Class 12th Chapter -7 Integrals | NCERT Maths Solution | NCERT Solution

It is important that you have a strong foundation in Maths Solutions. This can help you immensely in exams. Because there is no paper pattern set for the board exams. Every year there are different twists and turns in questions asked in board exams. While preparing for maths it becomes to go through every bit of detail. Thus, NCERT Solutions Class 12 Maths can help you lay a strong foundation. There solutions of each and every question given in the NCERT 12 Textbook in this article. There are crucial chapters like integration, algebra, and calculus which you can solve easily using these Class 12 Maths solutions. Using these NCERT Maths Class 12 Solutions can give you an edge over the other students.

Table of Contents

NCERT Solutions for Class 12 Maths Chapter : 7 Integrals

Class 12 Maths Chapter 7 Integrals Ex 7.1

Find an antiderivative (or integral) of the following by the method of inspection:

Ex 7.1 Class 12 Maths Question 1.
sin 2x
Solution:
$\int { sin2x\quad dx=-\frac { cos2x }{ 2 } +C }$

Ex 7.1 Class 12 Maths Question 2.
cos 3x
Solution:
$\int { cos3x\quad dx=\frac { sin3x }{ 3 } +C }$

Ex 7.1 Class 12 Maths Question 3.
${ e }^{ 2x }$
Solution:
$\int { { e }^{ 2x }dx=\frac { { e }^{ 2x } }{ 2 } +C }$

Ex 7.1 Class 12 Maths Question 4.
(ax + c)²
Solution:
$\int { { (ax+b) }^{ 2 }dx=\frac { { (ax+b) }^{ 3 } }{ 3a } } +C$

Ex 7.1 Class 12 Maths Question 5.
${ sin\quad 2x-4e }^{ 3x }$
Solution:
$\int { \left( { sin2x-4e }^{ 3x } \right) dx=-\frac { cos2x }{ 2 } -\frac { { 4e }^{ 3x } }{ 3 } +C }$

Find the following integrals in Exercises 6 to 20 :

Ex 7.1 Class 12 Maths Question 6.
$\int { \left( { 4e }^{ 3x }+1 \right) dx }$
Solution:
$=\int { { 4e }^{ 3x }dx+\int { dx=\frac { 4 }{ 3 } { e }^{ 3x }+x+c } }$

Ex 7.1 Class 12 Maths Question 7.
$\int { { x }^{ 2 }\left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx }$
Solution:
$=\int { { x }^{ 2 }\left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx } =\frac { { x }^{ 3 } }{ 3 } -x+C$

Ex 7.1 Class 12 Maths Question 8.
$\int { { (ax }^{ 2 }+bx+c)dx }$
Solution:
$=\frac { { ax }^{ 3 } }{ 3 } +\frac { { bx }^{ 2 } }{ 2 } +cx+d$

Ex 7.1 Class 12 Maths Question 9.
$\int { \left( { 2x }^{ 2 }+{ e }^{ x } \right) dx }$
Solution:
$=\frac { { 2x }^{ 3 } }{ 3 } +{ e }^{ x }+c$

Ex 7.1 Class 12 Maths Question 10.
$\int { { \left[ \sqrt { x } -\frac { 1 }{ \sqrt { x } } \right] }^{ 2 }dx }$
Solution:
$=\frac { { x }^{ 2 } }{ 2 } +logx-2x+C$

Ex 7.1 Class 12 Maths Question 11.
$\int { \frac { { x }^{ 3 }+{ 5x }^{ 2 }-4 }{ { x }^{ 2 } } dx }$
Solution:
$\int { \left( \frac { { x }^{ 3 } }{ { x }^{ 2 } } +\frac { { 5x }^{ 2 } }{ { x }^{ 2 } } -\frac { 4 }{ { x }^{ 2 } } \right) }$
$=\int { xdx+5\int { 1dx-4 } \int { { x }^{ 2 }dx } }$
$=\frac { { x }^{ 2 } }{ 2 } +5x+\frac { 4 }{ x } +c$

Ex 7.1 Class 12 Maths Question 12.
$\int { \frac { { x }^{ 3 }+3x+4 }{ \sqrt { x } } dx }$
Solution:
$=\int { \left( { x }^{ \frac { 5 }{ 2 } }+{ 3x }^{ \frac { 1 }{ 2 } }+4{ x }^{ -\frac { 1 }{ 2 } } \right) } dx$
$=\frac { 2 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+{ 2x }^{ \frac { 3 }{ 2 } }+8\sqrt { x } +c$

Ex 7.1 Class 12 Maths Question 13.
$\int { \frac { { x }^{ 3 }-{ x }^{ 2 }+x-1 }{ x-1 } dx }$
Solution:
$=\int { \frac { { x }^{ 2 }(x-1)+(x-1) }{ x-1 } dx }$
$=\int { \left( { x }^{ 2 }+1 \right) dx } =\frac { { x }^{ 3 } }{ 3 } +x+c$

Ex 7.1 Class 12 Maths Question 14.
$\int { \left( 1-x \right) \sqrt { x } dx }$
Solution:
$=\int { { x }^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 3 }{ 2 } }dx\quad =\quad \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { 2 }{ 5 } { x }^{ \frac { 5 }{ 2 } } }$

Ex 7.1 Class 12 Maths Question 15.
$\int { \sqrt { x } \left( { 3x }^{ 2 }+2x+3 \right) dx }$
Solution:
$=\int { \left( { 3x }^{ \frac { 5 }{ 2 } }+{ 2 }^{ \frac { 3 }{ 2 } }+{ 3x }^{ \frac { 1 }{ 2 } } \right) dx }$
$=\frac { 6 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+\frac { 4 }{ 5 } { x }^{ \frac { 5 }{ 2 } }+\frac { 6 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+c$

Ex 7.1 Class 12 Maths Question 16.
$\int { (2x-3cosx+{ e }^{ x })dx }$
Solution:
$=\frac { { 2x }^{ 2 } }{ 2 } -3sinx+{ e }^{ x }+c$
$={ x }^{ 2 }-3sinx+{ e }^{ x }+c$

Ex 7.1 Class 12 Maths Question 17.
$\int { \left( { 2x }^{ 2 }-3sinx+5\sqrt { x } \right) dx }$
Solution:
$=\frac { { 2x }^{ 3 } }{ 3 } +3cosx+5\frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } +c$
$=\frac { 2 }{ 3 } { x }^{ 3 }+3cosx+\frac { 10 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+c$

Ex 7.1 Class 12 Maths Question 18.
$\int { secx(secx+tanx)dx }$
Solution:
$=\int { { (sec }^{ 2 }x+secxtanx)dx }$
= tanx + secx + c

Ex 7.1 Class 12 Maths Question 19.
$\int { \frac { { sec }^{ 2 }x }{ { cosec }^{ 2 }x } dx }$
Solution:
$=\int { \frac { 1 }{ { cos }^{ 2 }x } } { sin }^{ 2 }xdx$
$=\int { tan } ^{ 2 }xdx\quad =\int { { (sec }^{ 2 }x-1)dx\quad =tanx-x+c }$

Ex 7.1 Class 12 Maths Question 20.
$\int { \frac { 2-3sinx }{ { cos }^{ 2 }x } dx }$
Solution:
$=\int { \left( \frac { 2 }{ { cos }^{ 2 }x } -3\frac { sinx }{ { cos }^{ 2 }x } \right) dx }$
$=\int { ({ 2sec }^{ 2 }x-3secxtanx)dx }$
= 2tanx – 3secx + c

Choose the correct answer in Exercises 21 and 22.

Ex 7.1 Class 12 Maths Question 21.
The antiderivative $\left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right)$ equals
(a) $\frac { 1 }{ 3 } { x }^{ \frac { 1 }{ 3 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c$
(b) $\frac { 2 }{ 3 } { x }^{ \frac { 2 }{ 3 } }+{ \frac { 1 }{ 2 } x }^{ 2 }+c$
(c) $\frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c$
(d) $\frac { 3 }{ 2 } { x }^{ \frac { 3 }{ 2 } }+\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } }+c$
Solution:
(c) $\int { \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) dx }$
$=\int { \left( { x }^{ \frac { 1 }{ 2 } }+{ x }^{ \frac { 1 }{ 2 } } \right) dx }$
$=\frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c$

Ex 7.1 Class 12 Maths Question 22.
If $\frac { d }{ dx } f(x)={ 4x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } }$ such that f(2)=0 then f(x) is
(a) ${ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } -\frac { 129 }{ 8 }$
(b) ${ x }^{ 3 }+\frac { 1 }{ { x }^{ 4 } } +\frac { 129 }{ 8 }$
(c) ${ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } +\frac { 129 }{ 8 }$
(d) ${ x }^{ 3 }+\frac { 1 }{ { x }^{ 4 } } -\frac { 129 }{ 8 }$
Solution:
(a) $f(x)=\int { \left( { 4x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } } \right) dx }$
$={ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } +c$
$\therefore f(2)={ (2) }^{ 4 }+\frac { 1 }{ { (2) }^{ 3 } } +c=0=-\frac { 129 }{ 8 }$

Class 12 Maths Chapter 7 Integrals Ex 7.2

Integrate the functions in Exercises 1 to 37:

Ex 7.1 Class 12 Maths Question 1.
$\frac { 2x }{ 1+{ x }^{ 2 } }$
Solution:
Let 1+x² = t
⇒ 2xdx = dt
$\therefore \int { \frac { 2x }{ 1+{ x }^{ 2 } } dx\quad = } \int { \frac { dt }{ t } \quad =logt+C\quad =log(1+{ x }^{ 2 })+C }$

Ex 7.2 Class 12 Maths Question 2.
$\frac { { \left( logx \right) }^{ 2 } }{ x }$
Solution:
Let logx = t
⇒ $\frac { 1 }{ x }dx=dt$
$\therefore \int { \frac { { (logx) }^{ 2 } }{ x } dx } \quad =\int { { t }^{ 2 }dt } \quad =\frac { { t }^{ 3 } }{ 3 } +c\quad =\frac { 1 }{ 3 } { (logx) }^{ 3 }+c$

Ex 7.2 Class 12 Maths Question 3.
$\frac { 1 }{ x+xlogx }$
Solution:
Put 1+logx = t
∴ $\frac { 1 }{ x }dx=dt$
$\int { \frac { 1 }{ x(1+logx) } dx } =\int { \frac { 1 }{ t } dt } =log|t|+c$
= log|1+logx|+c

Ex 7.2 Class 12 Maths Question 4.
sinx sin(cosx)
Solution:
Put cosx = t, -sinx dx = dt
$\int { sinx\quad sin(cosx)dx } =-\int { sin(cosx) } (-sinx)dx$
$=-\int { sint\quad dt } \quad =cost+c\quad =cos(cosx)+c$

Ex 7.2 Class 12 Maths Question 5.
sin(ax+b) cos(ax+b)
Solution:
let sin(ax+b) = t
⇒ cos(ax+b)dx = dt
$\therefore \int { sin(ax+b)cos(ax+b)dx } =\frac { 1 }{ a } \int { t\quad dt }$
$=\frac { 1 }{ a } .\frac { { t }^{ 2 } }{ 2 } +c\quad =\frac { 1 }{ 2a } { sin }^{ 2 }(ax+b)+C$

Ex 7.2 Class 12 Maths Question 6.
$\sqrt { ax+b }$
Solution:
$\int { \sqrt { ax+b } dx } \quad =\frac { 2 }{ 3a } { (ax+b) }^{ \frac { 3 }{ 2 } }+C$

Ex 7.2 Class 12 Maths Question 7.
$x\sqrt { x+2 }$
Solution:
Let x+2 = t²
⇒ dx = 2t dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q7.1

Ex 7.2 Class 12 Maths Question 8.
$x\sqrt { 1+{ 2x }^{ 2 } }$
Solution:
let 1+2x² = t²
⇒ 4x dx = 2t dt
$x\quad dx=\frac { t }{ 2 } dt\int { x\sqrt { 1+{ 2x }^{ 2 } } dx }$
$=\frac { 1 }{ 2 } \int { { t }^{ 2 }dt } =\frac { { t }^{ 3 } }{ 6 } +c=\frac { 1 }{ 6 } { ({ 1+2x }^{ 2 }) }^{ \frac { 3 }{ 2 } }+c$

Ex 7.2 Class 12 Maths Question 9.
$(4x+2)\sqrt { { x }^{ 2 }+x+1 }$
Solution:
let x²+x+1 = t
⇒(2x+1)dx = dt
$\therefore \int { (4x+1)\sqrt { { x }^{ 2 }+x+1 } dx } =2\int { \sqrt { t } dt }$
$=\frac { { 2t }^{ \frac { 3 }{ 2 } } }{ ^{ \frac { 3 }{ 2 } } } +c\quad =\frac { 4 }{ 3 } { t }^{ \frac { 3 }{ 2 } }+c\quad =\frac { 4 }{ 3 } { ({ x }^{ 2 }+x+1) }^{ \frac { 3 }{ 2 } }+c$

