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NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.3

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NCERT Solutions for Class 12 Maths Chapter 7 (Ex 7.3)

Our team of experts are focused on curating top-notch study materials for Chapter 7.3 Maths Class 12 that covers Integrals. The NCERT Solutions Class 12 Maths Chapter 7 Exercise 7.3 by Vedantu is focused on covering the different aspects of Integrals in a strategic manner, allowing students to maximise their learning opportunity about the topic. The Class 12 Maths Ch 7 Ex 7.3 has proven to be an essential chapter with repeated questions asked over the years and practising from this chapter gives you a better chance of scoring due to its easy mathematical solutions. Further, Ex 7.3 Maths Class 12 carries a significant weightage as it can be found over both the sections of the question paper. Therefore, the downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.3 will help you with the desired push for your examinations.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 7 - Integrals

Exercise:

Exercise - 7.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes


Download Free NCERT Solutions for Class 12 Math Chapter 7 Integers Ex 7.3 PDF. These solutions are extremely helpful while doing your homework and preparing for board examinations. The solutions are explained in a stepwise manner by our subject experts according to the guidelines issued by the CBSE Board. Download the solution for a comprehensive understanding of the chapter.

 

Topics Covered

  • Integration Using Trigonometric Identities

Some integrals involving trigonometric functions can be solved with the help of trigonometric identities. These allow the integrand to be written in an alternative form that may be more compliant with the integration.


The trigonometric identities used to solve the questions in the exercise are discussed below:

  • 2 sin A cos B = sin(A + B) + sin(A − B) 

  • 2 cos A cos B = cos(A − B) + cos(A + B) 

  • 2 sin A sin B = cos(A − B) − cos(A + B)

  •  Sin2 A + cos2 A = 1 

  • cos2A  = cos2 A − sin2

= 2 cos2 A − 1 

= 1 − 2 sin2

  • sin 2A = 2 sin A cos A

  •  1 + tan2 A = sec2


Some commonly needed trigonometric id

Access NCERT Solutions for Class 12 Mathematics Chapter 7- Integrals

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.


NCERT Solutions for Class 12 Maths PDF Download

 

NCERT Solution Class 12 Maths of Chapter 7 All Exercises

Chapter 7 - Integrals Exercises in PDF Format

Exercise 7.1

22 Questions & Solutions (21 Short Answers, 1 MCQs)

Exercise 7.2

39 Questions & Solutions (37 Short Answers, 2 MCQs)

Exercise 7.3

24 Questions & Solutions (22 Short Answers, 2 MCQs)

Exercise 7.4

25 Questions & Solutions (23 Short Answers, 2 MCQs)

Exercise 7.5

23 Questions & Solutions (21 Short Answers, 2 MCQs)

Exercise 7.6

24 Questions & Solutions (22 Short Answers, 2 MCQs)

Exercise 7.7

11 Questions & Solutions (9 Short Answers, 2 MCQs)

Exercise 7.8

6 Questions & Solutions (6 Short Answers)

Exercise 7.9

22 Questions & Solutions (20 Short Answers, 2 MCQs)

Exercise 7.10

10 Questions & Solutions (8 Short Answers, 2 MCQs)

Exercise 7.11

21 Questions & Solutions (19 Short Answers, 2 MCQs)

CBSE Class 12 Maths NCERT Solutions For Chapter 7 Exercise 7.3

Significance Of NCERT Solutions For Class 12 Maths Exercise 7.3, Integrals:

The Class 12 Maths Chapter 7 Exercise 7.3 Integrals, revolves around the application of using numbers that represent different functions, only to be placed in such a manner that they help us find and describe the relevant outputs behind different dynamics like area, displacement, volume, and other such concepts that revolve around infinitesimal data. Integration is a major wing that revolves around the concepts of calculus and can also be used in operations like inverse operation and differentiation.

Practical Problems For Exercise 7.3 Maths Class 12

Question 1:

Find the distance between the suggested pairs of given points:

i) (2,3), (4,1)

ii) (-5,7), (-1,3)

iii) (a,b), (-a,-b)

Solution 1: 

(i) Let the points be M (2,3) and N (4,1)

Therefore, 

x1= 2 

y1= 3

x2= 4

y2= 1

We know that the distance between the two points is given by the Distance Formula = \[\sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2}}\] ……….(1)

Hence, the distance between the two points M (2,3) and N (4,1) is shown by:

d = \[\sqrt{(2-4)^{2}+(3-1)^{2}}\]

 = \[\sqrt{(-2)^{2}+(2)^{2}}\]

= \[\sqrt{4+4}\] = \[\sqrt{8}\] 2\[\sqrt{2}\]

(ii) Distance between (-5,7) and (-1,3) can be presented by 

d = \[\sqrt{(-5-(-1))^{2}+(7-3)^{2}}\]

 = \[\sqrt{(-4)^{2}+(4)^{2}}\]

 = \[\sqrt{16 + 16}\]

= \[\sqrt{32}\]

 = 4\[\sqrt{2}\]

(iii) Distance between (a,b) (-a, -b) is given by:

d = \[\sqrt{(a-(-a))^{2}+(b-(-b))^{2}}\]

 = \[\sqrt{(2a)^{2}+(2b)^{2}}\]

= \[\sqrt{4a^{2} + 4b^{2}}\]

= 2 \[\sqrt{a^{2}+b^{2}}\]

Question 2: 

Find the value of y, implying that the points are P (2, -3) and Q (10y, y) is 10 units.

