How do I evaluate int cos^5(x) sin^4(x) dx?

1 Answer
Apr 20, 2018

I=sin^5x/5-(2sin^7x)/7+sin^9x/9+c

Explanation:

Here,

I=intcos^5xsin^4xdx

=intsin^4x(cos^2x)^2cosxdx

=intsin^4x(1-sin^2x)^2cosxdx

Let, sinx=t=>cosxdx=dt

So,

I=intt^4(1-t^2)^2dt

=int(t^4(1-2t^2+t^4)dt

=int(t^4-2t^6+t^8)dt

=t^5/5-2(t^7/7)+t^9/9+c..towhere,t=sinx

=sin^5x/5-(2sin^7x)/7+sin^9x/9+c