Question

How do you find the integral of `int cos^5xsin^4xdx`?

Answer 1

Refer to explanation

Explanation:

We have that

#int cos^5xsin^4xdx=intcosx*cos^4x*sin^4xdx= intcosx(1-sin^2x)*(1-sin^2x)*sin^4xdx#

We set `sinx=t` hence `dxcosx=dt` so we have

#int(1-t^2)(1-t^2)*t^4dt=int(t^4+t^8-2t^6)dt= t^5/5+t^9/9-2y^7/7+c#

Hence we replace with `sinx` to get

`int cos^5xsin^4xdx=sin^5x/5+sin^9x/9-2*sin^7x/7+c`