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NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3.

Board CBSE
Textbook NCERT
Class Class 12
Subject Maths
Chapter Chapter 7
Chapter Name Integrals
Exercise Ex 7.3
Number of Questions Solved 24
Category NCERT Solutions

NCERT Solutions for 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.xx)+c
(b) tan(xex)+c
(c) tan(ex)+c
(d) cot ex+c
Solution:
(b) \int { \frac { e^{ x }(1+x) }{ cos^{ 2 }({ e }^{ x }.{ x }) } } dx
= ∫sec²t dt
= tan t+c = tan(xex)+c

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