Ex 7.2 Class 12 Maths Question 10.
$\frac { 1 }{ x-\sqrt { x } }$
Solution:
$\int { \frac { 1 }{ x-\sqrt { x } } dx } =\int { \frac { 1 }{ \sqrt { x } (\sqrt { x-1 } ) } dx } =I$
Let √x-1 = t
$\frac { 1 }{ 2 } { x }^{ -\frac { 1 }{ 2 } }dx=dt$
$I=2\int { \frac { dt }{ t } }$
= 2logt + c
= 2log(√x-1)+c

Ex 7.2 Class 12 Maths Question 11.
$\frac { x }{ \sqrt { x+4 } } ,x>0$
Solution:
let x+4 = t
⇒ dx = dt, x = t-4
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q11.1

Ex 7.2 Class 12 Maths Question 12.
${ { (x }^{ 3 }-1) }^{ \frac { 1 }{ 3 } }.{ x }^{ 5 }$
Solution:
$\int { { { (x }^{ 3 }-1) }^{ \frac { 1 }{ 3 } }.{ x }^{ 5 }.dx } \quad =\frac { 1 }{ 7 } { { (x }^{ 3 }-1) }^{ \frac { 7 }{ 3 } }+\frac { 1 }{ 4 } { { (x }^{ 3 }-1) }^{ \frac { 4 }{ 3 } }+c$

Ex 7.2 Class 12 Maths Question 13.
$\frac { { x }^{ 2 } }{ { { (2+3x }^{ 3 }) }^{ 3 } }$
Solution:
Let 2+3x³ = t
⇒ 9x² dx = dt
$\therefore \int { \frac { { x }^{ 2 } }{ { { (2+3x }^{ 3 }) }^{ 3 } } dx } =\frac { 1 }{ 9 } \int { \frac { dt }{ { t }^{ 3 } } =\frac { 1 }{ 9 } \int { { t }^{ -3 }dt } }$
$=-\frac { 1 }{ { 18t }^{ 2 } } +c\quad =-\frac { 1 }{ 18(2+{ 3x }^{ 3 })^{ 2 } } +c$

Ex 7.2 Class 12 Maths Question 14.
$\frac { 1 }{ x(logx)^{ m } } ,x>0$
Solution:
Put log x = t, so that $\frac { 1 }{ x }dx=dt$
$\therefore \int { \frac { 1 }{ { x(logx) }^{ m } } dx } =\int { \frac { dt }{ { t }^{ m } } =\frac { { t }^{ -m+1 } }{ -m+1 } +c }$
$=\frac { { (logx) }^{ 1-m } }{ 1-m } +c$

Ex 7.2 Class 12 Maths Question 15.
$\frac { x }{ 9-4{ x }^{ 2 } }$
Solution:
put 9-4x² = t, so that -8x dx = dt
$\therefore \int { \frac { x }{ 9-{ 4x }^{ 2 } } dx } =-\frac { 1 }{ 8 } \int { \frac { dt }{ t } } =-\frac { 1 }{ 8 } log|t|+c$
$=\frac { 1 }{ 8 } log\frac { 1 }{ |9-{ 4x }^{ 2 }| } +c$

Ex 7.2 Class 12 Maths Question 16.
${ e }^{ 2x+3 }$
Solution:
put 2x+3 = t
so that 2dx = dt
$\int { { e }^{ 2x+3 } } dx\quad =\frac { 1 }{ 2 } \int { { e }^{ t }dt } \quad =\frac { 1 }{ 2 } { e }^{ t }+c\quad =\frac { 1 }{ 2 } { e }^{ 2x+3 }+c$

Ex 7.2 Class 12 Maths Question 17.
$\frac { x }{ { e }^{ { x }^{ 2 } } }$
Solution:
Let x² = t
⇒ 2xdx = dt ⇒ $xdx=\frac { dt }{ 2 }$
$\therefore \int { \frac { x }{ { e }^{ { x }^{ 2 } } } dx } \quad =\frac { 1 }{ 2 } \int { \frac { dt }{ { e }^{ t } } \quad =\frac { 1 }{ 2 } \int { { e }^{ -t } } dt }$
$=-\frac { 1 }{ 2 } { e }^{ { -x }^{ 2 } }+c$

Ex 7.2 Class 12 Maths Question 18.
$\frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } }$
Solution:
$let\quad { tan }^{ -1 }x=t\Rightarrow \frac { 1 }{ 1+{ x }^{ 2 } } dx=dt$
$\therefore \int { \frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } } dx } \quad =\int { { e }^{ t }dt\quad ={ e }^{ { tan }^{ -1 }x }+c }$

Ex 7.2 Class 12 Maths Question 19.
$\frac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 }$
Solution:
$\int { \frac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 } dx\quad =\int { \frac { { e }^{ x }({ e }^{ x }-{ e }^{ -x }) }{ { e }^{ x }({ e }^{ x }+{ e }^{ -x }) } dx=I } }$
put e^x+e^-x = t
so that (e^x-e^-x)dx = dt
$\therefore I=\int { \frac { dt }{ t } =log|t|+c } =log|{ e }^{ x }+{ e }^{ -x }|+c$

Ex 7.2 Class 12 Maths Question 20.
$\frac { { e }^{ 2x }-{ e }^{ 2x } }{ { e }^{ 2x }+{ e }^{ -2x } }$
Solution:
put e^2x-e^-2x = t
so that (2e^2x-2e^-2x)dx = dt
$\therefore \int { \frac { { e }^{ 2x }-{ e }^{ 2x } }{ { e }^{ 2x }+{ e }^{ -2x } } } dx=\frac { 1 }{ 2 } \int { \frac { 1 }{ t } dt } =\frac { 1 }{ 2 } log|t|+c$
$=\frac { 1 }{ 2 } log+|{ e }^{ 2x }+{ e }^{ -2x }|+c$

Ex 7.2 Class 12 Maths Question 21.
tan²(2x-3)
Solution:
∫tan²(2x-3)dx = ∫[sec²(2x-3)-1]dx = I
put 2x-3 = t
so that 2dx = dt
I = $\frac { 1 }{ 2 }$ ∫sec²t dt-x+c
= $\frac { 1 }{ 2 }t-x+c$
= $\frac { 1 }{ 2 }tan(2x-3)-x+c$

Ex 7.2 Class 12 Maths Question 22.
sec²(7-4x)
Solution:
∫sec²(7-4x)dx
= $\frac { tan(7-4x) }{ -4 }+c$

Ex 7.2 Class 12 Maths Question 23.
$\frac { { sin }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } }$
Solution:
$let\quad { sin }^{ -1 }x=t\quad \Rightarrow \frac { 1dx }{ \sqrt { 1-{ x }^{ 2 } } } =dt$
$\int { \frac { { sin }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } dx } =\int { t\quad dt } =\frac { 1 }{ 2 } { t }^{ 2 }+c=\frac { 1 }{ 2 } { { (sin }^{ -1 }x) }^{ 2 }+c$

Ex 7.2 Class 12 Maths Question 24.
$\frac { 2cosx-3sinx }{ 6cosx+4sinx }$
Solution:
put 2sinx+4cosx = t
⇒ (2cosx-3sinx)dx = dt
$\frac { 1 }{ 2 } \int { \frac { 2cosx-3sinx }{ 2sinx+3cosx } dx } =\frac { 1 }{ 2 } \int { \frac { dt }{ t } } =\frac { 1 }{ 2 } log|t|+c$
$\frac { 1 }{ 2 } log|2sinx+3cosx|+c$

Ex 7.2 Class 12 Maths Question 25.
$\frac { 1 }{ { cos }^{ 2 }x{ (1-tanx) }^{ 2 } }$
Solution:
put 1-tanx = t
so that -sec²x dx = dt
$\therefore \int { \frac { 1 }{ { cos }^{ 2 }x{ (1-tanx) }^{ 2 } } dx } =\int { \frac { { sec }^{ 2 }x }{ { (1-tanx) }^{ 2 } } dx }$
$=-\int { \frac { dt }{ { t }^{ 2 } } } =\frac { 1 }{ t } +c=\frac { 1 }{ (1-tanx) } +c$

Ex 7.2 Class 12 Maths Question 26.
$\frac { cos\sqrt { x } }{ \sqrt { x } }$
Solution:
$put\quad \sqrt { x } =t,so\quad that\frac { 1 }{ 2\sqrt { x } } dx=dt$
$\therefore \int { \frac { cos\sqrt { x } }{ \sqrt { x } } } dx\quad =\quad 2\quad =\int { cost\quad dt\quad = } 2sint+c$
= 2sin√x+c

Ex 7.2 Class 12 Maths Question 27.
$\sqrt { sin2x } cos2x$
Solution:
put sin2x = t²
⇒ cos2x dx = t dt
$\therefore \int { \sqrt { sin2x } .cos2x\quad dx } \quad =\int { t.tdt=\frac { { t }^{ 3 } }{ 3 } +c }$
$=\frac { { (sin2x) }^{ \frac { 3 }{ 2 } } }{ 3 } +c$

Ex 7.2 Class 12 Maths Question 28.
$\frac { cosx }{ \sqrt { 1+sinx } }$
Solution:
put 1+sinx = t²
⇒cosx dx = 2t dt
$\therefore \int { \frac { cosx }{ \sqrt { 1+sinx } } dx } =2\int { dt } =2t+c$
$=2\sqrt { 1+sinx } +c$

Ex 7.2 Class 12 Maths Question 29.
cotx log sinx
Solution:
put log sinx = t,
⇒ cot x dx = dt
$\therefore \int { cot\quad logsinx\quad dx } =\int { t } dt\quad =\frac { { t }^{ 2 } }{ 2 } +c$
$=\frac { 1 }{ 2 } { (log\quad sinx) }^{ 2 }+c$

Ex 7.2 Class 12 Maths Question 30.
$\frac { sinx }{ 1+cosx }$
Solution:
put 1+cosx = t
⇒ -sinx dx = dt
$\therefore \int { \frac { sinx }{ 1+cosx } dx } =\int { -\frac { dt }{ t } } =-logt+c$
=-log(1+cosx)+c

Ex 7.2 Class 12 Maths Question 31.
$\frac { sinx }{ { (1+cosx) }^{ 2 } }$
Solution:
put 1+cosx = t
so that -sinx dx = dt
$\therefore \int { \frac { sinx }{ { (1+cosx) }^{ 2 } } dx } =-\int { \frac { dt }{ { t }^{ 2 } } }$
$=\frac { 1 }{ t } +c=\frac { 1 }{ 1+cosx } +c$

Ex 7.2 Class 12 Maths Question 32.
$\frac { 1 }{ 1+cotx }$
Solution:
$\int { \frac { 1 }{ 1+\frac { cosx }{ sinx } } } dx=\frac { 1 }{ 2 } \int { \frac { 2sinx\quad dx }{ sinx+cosx } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q32.1

Ex 7.2 Class 12 Maths Question 33.
$\frac { 1 }{ 1-tanx }$
Solution:
$\int { \frac { 1 }{ 1-tanx } } dx=\frac { 1 }{ 2 } \int { \frac { 2cosx\quad dx }{ cosx-sinx } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q33.1

Ex 7.2 Class 12 Maths Question 34.
$\frac { \sqrt { tanx } }{ sinxcosx }$
Solution:
$\int { \frac { \sqrt { tanx } }{ sinxcosx } dx } =\int { \frac { \sqrt { tanx } }{ tanx } } .{ sec }^{ 2 }xdx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q34.1

Ex 7.2 Class 12 Maths Question 35.
$\frac { { (1+logx) }^{ 2 } }{ x }$
Solution:
let 1+logx = t
⇒ $\frac { 1 }{ x }dx=dt$
$\int { \frac { { (1+logx) }^{ 2 } }{ x } dx } =\int { { t }^{ 2 }dt } =\frac { { t }^{ 3 } }{ 3 } +c$
$=\frac { 1 }{ 3 } { (1+logx) }^{ 3 }+c$

Ex 7.2 Class 12 Maths Question 36.
$\frac { (x+1){ (x+logx) }^{ 2 } }{ x }$
Solution:
put x+logx = t
$\left( \frac { x+1 }{ x } \right) dx=dt$
$\therefore \int { \frac { { (x+1)(x+logx) }^{ 2 } }{ x } } dx=\int { { t }^{ 2 }dt }$
$=\frac { { (x+logx) }^{ 3 } }{ 3 } +c$

