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

Last Updated : 03 May, 2022

### Question 1. âˆ«sin4x cos3x dx

Solution:

Let I = âˆ« 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 = âˆ« t4 cosxdt/cosx
I = âˆ« t4 cos2 x dt

I = âˆ« t4 (1 – sin2 x) dt

I = âˆ« t4 (1 – t2) dt

I = âˆ« (t4– t2) dt

I  = t5/5 – t7/7 + c

I  = sin5/5 – sin7/7 + c

### Question 2. âˆ« sin5x dx

Solution:

Let I = âˆ« sin5x dx

I = âˆ«sin3xsin2x dx

= âˆ«sin3x(1 – cos2x)dx

= âˆ«(sin3x – sin3xcos2x)dx

= âˆ«[sinxsin2x – sin3xcos2x]dx

= âˆ«[sinx(1 – cos2x) – sin3xcos2x]dx

= âˆ«(sinx – sinxcos2x – sin3xcos2x)dx

I = âˆ«sinx dx – âˆ«sinxcos2x dx – âˆ«sin3xcos2x dx

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

I = âˆ«sinx dx + âˆ«t2dt + âˆ«sin2xt3dt/t

= âˆ«sinx dx + âˆ«t2 dt + âˆ«sin2xt2 dtâ€‹

= âˆ«sinx dx + âˆ«t2 dt + âˆ«(1 – cos2x)t2 dt

Putting value of t:

### Question 3.âˆ«cos5x dx

Solution:

Let I = âˆ«cos5x dx

I = âˆ«cos2xcos3x dx

= âˆ«(1 – sin2x)cos3x dx

= âˆ«(cos3xâˆ’sin2xcos3x)dx

= âˆ«(cos2xcosx – sin2xcos2xcosx)dx

= âˆ«[(1 – sin2x)cosx – sin2x(1 – sin2x)cosx]dx

= âˆ«(cosx – sin2xcosx – sin2xcosx + sin4xcosx)dx

= âˆ«cosx dx – 2âˆ«sin2xcosx dx + âˆ«sin4xcosx dx

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

I = âˆ«cos dx – 2âˆ«t2dt + âˆ«t4dt

= sinx – 2t3/3 + t5/5 + c

Putting value of t:

I = = sinx – 2sin3x/3 + cos5x/5 + c

### Question 4.âˆ«sin5xcosx dx

Solution:

Let I = âˆ«sin5xcosx dx         âˆ’(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 = âˆ«t5dt

= t6â€‹/6 + c

= sin6â€‹x/6 + c

### Question 5.âˆ«sin3xcos6x dx

Solution:

Let I = âˆ«sin3xcos6x dx           âˆ’(i)

Let cosx = t

On differentiating both sides w.r.tâ€²xâ€²:

-sinx = dt/dx

sinxdx = -dtâ€‹

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

I = -âˆ«sin2x t6dt

= -âˆ«(1 – cos2x)t6dt

= -âˆ«(1 – t2)t6dt

= -âˆ«(t6 – t8)dt

= -(t7/7â€‹ – t9/9â€‹) + c

Putting value of t:

I = -(cos7x/7â€‹ – cos9x/9â€‹) + c

### Question 6.âˆ«cos7x dx

Solution:

Let I = âˆ«cos7x dx

= âˆ«cos6xcosx dx

= âˆ«(cos2x)3cosx dx

= âˆ«(1 – sin2x)3cosx dx

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

= âˆ«(cosx – sin6xcosx – 3sin2xcosx + 3sin4xcosx)dx         âˆ’(i)

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

I = âˆ«cosx dx – âˆ«t6dt – 3âˆ«t2dt + 3âˆ«t4dt

= sinx – t7/7 â€‹- 3t3/3 â€‹ +3t5/5â€‹ + c

Putting value of t:

= sinx – sin7x/7 â€‹- 3sin3x/3 â€‹ +3sin5x/5â€‹ + c

### Question 7.âˆ«xcos3x2sinx2dx

Solution:

Let I = âˆ«xcos3x2sinx2dx          âˆ’(i)

Let cosx2 = t

On differentiating both sides:

-2xsinx2 = dt/dx

â€‹ xsinx2 dx = -dt/2

Putting values in (i):

= -t4â€‹/8 + c

Putting value of t:

### Question 8.âˆ«sin7x dx

Solution:

Let I = âˆ«sin7x dx

I = âˆ«sin6x sinx dx

= âˆ«(sin2x)3sinx dx

= âˆ«(1 – cos2x)3sinx dx

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

I = âˆ«sinx dx – âˆ«cos6xsinx dx + 3âˆ«cos4xsinx dx – 3âˆ«cos2xsinx dx

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

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

### Question 9.âˆ«sin3xcos5x dx

Solution:

Let I = âˆ«sin3xcos5x dx         âˆ’(i)

Let cosx = t

On differentiating both sides: -sinx = dt/dx

sinx dx = -dtâ€‹

Putting values in (i):

I = âˆ«sin2xt5(-dt)

= âˆ’âˆ«(1 – cos2x)t5 dt

= âˆ’âˆ«(1 – t2)t5 dt

= âˆ«(t7 – t5) dt

= t8/8 – t6/6â€‹ + 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):â€‹

### 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=tâŸ¹sec2x dx = dt: Putting value of t:

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