How to integrate sinx cos3x

cos 3x = 4 cos^3x - 3cosx. (And then sub u = cos x if you can't integrate it by sight).

Silly question though: I would say "best practice" would be to use the sum/product formula to rewrite as the sum of two trig terms (as Swayum has posted), but that won't give you the answer they are looking for without a lot of subsequent manipulation.

cos(3x) = 2 cos^2 (6x) - 1 ?

How do I get cos 3x = 4 cos^3x - 3cosx?

I did:
cos 3x sinx
= cos (x + 2x) sinx
= (cos x cos 2x - sin x sin 2x) sinx
= 1/2 sin 2x cos 2x - 2 sin^3 (x) cos x

and then I'm stuck.

DFranklin has just expanded out cos(3x) as cos(2x + x) and then expanded out the cos2x terms and got rid of the sines.

Nope. I think you mean is cos(6x) = 2cos^2(3x) - 1, which doesn't really help. DFranklin has just expanded out cos(3x) as cos(2x + x) and then expanded out the cos2x terms and got rid of the sines.

Use the trig identity sin 2x = 2sin(x)cos(x)

and the last bit is the differential outside the bracket, which is much easier.

Okay, I got sinxcosx cos 2x - 2 sin^3 x cos x
Now what?

Nooooooo.

0.25 * sin (4x) - 2sin^3 (x) cos (x) dx

Now solve.

Almost there, but you need to understand that when you differentiate 2 sin^4 (x) you get 8 sin^3 (x) cos (x)