Ex 7.2 Class 12 Maths Question 37.
$\frac { { x }^{ 3 }sin({ tan }^{ -1 }{ x }^{ 4 }) }{ 1+{ x }^{ 8 } } dx$
Solution:
$put\quad { tan }^{ -1 }{ x }^{ 4 }=t\quad so\quad that\frac { 1 }{ 1+{ x }^{ 8 } } .{ 4x }^{ 3 }dx=dt$
$\therefore \int { \frac { { x }^{ 3 }sin({ tan }^{ -1 }{ x }^{ 4 }) }{ 1+{ x }^{ 8 } } } dx=\frac { 1 }{ 4 } \int { sint\quad dt }$
$=\frac { 1 }{ 4 } (-cost)+c=-\frac { 1 }{ 4 } cos({ tan }^{ -1 }{ x }^{ 4 })+c$

Choose the correct answer in exercises 38 and 39

Ex 7.2 Class 12 Maths Question 38.
$\int { \frac { { 10x }^{ 9 }+{ 10 }^{ x }log{ e }^{ 10 } }{ { x }^{ 10 }+{ 10 }^{ x } } dx }$
(a) 10^x – x¹⁰ + C
(b) 10^x + x¹⁰ + C
(c) (10^x – x¹⁰) + C
(d) log (10^x + x¹⁰) + C
Solution:
(d) $\int { \frac { { 10x }^{ 9 }+{ 10 }^{ x }log{ e }^{ 10 } }{ { x }^{ 10 }+{ 10 }^{ x } } dx }$
= log (10^x + x¹⁰) + C

Ex 7.2 Class 12 Maths Question 39.
$\int { \frac { dx }{ { sin }^{ 2 }x{ \quad cos }^{ 2 }x } = }$
(a) tanx + cotx + c
(b) tanx – cotx + c
(c) tanx cotx + c
(d) tanx – cot2x + c
Solution:
(c) $\int { \frac { dx }{ { sin }^{ 2 }x{ \quad cos }^{ 2 }x } = }$
$=\int { \left( { sec }^{ 2 }x+{ cosec }^{ 2 }x \right) dx }$
= tanx – cotx + c

Class 12 Maths Chapter 7 Integrals Ex 7.3

Find the integrals of the functions in Exercises 1 to 22.

Ex 7.3 Class 12 Maths Question 1.
sin²(2x+5)
Solution:
∫sin²(2x+5)dx
= $\frac { 1 }{ 2 }$ ∫[1-cos2(2x+5)]dx
= $\frac { 1 }{ 2 }$ ∫[1-cos(4x+10)]dx
= $\frac { 1 }{ 2 } \left[ x-\frac { sin(4x+10) }{ 4 } \right] +c$

Ex 7.3 Class 12 Maths Question 2.
sin3x cos4x
Solution:
∫sin3x cos4x
= $\frac { 1 }{ 2 }$ ∫[sin(3x+4x)+cos(3x-4x)]dx
= $\frac { 1 }{ 2 }$ ∫[sin7x+sin(-x)]dx
= $-\frac { 1 }{ 14 } cos7x+\frac { 1 }{ 2 } cosx+c$

Ex 7.3 Class 12 Maths Question 3.
∫cos2x cos4x cos6x dx
Solution:
$\frac { 1 }{ 2 }$ ∫cos2x cos4x cos6x dx
= $\frac { 1 }{ 2 }$ ∫(cos6x+cos2x) cos6x dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q3.1

Ex 7.3 Class 12 Maths Question 4.
∫sin³(2x+1)dx
Solution:
= $\frac { 1 }{ 4 }$ ∫[3sin(2x+1)-sin3(2x+1)]dx
= $-\frac { 3 }{ 8 } cos(2x+1)+\frac { 1 }{ 24 } [4{ cos }^{ 3 }(2x+1)-3cos(2x+1)]+c$
= $-\frac { 1 }{ 2 } cos(2x+1)+\frac { 1 }{ 6 } { cos }^{ 3 }(2x+1)+c$

Ex 7.3 Class 12 Maths Question 5.
sin³x cos³x
Solution:
put sin x = t
⇒ cos x dx = dt
$\therefore \int { { sin }^{ 3 }x{ cos }^{ 3 }xdx } =\int { { t }^{ 3 }(1-{ t }^{ 2 })dt }$
$\frac { { t }^{ 4 } }{ 4 } -\frac { { t }^{ 6 } }{ 6 } +c=\frac { { (sinx) }^{ 4 } }{ 4 } -\frac { { (sinx) }^{ 6 } }{ 6 } +c$

Ex 7.3 Class 12 Maths Question 6.
sinx sin2x sin3x
Solution:
∫sinx sin2x sin3x dx
= $\frac { 1 }{ 2 }$ ∫ 2sin x sin 2x sin 3x dx
= $\frac { 1 }{ 2 }$ ∫ (cosx – cos3x)sin 3x dx
= $\frac { 1 }{ 2 }$ ∫ (sin 4x + sin 2x – sin 6x)dx
= $\frac { 1 }{ 4 } \left\{ \frac { -cos4x }{ 4 } -\frac { cos2x }{ 2 } +\frac { cos6x }{ 6 } \right\} +c$

Ex 7.3 Class 12 Maths Question 7.
sin 4x sin 8x
Solution:
$\frac { 1 }{ 2 }$ ∫sin 4x sin 8xdx
= $\frac { 1 }{ 2 }$ ∫(cos 4x – cos 12x)dx
= $\frac { 1 }{ 2 } \left[ \frac { sin4x }{ 4 } -\frac { sin12x }{ 12 } \right] +c$

Ex 7.3 Class 12 Maths Question 8.
$\frac { 1-cosx }{ 1+cosx }$
Solution:
$\int { \frac { 1-cosx }{ 1+cosx } dx }$
$\int { \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { 2cos }^{ 2 }\frac { x }{ 2 } } dx } =\int { { tan }^{ 2 }\frac { x }{ 2 } dx }$
$=\int { \left[ { sec }^{ 2 }\frac { x }{ 2 } -1 \right] } dx\quad =2tan\frac { x }{ 2 } -x+c$

Ex 7.3 Class 12 Maths Question 9.
$\frac { cosx }{ 1+cosx }$
Solution:
$\int { \frac { cosx }{ 1+cosx } dx }$
$=\int { 1 } dx-\int { \frac { 1 }{ 1+cosx } dx }$
$=x-\frac { 1 }{ 2 } \int { { sec }^{ 2 }\frac { x }{ 2 } dx+c\quad =x-tan\frac { x }{ 2 } +c }$

Ex 7.3 Class 12 Maths Question 10.
∫sinx⁴ dx
Solution:
$\int { { (\frac { 1-cos2x }{ 2 } ) }^{ 2 }dx } \quad =\frac { 1 }{ 4 } \int { \left( { 1+cos }^{ 2 }2x-2cos2x \right) dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q10.1

Ex 7.3 Class 12 Maths Question 11.
cos⁴ 2x
Solution:
∫ cos⁴ 2x dx
$\int { { \left( \frac { 1+cos4x }{ 2 } \right) }^{ 2 } } dx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q11.1

Ex 7.3 Class 12 Maths Question 12.
$\frac { { sin }^{ 2 }x }{ 1+cosx }$
Solution:
$\int { \frac { { sin }^{ 2 }x }{ 1+cosx } } dx\quad =\int { \frac { 1-{ cos }^{ 2 }x }{ 1+cosx } } dx$
$\int { (1-cosx) } dx\quad =x-sinx+c$

Ex 7.3 Class 12 Maths Question 13.
$\frac { cos2x-cos2\alpha }{ cosx-cos\alpha }$
Solution:
let I = $\int { \frac { \left( { 2cos }^{ 2 }x-1 \right) -\left( { 2cos }^{ 2 }\alpha -1 \right) }{ cosx-cos\alpha } } dx$
$\int { \frac { 2\left( { cos }x-cos\alpha \right) -\left( { cos }x+cos\alpha \right) }{ cosx-cos\alpha } } dx$
= 2∫cos x dx + 2cos α∫dx
= 2(sinx+xcosα)+c

Ex 7.3 Class 12 Maths Question 14.
$\frac { cosx-sinx }{ 1+sin2x }$
Solution:
let I = $\int { \frac { cosx-sinx }{ 1+sin2x } } dx=\int { \frac { cosx-sinx }{ { (cosx+sinx) }^{ 2 } } dx }$
put cosx+sinx = t
⇒ (-sinx+cosx)dx = dt
$I=\int { \frac { dt }{ { t }^{ 2 } } } =-\frac { 1 }{ t } +c\quad =\frac { -1 }{ cosx+sinx } +c$

Ex 7.3 Class 12 Maths Question 15.
$\int { { tan }^{ 3 }2x\quad sec2x\quad dx=I }$
Solution:
I = ∫(sec²2x-1)sec2x tan 2xdx
put sec2x=t,2 sec2x tan2x dx=dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q15.1

Ex 7.3 Class 12 Maths Question 16.
tan⁴x
Solution:
let I = ∫tan⁴ dx
= ∫(sec²x-1)²dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q16.1
Ex 7.3 Class 12 Maths Question 17.
$\frac { { sin }^{ 3 }x+{ cos }^{ 3 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x }$
Solution:
$\int { \left( \frac { { sin }^{ 3 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } +\frac { { cos }^{ 2 }x }{ sinx{ cos }^{ 2 }x } \right) dx }$
= secx-cosecx+c

Ex 7.3 Class 12 Maths Question 18.
$\frac { cos2x+{ 2sin }^{ 2 }x }{ { cos }^{ 2 }x }$
Solution:
$I=\int { \frac { \left( { cos }^{ 2 }x-{ sin }^{ 2 }x \right) +2{ sin }^{ 2 }x }{ { cos }^{ 2 }x } } dx$
$=\int { \frac { \left( { cos }^{ 2 }x-{ sin }^{ 2 }x \right) }{ { cos }^{ 2 }x } } dx\quad =\int { { sec }^{ 2 }xdx\quad =tanx+c }$

Ex 7.3 Class 12 Maths Question 19.
$\frac { 1 }{ sinx{ cos }^{ 3 }x }$
Solution:
$I=\int { \left( tanx+\frac { 1 }{ tanx } \right) } { sec }^{ 2 }xdx$
put tanx = t
so that sec²x dx = dt
$I=\int { \left( t+\frac { 1 }{ t } \right) } dt\quad =\frac { { t }^{ 2 } }{ 2 } +log|t|+c$
$=log|tanx|+\frac { 1 }{ 2 } { tan }^{ 2 }x+c$

Ex 7.3 Class 12 Maths Question 20.
$\frac { cos2x }{ { (cosx+sinx) }^{ 2 } }$
Solution:
$I=\int { \frac { { cos }^{ 2 }x-{ sin }^{ 2 }x }{ (cosx+sinx)^{ 2 } } dx } =\int { \frac { cosx-sinx }{ cosx+sinx } dx }$
put cosx+sinx=t
⇒(-sinx+cox)dx = dt
$I=\int { \frac { dt }{ t } } =log|t|+c\quad =log|cosx+sinx|+c$

Ex 7.3 Class 12 Maths Question 21.
sin^-1 (cos x)
Solution:
$\int { { sin }^{ -1 }(cosx)dx } \quad ={ sin }^{ -1 }\left[ sin\left( \frac { \pi }{ 2 } -x \right) \right] dx$
$\int { \left( \frac { \pi }{ 2 } -x \right) dx } \quad =\frac { \pi x }{ 2 } -\frac { { x }^{ 2 } }{ 2 } +c$

Ex 7.3 Class 12 Maths Question 22.
$\int { \frac { 1 }{ cos(x-a)cos(x-b) } dx }$
Solution:
$\frac { 1 }{ sin(a-b) } \int { \frac { sin[(x-b)-(x-a)] }{ cos(x-a)cos(x-b) } dx }$
$=\frac { 1 }{ sin(a-b) } \left[ \int { tan(x-b)dx-\int { tan(x-a)dx } } \right]$
$=\frac { 1 }{ sin(a-b) } log\left| \frac { cos(x-a) }{ cos(x-b) } \right| +c$

Ex 7.3 Class 12 Maths Question 23.
$\int { \frac { { sin }^{ 2 }x-{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx\quad is\quad equal\quad to$
(a) tanx+cotx+c
(b) tanx+cosecx+c
(c) -tanx+cotx+c
(d) tanx+secx+c
Solution:
(a) $\int { \frac { { sin }^{ 2 }x-{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx$
= ∫(sec²x-cosec²x)dx
= tanx+cotx+c

Ex 7.3 Class 12 Maths Question 24.
$\int { \frac { e^{ x }(1+x) }{ cos^{ 2 }({ e }^{ x }.{ x }) } } dx\quad is\quad equal\quad to$
(a) -cot(e.x^x)+c
(b) tan(xe^x)+c
(c) tan(e^x)+c
(d) cot e^x+c
Solution:
(b) $\int { \frac { e^{ x }(1+x) }{ cos^{ 2 }({ e }^{ x }.{ x }) } } dx$
= ∫sec²t dt
= tan t+c = tan(xe^x)+c

