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 half-angle formula to show that this is

1/2 (1-cos(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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