Ex 7.3 integration ncert maths solution class 12

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



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





=-\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}(\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


=\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


=\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


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

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

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

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

Question 23:\int{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx} is equal to

(A) \tan x+\cot x+C

(B) \tan x+cosecx+C

(C) -\tan x+\cot x+C

(D) \tan x+\sec x+C

Solution: option A is the correct answer.

Let I=\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}

=\int{{{\sec }^{2}}xdx}-\int{cose{{c}^{2}}xdx}

=\tan x+\cot x+C

Therefore, option (A )is the correct answer.

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

(A) -\cot \left( e{{x}^{x}} \right)+C

(B) \tan \left( x{{e}^{x}} \right)+C

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

(D) \cot \left( {{e}^{x}} \right)+C

Solution: option A is the correct answer.

Let I= \int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}.

Let {{e}^{x}}x=t

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

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

\Rightarrow dx =\frac{1}{e^x(x+1)}dt

Integration of given expression is

=\int{\frac{dt}{{{\cos }^{2}}t}}

=\int{{{\sec }^{2}}t}dt

=\tan t+C

=\tan \left( {{e}^{x}}x \right)+C

Therefore, option (B) is the correct answer.

Ex 7.2 integration ncert solution maths class 12

integration multiple choice question

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