Asked by Anonymous

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

Answers

Answered by Steve
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.
Answered by Steve
or, even easier, recall your half-angle formula to show that this is

1/2 (1-cos(32x))
Answered by Anonymous
If i use integration by parts can i take out 1/4 from the integral and make u=sin^2(16x) and dv=dx ?
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