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

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 cos^xdt/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: