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NCERT Solutions For Class 12 Maths Chapter 7 Integrals Ex 7.2

Here, Below you all know about NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Question Answer. I know many of you confuse about finding Chapter 7 Integrals Ex 7.2 Of Class 12 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 12
SubjectMaths
ChapterChapter 7
ExerciseClass 12 Maths Chapter 7 Integrals Exercise 7.2
Number of Questions Solved39
NCERT Solutions For Class 12 Maths Chapter 7 Integrals Ex 7.2

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

NCERT TEXTBOOK EXERCISES

Ex 7.1 Class 12 Maths Question 1.
$\frac { 2x }{ 1+{ x }^{ 2 } }$

Solution:

Let 1+x² = t
⇒ 2xdx = dt
Undefined control sequence \therefore

Ex 7.2 Class 12 Maths Question 2.
$\frac { { \left( logx \right) }^{ 2 } }{ x }$

Solution:

Let logx = t
⇒ $\frac { 1 }{ x }dx=dt$
Undefined control sequence \therefore

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
$=\frac { 1 }{ a } .\frac { { t }^{ 2 } }{ 2 } +c\quad =\frac { 1 }{ 2a } { sin }^{ 2 }(ax+b)+C$

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

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
$=\frac { { 2t }^{ \frac { 3 }{ 2 } } }{ ^{ \frac { 3 }{ 2 } } } +c\quad =\frac { 4 }{ 3 } { t }^{ \frac { 3 }{ 2 } }+c\quad =\frac { 4 }{ 3 } { ({ x }^{ 2 }+x+1) }^{ \frac { 3 }{ 2 } }+c$

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

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
Undefined control sequence \therefore
$=-\frac { 1 }{ { 18t }^{ 2 } } +c\quad =-\frac { 1 }{ 18(2+{ 3x }^{ 3 })^{ 2 } } +c$

Ex 7.2 Class 12 Maths Question 14.
$\frac { 1 }{ x(logx)^{ m } } ,x>0$

Solution:

Put log x = t, so that $\frac { 1 }{ x }dx=dt$
$=\frac { { (logx) }^{ 1-m } }{ 1-m } +c$

Ex 7.2 Class 12 Maths Question 15.
$\frac { x }{ 9-4{ x }^{ 2 } }$

Solution:

put 9-4x² = t, so that -8x dx = dt
Undefined control sequence \therefore
$=\frac { 1 }{ 8 } log\frac { 1 }{ |9-{ 4x }^{ 2 }| } +c$

Ex 7.2 Class 12 Maths Question 16.
e2x+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 ⇒ $xdx=\frac { dt }{ 2 }$
$=-\frac { 1 }{ 2 } { e }^{ { -x }^{ 2 } }+c$

Ex 7.2 Class 12 Maths Question 18.
$\frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } }$

Solution:

$let\quad { tan }^{ -1 }x=t\Rightarrow \frac { 1 }{ 1+{ x }^{ 2 } } dx=dt$
Undefined control sequence \therefore

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
Undefined control sequence \therefore

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
$=\frac { 1 }{ 2 } log+|{ e }^{ 2x }+{ e }^{ -2x }|+c$

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 = 1/2 ∫sec²t dt-x+c
= 1/2t−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
$=-\int { \frac { dt }{ { t }^{ 2 } } } =\frac { 1 }{ t } +c=\frac { 1 }{ (1-tanx) } +c$

Ex 7.2 Class 12 Maths Question 26.
$\frac { cos\sqrt { x } }{ \sqrt { x } }$

Solution:

$put\quad \sqrt { x } =t,so\quad that\frac { 1 }{ 2\sqrt { x } } dx=dt$
= 2sin√x+c

Ex 7.2 Class 12 Maths Question 27.
$\sqrt { sin2x } cos2x$

Solution:

put sin2x = t²
⇒ cos2x dx = t dt
$=\frac { { (sin2x) }^{ \frac { 3 }{ 2 } } }{ 3 } +c$

Ex 7.2 Class 12 Maths Question 28.
$\frac { cosx }{ \sqrt { 1+sinx } }$

Solution:

put 1+sinx = t²
⇒cosx dx = 2t dt
=$=2\sqrt { 1+sinx } +c$

Ex 7.2 Class 12 Maths Question 29.
cotx log sinx

Solution:

put log sinx = t,
⇒ cot x dx = dt
$=\frac { 1 }{ 2 } { (log\quad sinx) }^{ 2 }+c$

Ex 7.2 Class 12 Maths Question 30.
$\frac { sinx }{ 1+cosx }$

Solution:

put 1+cosx = t
⇒ -sinx dx = dt
=-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
$=\frac { 1 }{ t } +c=\frac { 1 }{ 1+cosx } +c$

Ex 7.2 Class 12 Maths Question 32.
$\frac { 1 }{ 1+cotx }$

Solution:

$\int { \frac { 1 }{ 1+\frac { cosx }{ sinx } } } dx=\frac { 1 }{ 2 } \int { \frac { 2sinx\quad dx }{ sinx+cosx } }$

Ex 7.2 Class 12 Maths Question 33.
$\frac { 1 }{ 1-tanx }$

Solution:

$\int { \frac { 1 }{ 1-tanx } } dx=\frac { 1 }{ 2 } \int { \frac { 2cosx\quad dx }{ cosx-sinx } }$

Ex 7.2 Class 12 Maths Question 34.
$\frac { \sqrt { tanx } }{ sinxcosx }$

Solution:

$\int { \frac { \sqrt { tanx } }{ sinxcosx } dx } =\int { \frac { \sqrt { tanx } }{ tanx } } .{ sec }^{ 2 }xdx$

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
$\left( \frac { x+1 }{ x } \right) dx=dt$
$=\frac { { (x+logx) }^{ 3 } }{ 3 } +c$

Ex 7.2 Class 12 Maths Question 37.
$\frac { { x }^{ 3 }sin({ tan }^{ -1 }{ x }^{ 4 }) }{ 1+{ x }^{ 8 } } dx$

Solution:

$put\quad { tan }^{ -1 }{ x }^{ 4 }=t\quad so\quad that\frac { 1 }{ 1+{ x }^{ 8 } } .{ 4x }^{ 3 }dx=dt$
$=\frac { 1 }{ 4 } (-cost)+c=-\frac { 1 }{ 4 } cos({ tan }^{ -1 }{ x }^{ 4 })+c$

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

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.2 PDF

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