Class 12 Maths Chapter 7 Integrals Ex 7.4

Integrate the functions in exercises 1 to 23

Ex 7.4 Class 12 Maths Question 1.
$\frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 }$
Solution:
Let x³ = t ⇒ 3x²dx = dt
$\int { \frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 } dx } =\int { \frac { dt }{ { t }^{ 2 }+1 } } ={ tan }^{ -1 }t+c$
= tan^-1 (x³)+c

Ex 7.4 Class 12 Maths Question 2.
$\frac { 1 }{ \sqrt { 1+{ 4x }^{ 2 } } }$
Solution:
$\frac { 1 }{ 2 } \int { \frac { dx }{ \sqrt { \frac { 1 }{ 4 } +{ x }^{ 2 } } } } =\frac { 1 }{ 2 } \int { \frac { dx }{ \sqrt { { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ x }^{ 2 } } } }$
$=\frac { 1 }{ 2 } log\left| 2x+\sqrt { 1+{ 4x }^{ 2 } } \right| +c$

Ex 7.4 Class 12 Maths Question 3.
$\frac { 1 }{ \sqrt { { (2-x) }^{ 2 }+1 } }$
Solution:
put (2-x)=t
so that -dx=dt
⇒ dx=-dt
$\int { \frac { dx }{ \sqrt { { (2-x) }^{ 2 }+1 } } } =-\int { \frac { dt }{ \sqrt { { t }^{ 2 }+1 } } } =-log|t+\sqrt { { t }^{ 2 }+1 } |+c$
$=log\left| \frac { 1 }{ (2-x)+\sqrt { { x }^{ 2 }-4x+5 } } \right| +c$

Ex 7.4 Class 12 Maths Question 4.
$\frac { 1 }{ \sqrt { 9-{ 25x }^{ 2 } } }$
Solution:
$\int { \frac { dx }{ \sqrt { 9-{ 25x }^{ 2 } } } } =\frac { 1 }{ 5 } \int { \frac { dx }{ \sqrt { { \left( \frac { 3 }{ 5 } \right) }^{ 2 }-{ x }^{ 2 } } } }$
$=\frac { 1 }{ 5 } { sin }^{ -1 }\left( \frac { x }{ \frac { 3 }{ 5 } } \right) +c\quad =\frac { 1 }{ 5 } { sin }^{ -1 }\left( \frac { 5x }{ 3 } \right) +c$

Ex 7.4 Class 12 Maths Question 5.
$\frac { 3x }{ 1+{ 2x }^{ 4 } }$
Solution:
Put x²=t,so that 2x dx=dt
⇒x dx = $\frac { dt }{ 2 }$
$\therefore \int { \frac { 3x }{ 1+{ 2x }^{ 4 } } dx } =\frac { 1 }{ 2 } \int { \frac { dt }{ 1+{ 2t }^{ 2 } } } =\frac { 3 }{ 4 } \int { \frac { dt }{ { \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+{ t }^{ 2 } } }$
$=\frac { 3 }{ 2\sqrt { 2 } } { tan }^{ -1 }(\sqrt { 2t } )+c\quad =\frac { 3 }{ 2\sqrt { 2 } } { tan }^{ -1 }(\sqrt { { 2x }^{ 2 } } )+c$

Ex 7.4 Class 12 Maths Question 6.
$\frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } }$
Solution:
put x³ = t,so that 3x²dx = dt
$\int { \frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } } dx } \quad =\frac { 1 }{ 3 } \int { \frac { dt }{ 1-{ t }^{ 2 } } \quad =\frac { 1 }{ 6 } log } \left| \frac { 1+t }{ 1-t } \right| +c$
$=\frac { 1 }{ 6 } log\left| \frac { 1+{ x }^{ 3 } }{ 1-{ x }^{ 3 } } \right| +c$

Ex 7.4 Class 12 Maths Question 7.
$\frac { x-1 }{ \sqrt { { x }^{ 2 }-1 } }$
Solution:
$I=\int { \frac { x-1 }{ \sqrt { { x }^{ 2 }-1 } } dx } -\int { \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } dx } ,I={ I }_{ 1 }-{ I }_{ 2 }$
put x²-1 = t,so that 2x dx = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q7.1

Ex 7.4 Class 12 Maths Question 8.
$\frac { { x }^{ 2 } }{ \sqrt { { x }^{ 6 }+{ a }^{ 6 } } }$
Solution:
put x³ = t
so that 3x²dx = dt
$I=\frac { 1 }{ 3 } \int { \frac { dt }{ { t }^{ 2 }+{ { (a }^{ 3 }) }^{ 2 } } =\frac { 1 }{ 3 } log\left| t+\sqrt { { t }^{ 2 }+{ a }^{ 6 } } \right| +c }$
$=\frac { 1 }{ 3 } log|{ x }^{ 3 }+\sqrt { { a }^{ 6 }+{ x }^{ 6 } } |+c$

Ex 7.4 Class 12 Maths Question 9.
$\frac { { sec }^{ 2 }x }{ \sqrt { { tan }^{ 2 }x+4 } }$
Solution:
let tanx = t
sec x²dx = dt
$I=\int { \frac { dt }{ \sqrt { { t }^{ 2 }+{ (2) }^{ 2 } } } } =log|t+\sqrt { { t }^{ 2 }+4 } |+c$
$=log|tanx+\sqrt { { tan }^{ 2 }x+4 } |+c$

Ex 7.4 Class 12 Maths Question 10.
$\frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } }$
Solution:
$\int { \frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } } dx } =\int { \frac { dx }{ \sqrt { { (x+1) }^{ 2 }+1 } } }$
$=log|(x+1)+\sqrt { { x }^{ 2 }+2x+2 } |+c$

Ex 7.4 Class 12 Maths Question 11.
$\frac { 1 }{ { 9x }^{ 2 }+6x+5 }$
Solution:
$\int { \frac { 1 }{ { 9x }^{ 2 }+6x+5 } } =\frac { 1 }{ 9 } \int { \frac { dx }{ { \left( x+\frac { 1 }{ 3 } \right) }^{ 2 }{ +\left( \frac { 2 }{ 3 } \right) }^{ 2 } } }$
$=\frac { 1 }{ 6 } { tan }^{ -1 }\left( \frac { 3x+1 }{ 2 } \right) +c$

Ex 7.4 Class 12 Maths Question 12.
$\frac { 1 }{ \sqrt { 7-6x-{ x }^{ 2 } } }$
Solution:
$I=\int { \frac { dx }{ \sqrt { { 4 }^{ 2 }-{ (x+3) }^{ 2 } } } } \quad ={ sin }^{ -1 }\left( \frac { x+3 }{ 4 } \right) +c$

Ex 7.4 Class 12 Maths Question 13.
$\frac { 1 }{ \sqrt { (x-1)(x-2) } }$
Solution:
$\int { \frac { 1 }{ \sqrt { (x-1)(x-2) } } dx } =\int { \frac { dx }{ \sqrt { { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } } }$
$=log\left| x-\frac { 3 }{ 2 } +\sqrt { { x }^{ 2 }-3x+2 } \right| +c$

Ex 7.4 Class 12 Maths Question 14.
$\frac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } } }$
Solution:
$\int { \frac { dx }{ \sqrt { 8+3x-{ x }^{ 2 } } } } =\int { \frac { dx }{ \sqrt { 8-\left( { x }^{ 2 }-3x \right) } } }$
$=\int { \frac { dx }{ \sqrt { { \left( \frac { \sqrt { 41 } }{ 2 } \right) }^{ 2 }-{ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 } } } } \quad ={ sin }^{ -1 }\left( \frac { 2x-3 }{ \sqrt { 41 } } \right) +c$

Ex 7.4 Class 12 Maths Question 15.
$\frac { 1 }{ \sqrt { (x-a)(x-b) } }$
Solution:
$\int { \frac { dx }{ \sqrt { (x-a)(x-b) } } } =\int { \frac { dx }{ { \left( x-\frac { a+b }{ 2 } \right) }^{ 2 }-{ \left( \frac { a-b }{ 2 } \right) }^{ 2 } } }$
$=log\left| \left( x-\frac { a+b }{ 2 } \right) +\sqrt { (x-a)(x-b) } \right| +c$

Ex 7.4 Class 12 Maths Question 16.
$\frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } }$
Solution:
$let\quad I=\int { \frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } } } dx$
put 2x²+x-3=t
so that (4x+1)dx=dt
$let\quad I=\int { \frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } } } dx$
$\therefore I=\int { \frac { dt }{ \sqrt { t } } } ={ 2t }^{ \frac { 1 }{ 2 } }+c\quad =2\sqrt { { 2x }^{ 2 }+x-3 } +c$

Ex 7.4 Class 12 Maths Question 17.
$\frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } }$
Solution:
$\int { \frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } } dx } \quad =\int { \frac { x }{ \sqrt { { x }^{ 2 }-1 } } dx } +\int { \frac { 2 }{ \sqrt { { x }^{ 2 }-1 } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q17.1

Ex 7.4 Class 12 Maths Question 18.
$\frac { 5x-2 }{ 1+2x+{ 3x }^{ 2 } }$
Solution:
put 5x-2=A $\frac { d }{ dx }$ (1+2x+3x²)+B
⇒ 6A=5, A= $\frac { 5 }{ 6 }-2=2A+B$ , B= $-\frac { 11 }{ 3 }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q18.1

Ex 7.4 Class 12 Maths Question 19.
$\frac { 6x+7 }{ \sqrt { (x-5)(x-4) } }$
Solution:
$\int { \frac { 6x+7 }{ \sqrt { (x-5)(x-4) } } dx } =\int { \frac { (6x+7)dx }{ \sqrt { { x }^{ 2 }-9x+20 } } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q19.1

Ex 7.4 Class 12 Maths Question 20.
$\frac { x+2 }{ \sqrt { 4x-{ x }^{ 2 } } }$
Solution:
$I=\int { \frac { x-2 }{ \sqrt { 4-{ (x-2) }^{ 2 } } } dx+4\int { \frac { dx }{ \sqrt { 4-{ (x-2) }^{ 2 } } } } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q20.1

Ex 7.4 Class 12 Maths Question 21.
$\frac { x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } }$
Solution:
$I=\frac { 1 }{ 2 } \int { \frac { 2x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q21.1

Ex 7.4 Class 12 Maths Question 22.
$\frac { x+3 }{ { x }^{ 2 }-2x-5 }$
Solution:
$I=\frac { 1 }{ 2 } \int { \frac { 2x-2 }{ { x }^{ 2 }-2x-5 } dx } +\int { \frac { dx }{ { x }^{ 2 }-2x-5 } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q22.1

Ex 7.4 Class 12 Maths Question 23.
$\frac { 5x+3 }{ \sqrt { { x }^{ 2 }+4x+10 } }$
Solution:
$I=\int { \frac { \frac { 5 }{ 2 } (2x+4)+(3-10) }{ \sqrt { { x }^{ 2 }+4x+10 } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q23.1

Ex 7.4 Class 12 Maths Question 24.
$\int { \frac { dx }{ { x }^{ 2 }+2x+2 } equals }$
(a) xtan^-1(x+1)+c
(b) (x+1)tan^-1x+c
(c) tan^-1(x+1)+c
(d) tan^-1x+c
Solution:
(b) $let\quad I=\int { \frac { dx }{ { x }^{ 2 }+2x+2 } } =\int { \frac { dx }{ (x+1)^{ 2 }+1 } }$
= (x+1)tan^-1x+c

Ex 7.4 Class 12 Maths Question 25.
$\int { \frac { dx }{ \sqrt { 9x-{ 4x }^{ 2 } } } equals }$
(a) $\frac { 1 }{ 9 } { sin }^{ -1 }\left( \frac { 9x-8 }{ 8 } \right) +c$
(b) $\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 8x-9 }{ 9 } \right) +c$
(c) $\frac { 1 }{ 3 } { sin }^{ -1 }\left( \frac { 9x-8 }{ 8 } \right) +c$
(d) ${ sin }^{ -1 }\left( \frac { 9x-8 }{ 9 } \right) +c$
Solution:
(b) $\int { \frac { dx }{ \sqrt { 9x-{ 4x }^{ 2 } } } } =\frac { 1 }{ 2 } \left[ \frac { dx }{ \sqrt { \left( \frac { 9 }{ 8 } \right) ^{ 2 }-\left[ { x }^{ 2 }-{ \frac { 9 }{ 4 } }x+\left( \frac { 9 }{ 8 } \right) ^{ 2 } \right] } } \right]$
$\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 8x-9 }{ 9 } \right) +c$

