integral[13,4]sin^2(8x)cos^2(8x)dx

Answer 1

sin^2(8x)cos^2(8x)
= 1/4 (2sin(8x)cos(8x))^2
= 1/4 sin^2(16x)
now use integration by parts twice to evaluate.

Answer 2

or, even easier, recall your half-angle formula to show that this is

1/2 (1-cos(32x))

Answer 3

If i use integration by parts can i take out 1/4 from the integral and make u=sin^2(16x) and dv=dx ?