# NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3

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## 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 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: $int { frac { 1-cosx }{ 1+cosx } dx }$ $int { frac { { 2sin }^{ 2 }frac { x }{ 2 } }{ { 2cos }^{ 2 }frac { x }{ 2 } } dx } =int { { tan }^{ 2 }frac { x }{ 2 } dx }$ $=int { left[ { sec }^{ 2 }frac { x }{ 2 } -1 right] } dxquad =2tanfrac { x }{ 2 } -x+c$

Ex 7.3 Class 12 Maths Question 9. $frac { cosx }{ 1+cosx }$
Solution: $int { frac { cosx }{ 1+cosx } dx }$ $=int { 1 } dx-int { frac { 1 }{ 1+cosx } dx }$ $=x-frac { 1 }{ 2 } int { { sec }^{ 2 }frac { x }{ 2 } dx+cquad =x-tanfrac { x }{ 2 } +c }$

Ex 7.3 Class 12 Maths Question 10.
∫sinx4 dx
Solution: $int { { (frac { 1-cos2x }{ 2 } ) }^{ 2 }dx } quad =frac { 1 }{ 4 } int { left( { 1+cos }^{ 2 }2x-2cos2x right) dx }$ Ex 7.3 Class 12 Maths Question 11.
cos4 2x
Solution:
∫ cos4 2x dx $int { { left( frac { 1+cos4x }{ 2 } right) }^{ 2 } } dx$ Ex 7.3 Class 12 Maths Question 12. $frac { { sin }^{ 2 }x }{ 1+cosx }$
Solution: $int { frac { { sin }^{ 2 }x }{ 1+cosx } } dxquad =int { frac { 1-{ cos }^{ 2 }x }{ 1+cosx } } dx$ $int { (1-cosx) } dxquad =x-sinx+c$

Ex 7.3 Class 12 Maths Question 13. $frac { cos2x-cos2alpha }{ cosx-cosalpha }$
Solution:
let I = $int { frac { left( { 2cos }^{ 2 }x-1 right) -left( { 2cos }^{ 2 }alpha -1 right) }{ cosx-cosalpha } } dx$ $int { frac { 2left( { cos }x-cosalpha right) -left( { cos }x+cosalpha right) }{ cosx-cosalpha } } dx$
= 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 $I=int { frac { dt }{ { t }^{ 2 } } } =-frac { 1 }{ t } +cquad =frac { -1 }{ cosx+sinx } +c$

Ex 7.3 Class 12 Maths Question 15. $int { { tan }^{ 3 }2xquad sec2xquad dx=I }$
Solution:
I = ∫(sec22x-1)sec2x tan 2xdx
put sec2x=t,2 sec2x tan2x dx=dt Ex 7.3 Class 12 Maths Question 16.
tan4x
Solution:
let I = ∫tan4 dx
= ∫(sec²x-1)²dx 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: $I=int { frac { left( { cos }^{ 2 }x-{ sin }^{ 2 }x right) +2{ sin }^{ 2 }x }{ { cos }^{ 2 }x } } dx$ $=int { frac { left( { cos }^{ 2 }x-{ sin }^{ 2 }x right) }{ { cos }^{ 2 }x } } dxquad =int { { sec }^{ 2 }xdxquad =tanx+c }$

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 $I=int { left( t+frac { 1 }{ t } right) } dtquad =frac { { t }^{ 2 } }{ 2 } +log|t|+c$ $=log|tanx|+frac { 1 }{ 2 } { tan }^{ 2 }x+c$

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 $I=int { frac { dt }{ t } } =log|t|+cquad =log|cosx+sinx|+c$

Ex 7.3 Class 12 Maths Question 21.
sin-1 (cos x)
Solution: $int { { sin }^{ -1 }(cosx)dx } quad ={ sin }^{ -1 }left[ sinleft( frac { pi }{ 2 } -x right) right] dx$ $int { left( frac { pi }{ 2 } -x right) dx } quad =frac { pi x }{ 2 } -frac { { x }^{ 2 } }{ 2 } +c$

Ex 7.3 Class 12 Maths Question 22. $int { frac { 1 }{ cos(x-a)cos(x-b) } dx }$
Solution: $frac { 1 }{ sin(a-b) } int { frac { sin[(x-b)-(x-a)] }{ cos(x-a)cos(x-b) } dx }$ $=frac { 1 }{ sin(a-b) } left[ int { tan(x-b)dx-int { tan(x-a)dx } } right]$ $=frac { 1 }{ sin(a-b) } logleft| frac { cos(x-a) }{ cos(x-b) } right| +c$

Ex 7.3 Class 12 Maths Question 23. $int { frac { { sin }^{ 2 }x-{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dxquad isquad equalquad to$
(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. $int { frac { e^{ x }(1+x) }{ cos^{ 2 }({ e }^{ x }.{ x }) } } dxquad isquad equalquad to$
(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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