What is the integral of sin4xcos4x dx?

Answer 1

In this problem, it is easiest to substitute twice.
First, substitute 4x for U and 4dx for dU (remember to balance it out by adding a 1/4 to the resulting integral).

Afterwards, substitute sin(U) for W and cos(U)dU for dW and you should get the integral of 1/4*W*dW, after which solve like a normal integral, which should be
(1/8)*(W^2) + C and replace the W's and U's for the substitutions made above and you should end up with
(1/8)*((sin4x)^2) + C.

Answer 2

simpler way:
recognize that sin4xcos4x = (1/2)sin8x
and the integral of that is

(-1/16)cos(8x) + c