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NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2.

Board CBSE
Textbook NCERT
Class Class 12
Subject Maths
Chapter Chapter 7
Chapter Name Integrals
Exercise Ex 7.2
Number of Questions Solved 39
Category NCERT Solutions

NCERT Solutions for 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 ex+e-x = t
so that (ex-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 e2x-e-2x = t
so that (2e2x-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) 10x – x10 + C
(b) 10x + x10 + C
(c) (10x – x10) + C
(d) log (10x + x10) + C
Solution:
(d) \(\int { \frac { { 10x }^{ 9 }+{ 10 }^{ x }log{ e }^{ 10 } }{ { x }^{ 10 }+{ 10 }^{ x } } dx } \)
= log (10x + 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

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