Class 12 Maths Chapter 7 Integrals Ex 7.5

Integrate the rational function in exercises 1 to 21

Ex 7.5 Class 12 Maths Question 1.
$\frac { x }{ (x+1)(x+2) }$
Solution:
let $\frac { x }{ (x+1)(x+2) }$ ≡ $\frac { A }{ x+1 } +\frac { B }{ x+2 }$
⇒ x ≡ A(x+2)+B(x+1)….(i)
putting x = -1 & x = -2 in (i)
we get A = 1,B = 2
$\therefore \int { \frac { 1 }{ (x+1)(x+2) } dx } =\int { \frac { -1 }{ x+1 } dx } +\int { \frac { 2 }{ x+2 } dx }$
=-log|x+1| + 2log|x+2|+c

Ex 7.5 Class 12 Maths Question 2.
$\frac { 1 }{ { x }^{ 2 }-9 }$
Solution:
let $\frac { 1 }{ { x }^{ 2 }-9 } =\frac { 1 }{ (x-3)(x+3) } \equiv \frac { A }{ x-3 } +\frac { B }{ x+3 }$
⇒ x ≡ A(x+3)+B(x-3)…(i)
put x = 3, -3 in (i)
we get $A=\frac { 1 }{ 6 }$ & $B=-\frac { 1 }{ 6 }$
$\therefore \int { \frac { 1 }{ { x }^{ 2 }-9 } dx } =\frac { 1 }{ 6 } \int { \left[ \frac { 1 }{ x-3 } -\frac { 1 }{ x+3 } \right] dx }$
$=\frac { 1 }{ 6 } log\left| \frac { x-3 }{ x+3 } \right| +c$

Ex 7.5 Class 12 Maths Question 3.
$\frac { 3x-1 }{ (x-1)(x-2)(x-3) }$
Solution:
Let $\frac { 3x-1 }{ (x-1)(x-2)(x-3) } =\frac { A }{ x-1 } +\frac { B }{ x-2 } +\frac { C }{ x-3 }$
⇒ 3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(-2)…..(i)
put x = 1,2,3 in (i)
we get A = 1,B = -5 & C = 4
$\therefore I=\int { \frac { 1 }{ x-1 } dx } -5\int { \frac { 1 }{ x-2 } dx } +4\int { \frac { 1 }{ x-3 } dx }$
=log|x-1| – 5log|x-2| + 4log|x+3| + C

Ex 7.5 Class 12 Maths Question 4.
$\frac { x }{ (x-1)(x-2)(x-3) }$
Solution:
let $\frac { x }{ (x-1)(x-2)(x-3) } =\frac { A }{ x-1 } +\frac { B }{ x-2 } +\frac { C }{ x-3 }$
⇒ x ≡ A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)…(i)
put x = 1,2,3 in (i)
$A=\frac { 1 }{ 2 } ,B=-2,C=\frac { 3 }{ 2 }$
$\therefore I=\frac { 1 }{ 2 } \int { \frac { dx }{ x-1 } } -2\int { \frac { dx }{ x-2 } } +\frac { 3 }{ 2 } \int { \frac { dx }{ x-3 } }$
$=\frac { 1 }{ 2 } log|x-1|-2log|x-2|+\frac { 3 }{ 2 } log|x-3|+c$

Ex 7.5 Class 12 Maths Question 5.
$\frac { 2x }{ { x }^{ 2 }+3x+2 }$
Solution:
let $\frac { 2x }{ { x }^{ 2 }+3x+2 } =\frac { 2x }{ (x+1)(x+2) } =\frac { A }{ x+1 } +\frac { B }{ x+2 }$
⇒ 2x = A(x+2)+B(x+1)…(i)
put x = -1, -2 in (i)
we get A = -2, B = 4
$\therefore \int { \frac { 2x }{ { x }^{ 2 }+3x+2 } dx } =-2\int { \frac { dx }{ x+1 } } +4\int { \frac { dx }{ x+2 } }$
=-2log|x+1|+4log|x+2|+c

Ex 7.5 Class 12 Maths Question 6.
$\frac { 1-{ x }^{ 2 } }{ x(1-2x) }$
Solution:
$\frac { 1-{ x }^{ 2 } }{ (x-2{ x }^{ 2 }) }$ is an improper fraction therefore we
convert it into a proper fraction. Divide 1 – x² by x – 2x² by long division.
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q6.1

Ex 7.5 Class 12 Maths Question 7.
$\frac { x }{ \left( { x }^{ 2 }+1 \right) \left( x-1 \right) }$
Solution:
let $\frac { x }{ \left( { x }^{ 2 }+1 \right) \left( x-1 \right) } =\frac { A }{ x-1 } +\frac { Bx+C }{ { x }^{ 2 }+1 }$
⇒ x = A(x²+1)+(Bx+C)(x-1)
Put x = 1,0
⇒ $A=\frac { 1 }{ 2 } C=\frac { 1 }{ 2 } \Rightarrow B=-\frac { 1 }{ 2 }$
$\therefore I=\frac { 1 }{ 2 } \int { \frac { dx }{ x-1 } } -\frac { 1 }{ 2 } \int { \frac { x }{ { x }^{ 2 }+1 } dx } +\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+1 } }$
$=\frac { 1 }{ 2 } log(x-1)-\frac { 1 }{ 4 } log({ x }^{ 2 }+1)+\frac { 1 }{ 2 } { tan }^{ -1 }x+c$

Ex 7.5 Class 12 Maths Question 8.
$\frac { x }{ { \left( x-1 \right) }^{ 2 }\left( x+2 \right) }$
Solution:
$\frac { x }{ { \left( x-1 \right) }^{ 2 }\left( x+2 \right) } =\frac { A }{ x-1 } +\frac { B }{ { \left( x-1 \right) }^{ 2 } } +\frac { C }{ x+2 }$
⇒ x ≡ A(x-1)(x+2)+B(x+2)+C(x-1)² …(i)
put x = 1, -2
we get $B=\frac { 1 }{ 3 } ,C=\frac { -2 }{ 9 }$
$\therefore I=\frac { 2 }{ 9 } \int { \frac { 1 }{ x-1 } dx } +\frac { 1 }{ 3 } \int { \frac { 1 }{ { (x-1) }^{ 2 } } dx } -\frac { 2 }{ 9 } \int { \frac { 1 }{ x+2 } dx }$
$=\frac { 2 }{ 9 } log\left| \frac { x-1 }{ x+2 } \right| -\frac { 1 }{ 3\left( x-1 \right) } +c$

Ex 7.5 Class 12 Maths Question 9.
$\frac { 3x+5 }{ { x }^{ 3 }-{ x }^{ 2 }-x+1 }$
Solution:
let $\frac { 3x+5 }{ { x }^{ 2 }(x-1)-1(x-1) }$
$\frac { 3x+5 }{ (x-1)^{ 2 }(x+1) } =\frac { A }{ x-1 } +\frac { B }{ { (x-1) }^{ 2 } } +\frac { C }{ x+1 }$
⇒ 3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)
put x = 1,-1,0
we get $B=4,C=\frac { 1 }{ 2 } ,A=-\frac { 1 }{ 2 }$
$\therefore I=-\frac { 1 }{ 2 } \int { \frac { dx }{ (x-1) } } +4\frac { dx }{ { (x-1) }^{ 2 } } +\frac { 1 }{ 2 } \int { \frac { dx }{ x+1 } }$
$=\frac { 1 }{ 2 } log\left| \frac { x+1 }{ x-1 } \right| -\frac { 4 }{ x-1 } +c$

Ex 7.5 Class 12 Maths Question 10.
$\frac { 2x-3 }{ ({ x }^{ 2 }-1)(2x+3) }$
Solution:
$\frac { 2x-3 }{ ({ x }^{ 2 }-1)(2x+3) } =\frac { 2x-3 }{ (x-1)(x+1)(2x+3) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q10.1

Ex 7.5 Class 12 Maths Question 11.
$\frac { 5x }{ (x-1)({ x }^{ 2 }-4) }$
Solution:
let $\frac { 5x }{ (x-1)({ x }^{ 2 }-4) } =\frac { 5x }{ (x+1)(x+2)(x-2) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q11.1

Ex 7.5 Class 12 Maths Question 12.
$\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 }$
Solution:
$\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 } =x+\frac { 2x+1 }{ (x+1)(x-1) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q12.1

Ex 7.5 Class 12 Maths Question 13.
$\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) }$
Solution:
$\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } =\frac { A }{ 1-x } +\frac { Bx+C }{ 1+{ x }^{ 2 } }$
⇒ 2 = A(1+x²) + (Bx+C)(1 -x) …(i)
Putting x = 1 in (i), we get; A = 1
Also 0 = A – B and 2 = A + C ⇒B = A = 1 & C = 1
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q13.1

Ex 7.5 Class 12 Maths Question 14.
$\frac { 3x-1 }{ { (x+2) }^{ 2 } }$
Solution:
$\frac { 3x-1 }{ { (x+2) }^{ 2 } } \equiv \frac { A }{ x+1 } +\frac { B }{ { (x+2) }^{ 2 } }$
=>3x – 1 = A(x + 2) + B …(i)
Comparing coefficients A = -1 and B = -7
$\therefore \int { \frac { 3x-1 }{ { (x+2) }^{ 2 } } dx } =3\int { \frac { dx }{ x+2 } } -7\int { \frac { dx }{ { (x+2) }^{ 2 } } }$
$=3log|x+2|+\frac { 7 }{ x+2 } +c$

Ex 7.5 Class 12 Maths Question 15.
$\frac { 1 }{ { x }^{ 4 }-1 }$
Solution:
$\frac { 1 }{ { x }^{ 4 }-1 } =\frac { A }{ x+1 } +\frac { B }{ x-1 } +\frac { Cx+D }{ { x }^{ 2 }+1 }$
⇒ 1 ≡ A(x-1)(x²+1) + B(x+1)(x²+1) + (Cx+D)(x+1)(x-1) ….(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q15.1

Ex 7.5 Class 12 Maths Question 16.
$\frac { 1 }{ x({ x }^{ n }+1) }$
[Hint : multiply numerator and denominator by x^n-1 and put xⁿ = t ]
Solution:
$\frac { { x }^{ n-1 } }{ x.{ x }^{ n-1 }({ x }^{ n }+1) } =\frac { { x }^{ n-1 } }{ { x }^{ n }({ x }^{ n }+1) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q16.1

Ex 7.5 Class 12 Maths Question 17.
$\frac { cosx }{ (1-sinx)(2-sinx) }$
Solution:
put sinx = t
so that cosx dx = dt
$\therefore I=\int { \frac { 1 }{ (1-t)(2-t) } dt }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q17.1

Ex 7.5 Class 12 Maths Question 18.
$\frac { \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+2 \right) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }+4 \right) }$
Solution:
put x²=y
$I=1-\frac { 2(2y+5) }{ (y+3)(y+4) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q18.1

Ex 7.5 Class 12 Maths Question 19.
$\frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) }$
Solution:
put x²=y
so that 2xdx = dy
$\therefore \int { \frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) } dx } =\int { \frac { dy }{ (y+1)(y+3) } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q19.1

Ex 7.5 Class 12 Maths Question 20.
$\frac { 1 }{ x({ x }^{ 4 }-1) }$
Solution:
put x⁴ = t
so that 4x³ dx = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q20.1

Ex 7.5 Class 12 Maths Question 21.
$\frac { 1 }{ { e }^{ x }-1 }$
Solution:
Let e^x = t ⇒ e^x dx = dt
⇒ $dx=\frac { dt }{ t }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q21.1

Ex 7.5 Class 12 Maths Question 22.
choose the correct answer in each of the following :
$\int { \frac { xdx }{ (x-1)(x-2) } equals }$
(a) $log\left| \frac { { (x-1) }^{ 2 } }{ x-2 } \right| +c$
(b) $log\left| \frac { { (x-2) }^{ 2 } }{ x-1 } \right| +c$
(c) $log\left| \left( \frac { x-{ 1 }^{ 2 } }{ x-2 } \right) \right| +c$
(d) log|(x-1)(x-2)|+c
Solution:
(b) $\int { \frac { x }{ (x-1)(x-2) } dx } =\int { \left[ \frac { -1 }{ x-1 } +\frac { 2 }{ x-2 } \right] dx }$
$log\left| \frac { { (x-2) }^{ 2 } }{ x-1 } \right| +c$

Ex 7.5 Class 12 Maths Question 23.
$\int { \frac { dx }{ x({ x }^{ 2 }+1) } equals }$
(a) $log|x|-\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c$
(b) $log|x|+\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c$
(c) $-log|x|+\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c$
(d) $\frac { 1 }{ 2 } log|x|+log({ x }^{ 2 }+1)+c$
Solution:
(a) let $\frac { 1 }{ x\left( { x }^{ 2 }+1 \right) } =\frac { A }{ x } +\frac { Bx+C }{ { x }^{ 2 }+1 }$
⇒ 1 = A(x²+1)+(Bx+C)(x)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q23.1

Class 12 Maths Chapter 7 Integrals Ex 7.6

Integrate the functions in Exercises 1 to 22.

