Help evaluating an integral calc

How do I go about evaluating this integral? Please explain to me how... this is quite hard for me to do and it'll help me understand better

the integral: sin^4*3xcos^3*3x dx

sin^4(3x)cos^3(3x)dx = sin^4(3x)cos^2(3x)cos(3x)dx = sin^4(3x)[1 - sin^2(3x)]cos(3x)dx =
[sin^4(3x)cos(3x) - sin^6(3x)cos(3x)]dx . Now substitute u = sin(3x) and du = 3cos(3x)dx. The integral becomes (1/3)(u^4 - u^6)du.

What do I do from here?

∫ 1/3 (u^4 - u^6) du
= 1/3 ∫ u^4 du - 1/3 ∫ u^6 du
= 1/15 u^5 - 1/21 u^7 + C

Now substitute u = sin 3x and you're done.

Incidentally, another way to attack this is using a Risch-like method:

sin^p x cos^q x

If p+q is odd, then there are constants a(1), a(3) ... a(p+q) (i.e. the subscript is on odd numbers only) and C such that:

∫ sin^p x cos^q x dx = a(1) sin x + a(3) sin 3x + ... + a(p+q) sin ((p+q)x) + C

If p and q are both odd, then:

∫ sin^p x cos^q x dx = a(2) cos 2x + a(4) cos 4x + ... + a(p+q) cos ((p+q)x) + C

and finally, if p and q are both even, then:

∫ sin^p x cos^q x dx = A x + a(2) sin 2x + a(4) sin 4x + ... + a(p+q) sin ((p+q)x) + C

Note the extra x term in that last equation. This is because of Pythagoras' theorem:

sin² x + cos² x = 1

...the integral of which is, obviously, x.