Question

How do you find the antiderivative of `sin^3(x) cos^5(x) dx`?

Answer 1

`int sin^3xcos^5xdx = 1/4sin^4x-1/3sin^6x+1/8sin^8x + C`

Explanation:

Let `u = sinx => (du)/dx = cosx`, so `int ...du=int ...cosxdx`

Using the trig identity `sin^2A+cos^2A-=1` we have;
`u^2+cos^2x=1 => cos^2x=1-u^2`

So we can substitute into `int sin^3xcos^5xdx` to get

`int sin^3xcos^5xdx = int sin^3x(cos^2x)^2cosxdx`
`:. int sin^3xcos^5xdx = int u^3(1-u^2)^2du`
`:. int sin^3xcos^5xdx = int u^3(1-2u^2+u^4)du`
`:. int sin^3xcos^5xdx = int u^3-2u^5+u^7du`
`:. int sin^3xcos^5xdx = 1/4u^4-1/3u^6+1/8u^8 + C`
`:. int sin^3xcos^5xdx = 1/4(sinx)^4-1/3(sinx)^6+1/8(sinx)^8 + C`
`:. int sin^3xcos^5xdx = 1/4sin^4x-1/3sin^6x+1/8sin^8x + C`

Substituting any other trig function would also yield the same result.