Ex 7.6 Class 12 Maths Question 1.
x sinx
Solution:
By part integration
∫x sinx dx = x(-cosx) – ∫1(-cosx)dx
=-x cosx + ∫cosxdx
=-x cosx + sinx + c

Ex 7.6 Class 12 Maths Question 2.
x sin3x
Solution:
∫x sin3x dx = $x\left( -\frac { cos3x }{ 3 } \right) -\int { 1 } .\left( \frac { -cos3x }{ 3 } \right) dx$
$=-\frac { 1 }{ 3 } x\quad cos3x+\frac { 1 }{ 9 } sin3x+c$

Ex 7.6 Class 12 Maths Question 3.
${ x }^{ 2 }{ e }^{ x }$
Solution:
$\int { { x }^{ 2 }{ e }^{ x } } dx={ x }^{ 2 }{ e }^{ x }-2{ x }{ e }^{ x }+2{ e }^{ x }+c$
$={ e }^{ x }\left( { x }^{ 2 }-2x+2 \right) +c$

Ex 7.6 Class 12 Maths Question 4.
x logx
Solution:
$\int { xlogx\quad dx } =logx\int { xdx } -\int { \left[ \frac { d }{ dx } (logx)\int { xdx } \right] dx }$
$=\frac { { x }^{ 2 } }{ 2 } logx-\frac { 1 }{ 2 } \int { x\quad dx } =\frac { { x }^{ 2 } }{ 2 } logx-\frac { 1 }{ 4 } { x }^{ 2 }+c$

Ex 7.6 Class 12 Maths Question 5.
x log2x
Solution:
$\int { x\quad log2xdx } =(log2x)\frac { { x }^{ 2 } }{ 2 } -\int { \frac { 1 }{ 2x } } .2\left( \frac { { x }^{ 2 } }{ 2 } \right) dx$
$=\frac { { x }^{ 2 } }{ 2 } log|2x|-\frac { 1 }{ 2 } \int { xdx } =\frac { { x }^{ 2 } }{ 2 } log|2x|-\frac { { x }^{ 2 } }{ 4 } +c$

Ex 7.6 Class 12 Maths Question 6.
${ x }^{ 2 }logx$
Solution:
$\int { { x }^{ 2 }logxdx } =log|x|\left( \frac { { x }^{ 3 } }{ 3 } \right) -\int { \frac { 1 }{ x } } \left( \frac { { x }^{ 3 } }{ 3 } \right) dx$
$=\frac { { x }^{ 3 } }{ 3 } log|x|-\frac { 1 }{ 3 } \int { { x }^{ 2 }dx } =\frac { { x }^{ 3 } }{ 3 } log|x|-\frac { { x }^{ 3 } }{ 9 } +c$

Ex 7.6 Class 12 Maths Question 7.
$x\quad { sin }^{ -1 }x$
Solution:
$I=x\quad { sin }^{ -1 }x.\left( \frac { { x }^{ 2 } }{ 2 } \right) -\int { \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } } .\frac { { x }^{ 2 } }{ 2 } dx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q7.1

Ex 7.6 Class 12 Maths Question 8.
$x\quad { tan }^{ -1 }x$
Solution:
$I=x\quad { tan}^{ -1 }x.\left( \frac { { x }^{ 2 } }{ 2 } \right) -\int { \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } .\frac { { x }^{ 2 } }{ 2 } dx$
$=\frac { { x }^{ 2 } }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 2 } \int { \left( 1-\frac { 1 }{ 1+{ x }^{ 2 } } \right) dx }$
$=\frac { { x }^{ 2 } }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 2 } x+\frac { 1 }{ 2 } { tan }^{ -1 }x+c$

Ex 7.6 Class 12 Maths Question 9.
$x\quad { cos }^{ -1 }x$
Solution:
let I = $\int { x } { cos }^{ -1 }xdx=\int { { cos }^{ -1 }x } .xdx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q9.1

Ex 7.6 Class 12 Maths Question 10.
${ (sin }^{ -1 }{ x })^{ 2 }$
Solution:
$put\quad { sin }^{ -1 }x=\theta \Rightarrow x=sin\theta \Rightarrow dx=cos\theta d\theta$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q10.1

Ex 7.6 Class 12 Maths Question 11.
$\frac { x\quad { cos }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } }$
Solution:
$put\quad { cos }^{ -1 }x=t\quad so\quad that\frac { x\quad { cos }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } dx=dt$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q11.1

Ex 7.6 Class 12 Maths Question 12.
x sec²x
Solution:
∫x sec²x dx =x(tanx)-∫1.tanx dx
= x tanx+log cosx+c

Ex 7.6 Class 12 Maths Question 13.
${ ta }n^{ -1 }x$
Solution:
$\int { { tan }^{ -1 }xdx } =x{ tan }^{ -1 }x-\frac { 1 }{ 2 } \int { \frac { 2x }{ 1+{ x }^{ 2 } } dx }$
$=x{ tan }^{ -1 }x-\frac { 1 }{ 2 } log|1+{ x }^{ 2 }|+c$

Ex 7.6 Class 12 Maths Question 14.
x(logx)²
Solution:
∫x(logx)² dx
$=\frac { { x }^{ 2 } }{ 2 } { (logx) }^{ 2 }-\left[ (logx)\frac { { x }^{ 2 } }{ 2 } -\int { \frac { 1 }{ x } \frac { { x }^{ 2 } }{ 2 } dx } \right]$
$=\frac { { x }^{ 2 } }{ 2 } { (logx) }^{ 2 }-\frac { { x }^{ 2 } }{ 2 } logx+\frac { 1 }{ 4 } { x }^{ 2 }+c$

Ex 7.6 Class 12 Maths Question 15.
(x²+1)logx
Solution:
∫(x²+1)logx dx
$=logx\left( \frac { { x }^{ 3 } }{ 3 } +x \right) -\int { \frac { 1 }{ x } \left( \frac { { x }^{ 3 } }{ 3 } +x \right) dx }$
$=\left( \frac { { x }^{ 3 } }{ 3 } +x \right) logx-\frac { { x }^{ 3 } }{ 9 } -x+c$

Ex 7.6 Class 12 Maths Question 16.
${ e }^{ x }(sinx+cosx)$
Solution:
$put\quad { e }^{ x }sinx=t\Rightarrow { e }^{ x }(sinx+cosx)dx=dt$
$\therefore \int { { e }^{ x }(sinx+cosx)dx } =\int { dt } =t+c$
$={ e }^{ x }sinx+c$

Ex 7.6 Class 12 Maths Question 17.
$\frac { { xe }^{ x } }{ { (1+x) }^{ 2 } }$
Solution:
$\int { \frac { { xe }^{ x } }{ { (1+x) }^{ 2 } } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q17.1

Ex 7.6 Class 12 Maths Question 18.
$\frac { { e }^{ x }(1+sinx) }{ 1+cosx }$
Solution:
$I=\int { { e }^{ x } } \left[ \frac { 1+2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ 2{ cos }^{ 2 }\frac { x }{ 2 } } \right] dx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q18.1

Ex 7.6 Class 12 Maths Question 19.
${ e }^{ x }\left( \frac { 1 }{ x } -\frac { 1 }{ { x }^{ 2 } } \right)$
Solution:
put $\frac { { e }^{ x } }{ x } =t\Rightarrow { e }^{ x }\left( \frac { 1 }{ x } -\frac { 1 }{ { x }^{ 2 } } \right) dx=dt$
$\therefore I=\int { dt } =t+c=\frac { { e }^{ x } }{ x } +c$

Ex 7.6 Class 12 Maths Question 20.
$\frac { { (x-2)e }^{ x } }{ { (x-1) }^{ 3 } }$
Solution:
$I=\int { { e }^{ x }\left[ \frac { 1 }{ { (x-1) }^{ 2 } } -\frac { 2 }{ { (x-1) }^{ 3 } } \right] dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q20.1

Ex 7.6 Class 12 Maths Question 21.
${ e }^{ 2x }sinx$
Solution:
let $I=\int { { e }^{ 2x }sinx }$
$={ e }^{ 2x }(-cosx)-\int { 2{ e }^{ 2x }(-cosx)dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q21.1

Ex 7.6 Class 12 Maths Question 22.
${ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right)$
Solution:
Put x = tan t
so that dx = sec² t dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q22.1

Choose the correct answer in exercise 23 and 24

Ex 7.6 Class 12 Maths Question 23.
$\int { { x }^{ 2 }{ e }^{ { x }^{ 3 } } } dx\quad equals$
(a) $\frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+c$
(b) $\frac { 1 }{ 3 } +{ e }^{ { x }^{ 2 } }+c$
(c) $\frac { 1 }{ 2 } { e }^{ { x }^{ 3 } }+c$
(d) $\frac { 1 }{ 2 } { e }^{ { x }^{ 2 } }+c$
Solution:
(a) let x³ = t
⇒3x² dx = dt
$\therefore \int { { x }^{ 2 }{ e }^{ { x }^{ 3 } }dx } =\frac { 1 }{ 3 } \int { { e }^{ t }dt } =\frac { 1 }{ 3 } { e }^{ t }+c=\frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+c$

Ex 7.6 Class 12 Maths Question 24.
$\int { { e }^{ x }secx(1+tanx) } dx\quad equals$
(a) ${ e }^{ x }cosx+c$
(b) ${ e }^{ x }secx+c$
(c) ${ e }^{ x }sinx+c$
(d) ${ e }^{ x }tanx+c$
Solution:
(b) $\int { { e }^{ x }(secx+secx\quad tanx)dx } ={ e }^{ x }secx+c$

Class 12 Maths Chapter 7 Integrals Ex 7.7

Integral the function in exercises 1 to 9

Ex 7.7 Class 12 Maths Question 1.
$\sqrt { 4-{ x }^{ 2 } }$
Solution:
$let\quad I=\int { \sqrt { 4-{ x }^{ 2 } } } dx=\int { \sqrt { { (2) }^{ 2 }-{ x }^{ 2 } } dx }$
$=\frac { x\sqrt { 4-{ x }^{ 2 } } }{ 2 } +2{ sin }^{ -1 }\left( \frac { x }{ 2 } \right) +c$

Ex 7.7 Class 12 Maths Question 2.
$\sqrt { 1-{ 4x }^{ 2 } }$
Solution:
$\int { \sqrt { 1-{ 4x }^{ 2 } } } dx=2\int { \sqrt { { \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ x }^{ 2 } } } dx$
$=\frac { x\sqrt { 1-{ 4x }^{ 2 } } }{ 2 } +\frac { 1 }{ 4 } { sin }^{ -1 }(2x)+c$

Ex 7.7 Class 12 Maths Question 3.
$\sqrt { { x }^{ 2 }+4x+6 }$
Solution:
$\int { \sqrt { { x }^{ 2 }+4x+6 } } dx=\int { \sqrt { { (x+2) }^{ 2 }+{ (\sqrt { 2 } ) }^{ 2 } } } dx$
$=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x+6 } +log\left| (x+2)+\sqrt { { x }^{ 2 }+4x+6 } \right| +c$

Ex 7.7 Class 12 Maths Question 4.
$\sqrt { { x }^{ 2 }+4x+1 }$
Solution:
$\int { \sqrt { { x }^{ 2 }+4x+1 } } dx=\int { \sqrt { { (x+2) }^{ 2 }-{ (\sqrt { 3 } ) }^{ 2 } } } dx$
$=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x+1 } -\frac { 3 }{ 2 } log\left| (x+2)+\sqrt { { x }^{ 2 }+4x+1 } \right| +c$

Ex 7.7 Class 12 Maths Question 5.
$\sqrt { 1-4x-{ x }^{ 2 } }$
Solution:
$\int { \sqrt { 1-4x-{ x }^{ 2 } } } dx=\int { \sqrt { { (5) }^{ 2 }-{ (x+2) }^{ 2 } } dx }$
$=\frac { x+2 }{ 2 } \sqrt { 5-{ (x+2) }^{ 2 } } dx$

Ex 7.7 Class 12 Maths Question 6.
$\sqrt { { x }^{ 2 }+4x-5 }$
Solution:
$\int { \sqrt { { x }^{ 2 }+4x-5 } } dx=\int { \sqrt { { (x+2) }^{ 2 }-{ (3) }^{ 2 } } } dx$
$=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x-5 } -\frac { 9 }{ 2 } log|x+2+\sqrt { { x }^{ 2 }+4x-5 } |+c$

