How would you do the integral for (cos4x)^5?

Answer 1

In general for the integral of an odd power of a trig function, factor out one copy and rewrite the rest in terms of the complementary function: sin^(2n+1) x = (1 - cos^2 x)^n . sin x
Then you can substiute for the complementary function, and the left over term gets absorbed into the differential (i,.e. in the case above you'd substitute y = cos x, dy = - sin x dx to get ∫-(1-y^2)^n dy). Then you just have a polynomial to expand and integrate term by term.

Here's how it works for your specific question, which will probably illuminate the general technique a bit better:

∫(cos 4x)^5 dx
= ∫(1 - sin^2 4x)^2 . (cos 4x) dx; let u = sin 4x, du = 4 cos 4x dx
= ∫(1 - u^2)^2 (1/4) du
= (1/4) ∫(1 - 2u^2 + u^4) du
= (1/4) [u - 2u^3 / 3 + u^5 / 5] + c
= (sin 4x) / 4 - (sin^3 4x) / 6 + (sin^5 4x) / 20 + c.