Solution 2:

Given, 

Distance between points A(2,-3) and B(10,y) is 10.

We know that the distance between the two points given by the Distance Formula,

\[\sqrt{(-4)^{2} + (4)^{2}}\] …..(1)

By Substituting the values of points A (2, -3) and B (10, y) in Equation (1)

Therefore, \[\sqrt{(2-10)^{2}+(-3-y)^{2}}\] = 10 

= \[\sqrt{(-8)^{2}+(3+y)^{2}}\] = 100

= 64 + (y+3)2 = 100

= (y+3)2 = 36

= y=3 = 6 or y+3= -6

Did You Know? 

If any given solution is derivable at a given function of an, i.e., its derivative f’ is existing in a given point over each of the suggested points of I, then the suggested functions are called antiderivatives. This is explained further in Exercise 7.3 Class 12 Maths Ncert Solutions.

FAQs on NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.3

1. What is the Concept of Integral Calculus in Class 12 Maths Chapter 7 Exercise 7.3?

The concept around the development of integral calculus tackles different problems that revolve around mathematics. This concept saves the efforts of users who find it difficult to tackle the different solutions of the given type:

  • Finding a function when they are provided with a derivative

  • Problems in finding areas when it’s bounded by graphs and bound by certain conditions.

These two problems are the bases that lead to providing forms that allow integrals, e.g., indefinite and definite integrals when put together develop Integral Calculus. This part is widely covered and explained in Exercise 7.3 Class 12 Maths.

2. Is Studying from Exercise 7.3 Class 12 NCERT Solutions helpful?

Studying from the NCERT solutions that are based out of Exercise 7.3 Maths Class 12, helps students in ways more than one. To begin with, the solutions are completely focused on explaining the toughest and important questions of the integrals from its roots and initiate an easier learning experience of the Ex 7.3 Class 12 Maths. The solutions have been demonstrated by a dedicated pool of teachers and are focused on helping students grasp the different concepts that are included in the intervals section. The Ex 7.3 Class 12 Maths Solutions provided are mainly developed for the students appearing in Class 12 Board exams for the current year as per the CBSE curriculum.

3. Which are the most important topics covered in Chapter 7 Exercise 7.3 of Class 12 Maths?

Chapter 7 of Class 12 Maths NCERT is known as “Integrals”. This is a crucial chapter in the syllabus of Class 12 Maths and holds a significant weightage in terms of marks. While the chapter covers various important topics, the most important ones that students shall prepare well and practise regularly include Exercise 7.3, which includes Integration by Substitution, Integration using Partial, and Fractions Integration by Parts.

4. How many exercises and questions are there in Chapter 7 of Class 12 Maths apart from Exercise 7.3 in Chapter 7?

Chapter 7-Integrals includes a total of 11 exercises. The following are the number of questions in each exercise:

  • Exercise 7.1 - 22 Questions

  • Exercise 7.2 - 39 Questions

  • Exercise 7.3 - 24 Questions

  • Exercise 7.4 - 25 Questions

  • Exercise 7.5 - 23 Questions

  • Exercise 7.6 - 24 Questions

  • Exercise 7.7 - 11 Questions

  • Exercise 7.8 - Six Questions

  • Exercise 7.9 - 22 Questions

  • Exercise 7.10 - 10 Questions

  • Exercise 7.11 - 21 Questions

5. Do I need to practise all the questions provided in Exercise 7.3 of Chapter 7  of Class 12 Maths NCERT Solutions?

Questions in your Maths exams can be picked up from anywhere from your Class 12 Maths NCERT. It is not possible to be certain about which questions may or may not appear in your exam. Hence, it is strongly advised that students practice all the questions provided in NCERT Solutions for Exercise 7.3 of Chapter 7 Integrals of Class 12 Maths. Practising all the questions will reduce the possibility of losing any marks in the exam.

6. Where can I find NCERT Solutions for Exercise 7.3 of Chapter 7 Integrals of Class 12 Maths?

Referring to NCERT Solutions can help students develop a proper understanding of the questions while preparing for their Class 12 Maths exam. At Vedantu, subject experts provide NCERT Solutions that depict each step of the answer for students to be able to grasp the methods and calculations done in each answer. You can find NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.3 on Vedantu’s website. These solutions are available at free of cost on Vedantu website(vedantu.com) and mobile app as well.