Ex 7.7 Class 12 Maths Question 7.
$\sqrt { 1+3x-{ x }^{ 2 } }$
Solution:
$\int { \sqrt { 1-\left( { x }^{ 2 }-3x \right) } } dx$
$=\int { \sqrt { { \left( \frac { \sqrt { 13 } }{ 2 } \right) }^{ 2 }-{ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 } } } dx$
$=\frac { 2x-3 }{ 4 } \sqrt { 1+3x-{ x }^{ 2 } } +\frac { 13 }{ 8 } { sin }^{ -1 }\left[ \frac { 2x-3 }{ \sqrt { 3 } } \right] +c$

Ex 7.7 Class 12 Maths Question 8.
$\sqrt { { x }^{ 2 }+3x }$
Solution:
$\int { \sqrt { { x }^{ 2 }+3x } } dx=\int { \sqrt { { \left( x+\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 } } } dx$
$=\frac { 2x+3 }{ 4 } \sqrt { { x }^{ 2 }+3x } -\frac { 9 }{ 8 } log\left| x+\frac { 3 }{ 2 } +\sqrt { { x }^{ 2 }+3x } \right| +c$

Ex 7.7 Class 12 Maths Question 9.
$\sqrt { 1+\frac { { x }^{ 2 } }{ 9 } }$
Solution:
$\int { \sqrt { 1+\frac { { x }^{ 2 } }{ 9 } } } dx=\frac { 1 }{ 3 } \int { \sqrt { { x }^{ 2 }+{ 3 }^{ 2 } } }$
$=\frac { 1 }{ 6 } \left[ x\sqrt { { x }^{ 2 }+9 } +9log|x+\sqrt { { x }^{ 2 }+9 } | \right] +c$

Choose the correct answer in the Exercises 10 to 11:

Ex 7.7 Class 12 Maths Question 10.
$\int { \sqrt { 1+{ x }^{ 2 } } } dx\quad is\quad equal\quad to$
(a) $\frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log|x+\sqrt { 1+{ x }^{ 2 } } |+c$
(b) $\frac { 2 }{ 3 } { \left( 1+{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+c$
(c) $\frac { 2 }{ 3 } x{ \left( 1+{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+c$
(d) $\frac { { x }^{ 2 } }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } { x }^{ 2 }log\left| x+\sqrt { 1+{ x }^{ 2 } } \right| +c$
Solution:
(a) $\int { \sqrt { 1+{ x }^{ 2 } } } dx$
$=\frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log(x+\sqrt { 1+{ x }^{ 2 } } )+c$

Ex 7.7 Class 12 Maths Question 11.
$\int { \sqrt { { x }^{ 2 }-8x+7 } } dx\quad is\quad equal\quad to$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.7 Q11.1
Solution:
(d) $\int { \sqrt { { x }^{ 2 }-8x+7 } } dx=\int { \sqrt { { (x-4) }^{ 2 }-{ (3) }^{ 2 } } } dx$
$=\frac { x-4 }{ 2 } \sqrt { { x }^{ 2 }-8x+7 } -\frac { 9 }{ 2 } log\left| (x-4)+\sqrt { { x }^{ 2 }+8x+7 } \right| +c$

Class 12 Maths Chapter 7 Integrals Ex 7.8

Evaluate the following definite integral as limit of sums.

Ex 7.8 Class 12 Maths Question 1.
$\int _{ a }^{ b }{ x\quad dx }$
Solution:
on comparing
$\int _{ a }^{ b }{ x\quad dx } \quad with\quad \int _{ a }^{ b }{ f(x)dx }$
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q1.1

Ex 7.8 Class 12 Maths Question 2.
$\int _{ 0 }^{ 5 }{ (x+1)dx }$
Solution:
on comparing
$\int _{ 0 }^{ 5 }{ (x+1)dx } \quad with\quad \int _{ 0 }^{ 5 }{ f(x)dx }$
we have f(x) = x+1, a = 0, b = 5
and nh = b-a = 5-0 = 5
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q2.1

Ex 7.8 Class 12 Maths Question 3.
$\int _{ 2 }^{ 3 }{ { x }^{ 2 } } dx$
Solution:
compare
$\int _{ 2 }^{ 3 }{ { x }^{ 2 } } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx$
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q3.1

Ex 7.8 Class 12 Maths Question 4.
$\int _{ 1 }^{ 4 }{ ({ x }^{ 2 }-x) } dx$
Solution:
compare
$\int _{ 1 }^{ 4 }{ ({ x }^{ 2 }-x) } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx$
we have f(x) = x²-x and a = 1, b = 4
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q4.1

Ex 7.8 Class 12 Maths Question 5.
$\int _{ -1 }^{ 1 }{ { e }^{ x } } dx\quad$
Solution:
compare
$\int _{ -1 }^{ 1 }{ { e }^{ x } } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx$
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q5.1

Ex 7.8 Class 12 Maths Question 6.
$\int _{ 0 }^{ 4 }{ { (x+e }^{ 2x }) } dx\quad$
Solution:
let f(x) = x + e^2x,
a = 0, b = 4
and nh = b – a = 4 – 0 = 4

Class 12 Maths Chapter 7 Integrals Ex 7.9

Evaluate the definite integrals in Exercise 1 to 20.

Ex 7.9 Class 12 Maths Question 1.
$\int _{ -1 }^{ 1 }{ { (x+1 }) } dx\quad$
Solution:
${ =\left[ \frac { { x }^{ 2 } }{ 2 } +x \right] }_{ -1 }^{ 1 }=\frac { 1 }{ 2 } (1-1)+(1+1)\quad =2$

Ex 7.9 Class 12 Maths Question 2.
$\int _{ 2 }^{ 3 }{ \frac { 1 }{ x } dx }$
Solution:
$={ \left[ log\quad x \right] }_{ 2 }^{ 3 }\quad =log3-log2\quad =log\frac { 3 }{ 2 }$

Ex 7.9 Class 12 Maths Question 3.
$\int _{ 1 }^{ 2 }{ \left( { 4x }^{ 3 }-{ 5x }^{ 2 }+6x+9 \right) dx }$
Solution:
$={ \left[ \frac { { 4x }^{ 4 } }{ 4 } -\frac { { 5x }^{ 3 } }{ 3 } +\frac { { 6x }^{ 2 } }{ 2 } +9x \right] }_{ 1 }^{ 2 }$
$={ \left[ { x }^{ 4 }-\frac { 5 }{ 3 } { x }^{ 3 }+{ 3x }^{ 2 }+9x \right] }_{ 1 }^{ 2 }\quad =\frac { 64 }{ 3 }$

Ex 7.9 Class 12 Maths Question 4.
$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ sin2x\quad dx }$
Solution:
$={ \left[ -\frac { 1 }{ 2 } cos2x \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\quad =\frac { 1 }{ 2 }$

Ex 7.9 Class 12 Maths Question 5.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cos2x\quad dx }$
Solution:
$={ \left[ \frac { 1 }{ 2 } sin2x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\quad =0$

Ex 7.9 Class 12 Maths Question 6.
$\int _{ 4 }^{ 5 }{ { e }^{ x }dx }$
Solution:
$={ \left[ { e }^{ x } \right] }_{ 4 }^{ 5 }\quad ={ e }^{ 5 }-{ e }^{ 4 }$

Ex 7.9 Class 12 Maths Question 7.
$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tanx\quad dx }$
Solution:
$={ \left[ log\quad secx \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\quad =\frac { 1 }{ 2 } log2$

Ex 7.9 Class 12 Maths Question 8.
$\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 4 } }{ cosec\quad xdx }$
Solution:
$=log{ \left( cosecx-cotx \right) }_{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 4 } }$
$=log(\sqrt { 2 } -1)-log(2-\sqrt { 3 } )\quad =log\left( \frac { \sqrt { 2 } -1 }{ 2-\sqrt { 3 } } \right)$

Ex 7.9 Class 12 Maths Question 9.
$\int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } }$
Solution:
$={ sin }^{ -1 }(1)-{ sin }^{ -1 }(0)\quad =\frac { \pi }{ 2 }$

Ex 7.9 Class 12 Maths Question 10.
$\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } }$
Solution:
$={ \left[ { tan }^{ -1 }x \right] }_{ 0 }^{ 1 }\quad ={ tan }^{ -1 }(1)-{ ta }n^{ -1 }(0)\quad =\frac { \pi }{ 4 }$

Ex 7.9 Class 12 Maths Question 11.
$\int _{ 2 }^{ 3 }{ \frac { dx }{ { x }^{ 2 }-1 } }$
Solution:
$={ \left[ \frac { 1 }{ 2 } log\left( \frac { x-1 }{ x+1 } \right) \right] }_{ 2 }^{ 3 }\quad =\frac { 1 }{ 2 } log\frac { 3 }{ 2 }$

Ex 7.9 Class 12 Maths Question 12.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2 } } xdx$
Solution:
$=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \frac { 1+cos2x }{ 2 } } } dx=\frac { 1 }{ 2 } { \left[ x+\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { \pi }{ 4 }$

Ex 7.9 Class 12 Maths Question 13.
$\int _{ 2 }^{ 3 }{ \frac { x }{ { x }^{ 2 }+1 } } dx$
Solution:
$=\frac { 1 }{ 2 } \int _{ 2 }^{ 3 }{ \frac { 2x }{ { x }^{ 2 }+1 } } dx\quad =\frac { 1 }{ 2 } { \left[ log\left( { x }^{ 2 }+1 \right) \right] }_{ 2 }^{ 3 }\quad =\frac { 1 }{ 2 } log2$

Ex 7.9 Class 12 Maths Question 14.
$\int _{ 0 }^{ 1 }{ \frac { 2x+3 }{ { 5x }^{ 2 }+1 } dx }$
Solution:
$=\frac { 1 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { 10x }{ { 5x }^{ 2 }+1 } dx } +\frac { 3 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { dx }{ { { x }^{ 2 }+\left[ \frac { 1 }{ \sqrt { 5 } } \right] }^{ 2 } } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q14.1

Ex 7.9 Class 12 Maths Question 15.
$\int _{ 0 }^{ 1 }{ { xe }^{ { x }^{ 2 } }dx }$
Solution:
let x² = t ⇒ 2xdx = dt
when x = 0, t = 0 & when x = 1,t = 1
$\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ { e }^{ t }dt } \quad =\frac { 1 }{ 2 } { \left( { e }^{ t } \right) }_{ 0 }^{ 1 }\quad =\frac { 1 }{ 2 } [e-1]$

Ex 7.9 Class 12 Maths Question 16.
$\int _{ 1 }^{ 2 }{ \frac { { 5x }^{ 2 } }{ { x }^{ 2 }+4x+3 } dx }$
Solution:
$\int _{ 1 }^{ 2 }{ \left( 5-\frac { 20x+15 }{ { x }^{ 2 }+4x+3 } \right) dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q16.1

Ex 7.9 Class 12 Maths Question 17.
$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( { 2sec }^{ 2 }x+{ x }^{ 3 }+2 \right) dx }$
Solution:
$={ \left[ 2tanx+\frac { { x }^{ 4 } }{ 4 } +2x \right] }_{ 0 }^{ \frac { \pi }{ 4 } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q17.1

Ex 7.9 Class 12 Maths Question 18.
$\int _{ 0 }^{ \pi }{ \left( { sin }^{ 2 }\frac { x }{ 2 } -{ cos }^{ 2 }\frac { x }{ 2 } \right) } dx$
Solution:
$=-\int _{ 0 }^{ \pi }{ cosx } dx\quad =-{ \left[ sinx \right] }_{ 0 }^{ \pi }-(0-0)\quad =0$

Ex 7.9 Class 12 Maths Question 19.
$\int _{ 0 }^{ 2 }{ \frac { 6x+3 }{ { x }^{ 2 }+4 } } dx$
Solution:
$=\int _{ 0 }^{ 2 }{ \frac { 6x }{ { x }^{ 2 }+4 } } dx+\int _{ 0 }^{ 2 }{ \frac { 3 }{ { x }^{ 2 }+4 } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q19.1

Ex 7.9 Class 12 Maths Question 20.
$\int _{ 0 }^{ 1 }{ \left( { xe }^{ x }+sin\frac { \pi x }{ 4 } \right) dx }$
Solution:
$=\int _{ 0 }^{ 1 }{ { xe }^{ x }dx } +\int _{ 0 }^{ 1 }{ sin\frac { \pi x }{ 4 } } dx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q20.1

Ex 7.9 Class 12 Maths Question 21.
$\int _{ 1 }^{ \sqrt { 3 } }{ \frac { dx }{ { 1+x }^{ 2 } } \quad equals }$
(a) $\frac { \pi }{ 3 }$
(b) $\frac { 2\pi }{ 3 }$
(c) $\frac { \pi }{ 6 }$
(d) $\frac { \pi }{ 12 }$
Solution:
(d) $\int _{ 1 }^{ \sqrt { 3 } }{ \frac { dx }{ { 1+x }^{ 2 } } } \quad ={ \left[ { tan }^{ -1 }x \right] }_{ 1 }^{ \sqrt { 3 } }\quad =\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \quad =\frac { \pi }{ 12 }$

Ex 7.9 Class 12 Maths Question 22.
$\int _{ 0 }^{ \frac { 2 }{ 3 } }{ \frac { dx }{ 4+{ 9x }^{ 2 } } equals }$
(a) $\frac { \pi }{ 6 }$
(b) $\frac { \pi }{ 12 }$
(c) $\frac { \pi }{ 24 }$
(d) $\frac { \pi }{ 4 }$
Solution:
(c) $\int _{ 0 }^{ \frac { 2 }{ 3 } }{ \frac { dx }{ 4+{ 9x }^{ 2 } } } \quad =\frac { 1 }{ 9 } \int _{ 0 }^{ \frac { 2 }{ 3 } }{ \frac { dx }{ { \left( \frac { 2 }{ 3 } \right) }^{ 2 }+{ x }^{ 2 } } }$
$=\frac { 1 }{ 6 } { \left[ { tan }^{ -1 }\left( \frac { 3x }{ 2 } \right) \right] }_{ 0 }^{ \frac { 2 }{ 3 } }\quad =\frac { 1 }{ 6 } \times \frac { \pi }{ 4 } \quad =\frac { \pi }{ 24 }$

Class 12 Maths Chapter 7 Integrals Ex 7.10

Evaluate the integrals in Exercises 1 to 8 using substitution.

