Calculus
posted by Anonymous on .
integral[13,4]sin^2(8x)cos^2(8x)dx

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. 
or, even easier, recall your halfangle formula to show that this is
1/2 (1cos(32x)) 
If i use integration by parts can i take out 1/4 from the integral and make u=sin^2(16x) and dv=dx ?