# Class 12 RD Sharma Solutions- Chapter 19 Indefinite Integrals – Exercise 19.12

• Last Updated : 03 May, 2022

### Question 1. ∫sin4x cos3x dx

Solution:

Let I = ∫ sin4x cos3x dx          -(i)

Let sinx = t

On differentiating with respect to x:

cosx = dt/dx

cosx dx = dt

dx = dt/cosx

Putting value of dx and sinx in equation (i):

I = ∫ t4 cosxdt/cosx
I = ∫ t4 cos2 x dt

I = ∫ t4 (1 – sin2 x) dt

I = ∫ t4 (1 – t2) dt

I = ∫ (t4– t2) dt

I  = t5/5 – t7/7 + c

I  = sin5/5 – sin7/7 + c

### Question 2. ∫ sin5x dx

Solution:

Let I = ∫ sin5x dx

I = ∫sin3xsin2x dx

= ∫sin3x(1 – cos2x)dx

= ∫(sin3x – sin3xcos2x)dx

= ∫[sinxsin2x – sin3xcos2x]dx

= ∫[sinx(1 – cos2x) – sin3xcos2x]dx

= ∫(sinx – sinxcos2x – sin3xcos2x)dx

I = ∫sinx dx – ∫sinxcos2x dx – ∫sin3xcos2x dx

Putting cosx = t and -sinxdx = dt in 2nd and 3rd integral:

I = ∫sinx dx + ∫t2dt + ∫sin2xt3dt/t

= ∫sinx dx + ∫t2 dt + ∫sin2xt2 dt​

= ∫sinx dx + ∫t2 dt + ∫(1 – cos2x)t2 dt

Putting value of t:

### Question 3.∫cos5x dx

Solution:

Let I = ∫cos5x dx

I = ∫cos2xcos3x dx

= ∫(1 – sin2x)cos3x dx

= ∫(cos3x−sin2xcos3x)dx

= ∫(cos2xcosx – sin2xcos2xcosx)dx

= ∫[(1 – sin2x)cosx – sin2x(1 – sin2x)cosx]dx

= ∫(cosx – sin2xcosx – sin2xcosx + sin4xcosx)dx

= ∫cosx dx – 2∫sin2xcosx dx + ∫sin4xcosx dx

Putting sinx = t and cosxdx = dt in 2nd and 3rd integral we get:

I = ∫cos dx – 2∫t2dt + ∫t4dt

= sinx – 2t3/3 + t5/5 + c

Putting value of t:

I = = sinx – 2sin3x/3 + cos5x/5 + c

### Question 4.∫sin5xcosx dx

Solution:

Let I = ∫sin5xcosx dx         −(i)

Let sinx = t:

On differentiating with respect to x:

-cosx = dt/dx

cosx dx = -dt

Putting cosxdx = -dt and sinx = t in eq (i):

I = ∫t5dt

= t6​/6 + c

= sin6​x/6 + c

### Question 5.∫sin3xcos6x dx

Solution:

Let I = ∫sin3xcos6x dx           −(i)

Let cosx = t

On differentiating both sides w.r.t′x′:

-sinx = dt/dx

sinxdx = -dt​

Putting cosx = t and sinxdx = -dt in eq (i):

I = -∫sin2x t6dt

= -∫(1 – cos2x)t6dt

= -∫(1 – t2)t6dt

= -∫(t6 – t8)dt

= -(t7/7​ – t9/9​) + c

Putting value of t:

I = -(cos7x/7​ – cos9x/9​) + c

### Question 6.∫cos7x dx

Solution:

Let I = ∫cos7x dx

= ∫cos6xcosx dx

= ∫(cos2x)3cosx dx

= ∫(1 – sin2x)3cosx dx

= ∫(1 – sin6x – 3sin2x + 3sin4x)cosx dx

= ∫(cosx – sin6xcosx – 3sin2xcosx + 3sin4xcosx)dx         −(i)

Putting sinx = t and cosx dx = t in 2nd,3rd and 4th integral in (i):

I = ∫cosx dx – ∫t6dt – 3∫t2dt + 3∫t4dt

= sinx – t7/7 ​- 3t3/3 ​ +3t5/5​ + c

Putting value of t:

= sinx – sin7x/7 ​- 3sin3x/3 ​ +3sin5x/5​ + c

### Question 7.∫xcos3x2sinx2dx

Solution:

Let I = ∫xcos3x2sinx2dx          −(i)

Let cosx2 = t

On differentiating both sides:

-2xsinx2 = dt/dx

​ xsinx2 dx = -dt/2

Putting values in (i):

= -t4​/8 + c

Putting value of t:

### Question 8.∫sin7x dx

Solution:

Let I = ∫sin7x dx

I = ∫sin6x sinx dx

= ∫(sin2x)3sinx dx

= ∫(1 – cos2x)3sinx dx

= ∫(1 – cos6x – 3cos2x + 3cos4x)sinx dx

I = ∫sinx dx – ∫cos6xsinx dx + 3∫cos4xsinx dx – 3∫cos2xsinx dx

Putting cosx = t and sinx dx = -dt in 2nd,3rd and 4th integral:

I = ∫sinx dx – ∫t6(-dt) + 3∫t4(-dt) – 3∫t2(-dt)

### Question 9.∫sin3xcos5x dx

Solution:

Let I = ∫sin3xcos5x dx         −(i)

Let cosx = t

On differentiating both sides: -sinx = dt/dx

sinx dx = -dt​

Putting values in (i):

I = ∫sin2xt5(-dt)

= ∫(1 – cos2x)t5 dt

= ∫(1 – t2)t5 dt

= ∫(t7 – t5) dt

= t8/8 – t6/6​ + c

Putting value of t:

### Question 10.

Solution:

Let I =

Dividing and multiplying the equation by cos6x:

Let tanx = t, then:

sec2x = dt/dx

sec2x dx = dt

Putting values in eq (ii):​

### Question 11.

Solution:

Dividing and multiplying by cos8x: Let tanx=t,then: Putting values in ii:

### Question 12.

Solution:

Dividing and multiplying by cos4x: Let tanx=t,then: sec2xdx = dt Putting values in i: Putting value of t:

### Question 13.

Solution:

Let tanx=t⟹sec2x dx = dt: Putting value of t:

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