Ex 7.10 Class 12 Maths Question 1.
$\int _{ 0 }^{ 1 }{ \frac { x }{ { x }^{ 2 }+1 } } dx=I$
Solution:
Let x² + 1 = t
⇒2xdx = dt
when x = 0, t = 1 and when x = 1, t = 2
$\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { dt }{ t } } ={ \left[ \frac { 1 }{ 2logt } \right] }_{ 1 }^{ 2 }\quad =\frac { 1 }{ 2 } log2$

Ex 7.10 Class 12 Maths Question 2.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { sin\phi } { cos }^{ 5 }\phi d\phi =I }$
Solution:
$I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { sin\phi } { (1-{ sin }^{ 2 }) }^{ 2 }cos\phi d\phi }$
put sinφ = t,so that cosφdφ = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q2.1

Ex 7.10 Class 12 Maths Question 3.
$\int _{ 0 }^{ 1 }{ { sin }^{ -1 } } \left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) dx=I$
Solution:
let x = tanθ =>dx = sec²θ dθ
when x = 0 => θ = 0
and when x = 1 => $\theta \frac { \pi }{ 4 }$
$\frac { 1 }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q3.1

Ex 7.10 Class 12 Maths Question 4.
$\int _{ 0 }^{ 2 }{ x\sqrt { x+2 } } dx=I(say)(put\quad x+2={ t }^{ 2 })$
Solution:
let x+2 = t =>dx = dt
when x = 0,t = 2 and when x = 2, t = 4
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q4.1

Ex 7.10 Class 12 Maths Question 5.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx\quad dx }{ 1+{ cos }^{ 2 }x } =I }$
Solution:
put cosx = t
so that -sinx dx = dt
when x = 0, t = 1; when $x=\frac { \pi }{ 2 }$ , t = 0
$\therefore I=\int _{ 1 }^{ 0 }{ \frac { -dt }{ 1+{ t }^{ 2 } } =-{ \left[ { tan }^{ -1 }t \right] }_{ 1 }^{ 0 } } =\frac { \pi }{ 4 }$

Ex 7.10 Class 12 Maths Question 6.
$\int _{ 0 }^{ 2 }{ \frac { dx }{ x+4-{ x }^{ 2 } } =I }$
Solution:
$\int _{ 0 }^{ 2 }{ \frac { dx }{ x+4-{ x }^{ 2 } } =I }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q6.1

Ex 7.10 Class 12 Maths Question 7.
$\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } =I }$
Solution:
$I=\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }+{ 2 }^{ 2 } } } =\frac { 1 }{ 2 } { \left[ { tan }^{ -1 }\frac { x+1 }{ 2 } \right] }_{ -1 }^{ 1 }\quad =\frac { \pi }{ 8 }$

Ex 7.10 Class 12 Maths Question 8.
$\int _{ 1 }^{ 2 }{ \left[ \frac { 1 }{ x } -\frac { 1 }{ { 2x }^{ 2 } } \right] { e }^{ 2x }dx } =I$
Solution:
let 2x = t ⇒ 2dx = dt
when x = 1, t = 2 and when x = 2, t = 4
$I=\int _{ 2 }^{ 4 }{ e } ^{ t }\left( \frac { 1 }{ t } -\frac { 1 }{ { t }^{ 2 } } \right) dt\quad ={ e }^{ t }{ \left[ \frac { 1 }{ t } \right] }_{ 2 }^{ 4 }\quad =\frac { e^{ 2 } }{ 2 } \left[ \frac { { e }^{ 2 } }{ 2 } -1 \right]$

Choose the correct answer in Exercises 9 and 10

Ex 7.10 Class 12 Maths Question 9.
The value of integral $\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { (x-x }^{ 3 }) }^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx }$ is
(a) 6
(b) 0
(c) 3
(d) 4
Solution:
(a) let I = $\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { (x-x }^{ 3 }) }^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx } \quad =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { x }^{ \frac { 1 }{ 3 } }(1-{ x }^{ 2 })^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q9.1

Ex 7.10 Class 12 Maths Question 10.
$If\quad f(x)=\int _{ 0 }^{ x }{ tsint,\quad then\quad { f }^{ \prime }(x)\quad is }$
(a) cosx+xsinx
(b) xsinx
(c) xcosx
(d) sinx+xcosx
Solution:
(b) $f(x)=\int _{ 0 }^{ x }{ tsint\quad dt }$
$=t(-cost)-\int { 1{ \left[ (-cost)dt \right] }_{ 0 }^{ x } }$
=-x cox+sinx

Class 12 Maths Chapter 7 Integrals Ex 7.11

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Ex 7.11 Class 12 Maths Question 1.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2 }x\quad dx } =I$
Solution:
$I=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ (1+cos2x)dx } =\frac { 1 }{ 2 } { \left[ x+\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\quad =\frac { \pi }{ 4 }$

Ex 7.11 Class 12 Maths Question 2.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { cosx } } dx }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { cosx } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q2.1

Ex 7.11 Class 12 Maths Question 3.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sin }^{ \frac { 3 }{ 2 } }xdx }{ { sin }^{ \frac { 3 }{ 2 } }x+{ cos }^{ \frac { 3 }{ 2 } }dx } dx }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sin }^{ \frac { 3 }{ 2 } }xdx }{ { sin }^{ \frac { 3 }{ 2 } }x+{ cos }^{ \frac { 3 }{ 2 } }dx } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q3.1

Ex 7.11 Class 12 Maths Question 4.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 5 }xdx }{ { sin }^{ 5 }x+{ cos }^{ 5 }x } }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 5 }xdx }{ { sin }^{ 5 }x+{ cos }^{ 5 }x } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q4.1

Ex 7.11 Class 12 Maths Question 5.
$\int _{ -5 }^{ 5 }{ \left| x+2 \right| dx=I }$
Solution:
$I=\int _{ -5 }^{ 5 }{ \left| x+2 \right| dx+\int _{ -2 }^{ 5 }{ \left| x+2 \right| dx } }$
at x = – 5, x + 2 < 0; at x = – 2, x + 2 = 0; at x = 5, x + 2>0;x + 2<0, x + 2 = 0, x + 2>0
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q5.1

Ex 7.11 Class 12 Maths Question 6.
$\int _{ 2 }^{ 8 }{ |x-5|dx } =I$
Solution:
$\int _{ 2 }^{ 8 }{ |x-5|dx } =I$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q6.1

Ex 7.11 Class 12 Maths Question 7.
$\int _{ 0 }^{ 1 }{ x(1-x)^{ n }dx } =I$
Solution:
$\int _{ 0 }^{ 1 }{ x(1-x)^{ n }dx } =I$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q7.1

Ex 7.11 Class 12 Maths Question 8.
$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tanx)dx }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tanx)dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q8.1

Ex 7.11 Class 12 Maths Question 9.
$\int _{ 0 }^{ 2 }{ x\sqrt { 2-x } dx=I }$
Solution:
let 2-x = t
⇒ – dx = dt
when x = 0, t = 2 and when x = 2,t = 0
$\frac { 1 }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q9.1

Ex 7.11 Class 12 Maths Question 10.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2logsinx-logsin2x \right) dx=I }$
Solution:
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2logsinx-logsin2x \right) dx=I }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q10.1

Ex 7.11 Class 12 Maths Question 11.
$\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2 } } xdx$
Solution:
Let f(x) = sin² x
f(-x) = sin² x = f(x)
∴ f(x) is an even function
$\therefore \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2 } } xdx\quad =2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left[ \frac { 1-cos2x }{ 2 } \right] dx }$
$={ \left[ x-\frac { sin2x }{ x } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\therefore I=\frac { \pi }{ 2 }$

Ex 7.11 Class 12 Maths Question 12.
$\int _{ 0 }^{ \pi }{ \frac { xdx }{ 1+sinx } }$
Solution:
let I = $\int _{ 0 }^{ \pi }{ \frac { xdx }{ 1+sinx } }$ …(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q12.1

Ex 7.11 Class 12 Maths Question 13.
$\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 7 } } xdx$
Solution:
Let f(x) = sin⁷ xdx
⇒ f(-x) = -sin⁷ x = -f(x)
⇒ f(x) is an odd function of x
⇒ $\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 7 } } xdx=0$

Ex 7.11 Class 12 Maths Question 14.
$\int _{ 0 }^{ 2\pi }{ { cos }^{ 5 } } xdx$
Solution:
let f(x) = cos⁵ x
⇒ f(2π – x) = cos⁵ x
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q14.1

Ex 7.11 Class 12 Maths Question 15.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx-cosx }{ 1+sinx\quad cosx } dx }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx-cosx }{ 1+sinx\quad cosx } dx }$ …(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q15.1

Ex 7.11 Class 12 Maths Question 16.
$\int _{ 0 }^{ \pi }{ log(1+cosx)dx }$
Solution:
let I = $\int _{ 0 }^{ \pi }{ log(1+cosx)dx }$
then I = $\int _{ 0 }^{ \pi }{ log[1+cos(\pi -x)]dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q16.1

Ex 7.11 Class 12 Maths Question 17.
$\int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { a-x } } dx }$
Solution:
let I = $\int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { a-x } } dx }$ …(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q17.1

Ex 7.11 Class 12 Maths Question 18.
$\int _{ 0 }^{ 4 }{ \left| x-1 \right| dx=I }$
Solution:
$I=-\int _{ 0 }^{ 1 }{ (x-1)dx } +\int _{ 1 }^{ 4 }{ (x-1)dx }$
$=-{ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 0 }^{ 1 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 1 }^{ 4 }=5$

Ex 7.11 Class 12 Maths Question 19.
show that $4\int _{ 0 }^{ a }{ f(x)g(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx }$ if f and g are defined as f(x)=f(a-x) and g(x)+g(a-x)=4
Solution:
let I = $\int _{ 0 }^{ a }{ f(x)g(x)dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q19.1

Ex 7.11 Class 12 Maths Question 20.
The value of $\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left( { x }^{ 3 }+xcosx+{ tan }^{ 5 }x+1 \right) dx }$ is
(a) 0
(b) 2
(c) π
(d) 1
Solution:
(c) let I = $\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left( { x }^{ 3 }+xcosx+{ tan }^{ 5 }x+1 \right) dx }$ is
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q20.1

Ex 7.11 Class 12 Maths Question 21.
The value of $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\left[ \frac { 4+3sinx }{ 4+3sinx } \right] dx }$ is
(a) 2
(b) $\frac { 3 }{ 4 }$
(c) 0
(d) -2
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\left[ \frac { 4+3sinx }{ 4+3sinx } \right] dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q21.1

Discover more from EduGrown School

Subscribe to get the latest posts sent to your email.

NCERT Solutions for Class 12 Maths Chapter : 7 Integrals

Like this:

Related

Discover more from EduGrown School

Author: Rohit

Leave a Reply Cancel reply

Class 12th Chapter -7 Integrals | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter : 7 Integrals

Share this:

Like this:

Related

Discover more from EduGrown School

Author: Rohit

Related posts

Leave a Reply Cancel reply

Discover more from EduGrown School