NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 pdf, free NCERT solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 book pdf download.

BoardCBSE
TextbookNCERT
ClassClass 12
SubjectMaths
ChapterChapter 7
Chapter NameIntegrals
ExerciseEx 7.2
Number of Questions Solved39
CategoryNCERT 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 } } dxquad = } int { frac { dt }{ t } quad =logt+Cquad =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 } +cquad =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 { sinxquad sin(cosx)dx } =-int { sin(cosx) } (-sinx)dx
=-int { sintquad dt } quad =cost+cquad =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 { tquad dt }
=frac { 1 }{ a } .frac { { t }^{ 2 } }{ 2 } +cquad =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.
xsqrt { x+2 }
Solution:
Let x+2 = t²
⇒ dx = 2t dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 7

Ex 7.2 Class 12 Maths Question 8.
xsqrt { 1+{ 2x }^{ 2 } }
Solution:
let 1+2x² = t²
⇒ 4x dx = 2t dt
xquad dx=frac { t }{ 2 } dtint { xsqrt { 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 } =2int { sqrt { t } dt }
=frac { { 2t }^{ frac { 3 }{ 2 } } }{ ^{ frac { 3 }{ 2 } } } +cquad =frac { 4 }{ 3 } { t }^{ frac { 3 }{ 2 } }+cquad =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=2int { frac { dt }{ t } }
= 2logt + c
= 2log(√x-1)+c

Ex 7.2 Class 12 Maths Question 11.
frac { x }{ sqrt { x+4 } } ,x>0″><br><strong>Solution:</strong><br>let x+4 = t<br>⇒ dx = dt, x = t-4<br><img loading=

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 } } +cquad =-frac { 1 }{ 18(2+{ 3x }^{ 3 })^{ 2 } } +c

Ex 7.2 Class 12 Maths Question 14.
frac { 1 }{ x(logx)^{ m } } ,x>0″><br><strong>Solution:</strong><br>Put log x = t, so that <img src=
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 } logfrac { 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 } } dxquad =frac { 1 }{ 2 } int { { e }^{ t }dt } quad =frac { 1 }{ 2 } { e }^{ t }+cquad =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:
letquad { tan }^{ -1 }x=tRightarrow frac { 1 }{ 1+{ x }^{ 2 } } dx=dt
therefore int { frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } } dx } quad =int { { e }^{ t }dtquad ={ 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 } dxquad =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:
letquad { sin }^{ -1 }x=tquad Rightarrow frac { 1dx }{ sqrt { 1-{ x }^{ 2 } } } =dt
int { frac { { sin }^{ -1 }x }{ sqrt { 1-{ x }^{ 2 } } } dx } =int { tquad 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 { cossqrt { x } }{ sqrt { x } }
Solution:
putquad sqrt { x } =t,soquad thatfrac { 1 }{ 2sqrt { x } } dx=dt
therefore int { frac { cossqrt { x } }{ sqrt { x } } } dxquad =quad 2quad =int { costquad dtquad = } 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 } .cos2xquad 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 } =2int { dt } =2t+c
=2sqrt { 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 { cotquad logsinxquad dx } =int { t } dtquad =frac { { t }^{ 2 } }{ 2 } +c
=frac { 1 }{ 2 } { (logquad 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 { 2sinxquad dx }{ sinx+cosx } }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 32

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

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 34

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:
putquad { tan }^{ -1 }{ x }^{ 4 }=tquad soquad thatfrac { 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 { sintquad 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

All Chapter NCERT Solutions For Class12 Maths

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All Subject NCERT Solutions For Class12

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