What's the integral of sin^5(2x)?

The integral of sin^5(2x) is (-1/10)cos(2x) + (1/6)cos^3(2x) + C.

To solve this integral, we can use the substitution u = cos(2x), du/dx = -2sin(2x)dx. We can then rewrite sin^5(2x) as (1-cos^2(2x))^2sin(2x) and substitute u into the integral:

∫sin^5(2x)dx = ∫(1-cos^2(2x))^2sin(2x)dx
Let u = cos(2x), du/dx = -2sin(2x)dx
= -1/2 ∫(1-u^2)^2du
= -1/2 ∫(1-2u^2+u^4)du
= -1/2 (u - 2/3u^3 + 1/5u^5) + C
= (-1/2)cos(2x) + (1/3)cos^3(2x) - (1/10)cos^5(2x) + C

We can simplify this further by using the identity cos^2(2x) = (1+cos(4x))/2:

(-1/2)cos(2x) + (1/3)cos^3(2x) - (1/10)cos^5(2x) + C
= (-1/2)cos(2x) + (1/3)cos^3(2x) - (1/10)(1+cos(4x))/2 * cos^3(2x) + C
= (-1/2)cos(2x) + (1/3)cos^3(2x) - (1/20)cos^3(2x) - (1/20)cos^7(2x) + C
= (-1/10)cos(2x) + (1/6)cos^3(2x) + C

Therefore, the integral of sin^5(2x) is (-1/10)cos(2x) + (1/6)cos^3(2x) + C.