Posted by Anonymous on Friday, November 30, 2012 at 8:20pm.
integral[13,4]sin^2(8x)cos^2(8x)dx

Calculus  Steve, Saturday, December 1, 2012 at 12:20am
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.

Calculus  Steve, Saturday, December 1, 2012 at 12:21am
or, even easier, recall your halfangle formula to show that this is
1/2 (1cos(32x))

Calculus  Anonymous, Saturday, December 1, 2012 at 12:33am
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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