Find the integral of sin^3(x)?

Question

I can get this far:

u = sin(x)
du = cos(x)

Now what?

Answer 1

You're going down precisely the wrong path.

∫sin³ x dx

Break up the sines:

∫sin x sin² x dx

Use the Pythagorean theorem:

∫sin x (1-cos² x) dx

Distribute:

∫sin x dx - ∫cos² x sin x dx

Now, integrate (using the inverse chain rule on the second integral):

-cos x + cos³ x/3 + C

And we are done.