Question

How do you find the integral of `int sin^5(x)cos^8(x) dx`?

Answer 1

`-cos^9x/9 + (2cos^11x)/11 - cos^13x/13 + C`

Explanation:

`I=int sin^5x cos^8xdx=intsinx sin^4x cos^8x dx`

`I=int sinx (sin^2x)^2 cos^8xdx=int (1-cos^2x)^2 cos^8x sinx dx`

`cosx=t => -sinxdx=dt => sinxdx=-dt`

`I=int (1-t^2)^2 t^8 (-dt) = -int (1-2t^2+t^4)t^8 dt`

`I=-int (t^8-2t^10+t^12) dt = -t^9/9+(2t^11)/11-t^13/13+C`

`I=-cos^9x/9 + (2cos^11x)/11 - cos^13x/13 + C`