NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 7
Chapter Name	Integrals
Exercise	Ex 7.10
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10

Evaluate the integrals in Exercises 1 to 8 using substitution.

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
$\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { dt }{ t } } ={ \left[ \frac { 1 }{ 2logt } \right] }_{ 1 }^{ 2 }\quad =\frac { 1 }{ 2 } log2$

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 2

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 3

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 4

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
$\therefore I=\int _{ 1 }^{ 0 }{ \frac { -dt }{ 1+{ t }^{ 2 } } =-{ \left[ { tan }^{ -1 }t \right] }_{ 1 }^{ 0 } } =\frac { \pi }{ 4 }$

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 6

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

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

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 9

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

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10

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