How to solve ∫ (cos^3x)(sin^5x) dx ?

1 Answer
Mar 9, 2018

#I=sin^6x/6-sin^8x/8+c#

Explanation:

#I=intcos^3xsin^5xdx#
#I=intcos^2xsin^5x*cosxdx#
#I=int(1-sin^2x)sin^5x*cosxdx#
Let, #sinx=trArrcosxdx=dt#
#I=int(1-t^2)t^5dt#
#=int(t^5-t^7)dt#
#=t^6/6-t^8/8+c#, where, #t=sinx#
#I=sin^6x/6-sin^8x/8+c#