t}^{-2}}dt}}$

$=-{{t}^{-1}}+C$

$=-\frac{1}{t}+C$

$=-\frac{1}{\sin x+\cos x}+C$

Question 15: ${{\tan }^{3}}2x\sec 2x$ .

Solution: Let $I=\int {{\tan }^{3}}2x\sec 2x\ dx$ .

$=\int {{\tan }^{2}}2x\tan 2x\sec 2x\ dx$

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

Let $\sec 2x=t$

$\Rightarrow 2\sec 2x\tan 2xdx=dt$

$\Rightarrow dx=\frac{1}{2\sec 2x\tan 2x}dt$

$I=\frac{1}{2}\int(\sec^2 2x-1)\tan 2x \sec 2xdx$

$=\frac{1}{2}\int(t^2-1)dt$

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

$=\frac{{{\left( \sec 2x \right)}^{3}}}{6}-\frac{\sec 2x}{2}+C$

Question 16: ${{\tan }^{4}}x$ .

Solution: Let $I=\int{{\tan }^{4}}xdx$ .

$=\int {{\tan }^{2}}x{{\tan }^{2}}xdx$

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

$=\int {{\sec }^{2}}x{{\tan }^{2}}x-{{\tan }^{2}}x\ dx$

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

$=\int {{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1\ dx$

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

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

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

Let $\tan x=t$

$\Rightarrow {{\sec }^{2}}xdx=dt$

$\Rightarrow dx = \frac{1}{\sec^2x}dt$

$I=\int{{{t}^{2}}}dt-\tan x+x+C$

$=\frac{{{t}^{3}}}{3}-\tan x+x+C$

$=\frac{1}{3}{{\tan }^{3}}x-\tan x+x+C$

Question 17: $\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$ .

Solution:Let $I = \int \frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx$ .

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

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

$=\int (\tan x\sec x+\cot xcosecx)\ dx$

$=\int{\tan x\sec xdx}+\int{\cot x\operatorname{cosecx}dx}$

$=\sec x-cosecx+C$

Question18: $\frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}$ .

Solution: Let $I = \int \frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}dx$ .

By applying the identity

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

$I=\int\frac{\cos 2x+1-\cos 2x}{{{\cos }^{2}}x}dx$

$=\int \frac{1}{{{\cos }^{2}}x}dx$

$=\int {{\sec }^{2}}xdx$

$=\tan x+C$

Question 19: $\frac{1}{\sin x{{\cos }^{3}}x}$ .

Solution: Let $I=\int \frac{1}{\sin x{{\cos }^{3}}x}dx$ .

We can apply the identity

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

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

$=\int (\frac{{{\sin }^{2}}x}{\sin x{{\cos }^{3}}x}+\frac{{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x})dx$

$=\int(\frac{\sin x}{{{\cos }^{3}}x}+\frac{1}{\sin x\cos x})dx$

$=\int(\tan x{{\sec }^{2}}x+\frac{{{\cos }^{2}}x}{\frac{\sin x\cos x}{{{\cos }^{2}}x}})dx$

$=\int (\tan x{{\sec }^{2}}x+\frac{{{\sec }^{2}}x}{\tan x})dx$

Let $\tan x=t$

$\Rightarrow {{\sec }^{2}}xdx=dt$

$\Rightarrow dx= \frac{1}{\sec^2 x}dt$

$=\int{t}dt+\int{\frac{1}{\operatorname{t}}}dt$

$=\frac{{{t}^{2}}}{2}+\log \left| t \right|+C$

$=\frac{1}{2}{{\tan }^{2}}x+\log \left| \tan x \right|+C$

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

Solution:Let $I=\int\frac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx$ .

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

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

$=\int \frac{\cos x-\sin x}{\cos x+\sin x}dx$

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

$\Rightarrow (-\sin x+\cos x)dx=dt$

$\Rightarrow dx = \frac{1}{\cos x-\sin x}dt$

$I=\int{\frac{1}{t}dt}$

$=\log \left| t \right|+C$

$=\log \left| \left( \cos x+\sin x \right) \right|+C$

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

Solution: Let $I= \int {{\sin }^{-1}}\left( \cos x \right)dx$ .

$=\int \sin^{-1}\sin(\frac{\pi}{2}-x)dx$

$=\int (\frac{\pi}{2}-x)dx$

$=\frac{\pi}{2}x-\frac{x^2}{2}+C$

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

Solution: Let $I= \int \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx$ .

Given expression can be written as

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

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

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