# Trigonometric Integrals

## Homework Statement

integral 31(cos^2x)(sin(2x)dx

## The Attempt at a Solution

I am so lost on this problem... Any suggestions would be great

phyzguy
Try looking up the double angle identities and use them to re-write sin(2x). Then try integration by substitution.

So if the double angle for sin(2x) is 2sinxcosx... the problem would be rewritten as integral 31cos^2x(2sinxcosx)dx? Then what??? Sorry im still confused

Thanks for the help!

phyzguy
Now collect the cos(x) terms together. Then what do you have?

So it would become integral 31(2sinxcos^3)... Then make u=sinx du=cosxdx... Which would lead to 31 integral 2sinxcos^2xcosxdx... Then 31 integral 2sinx(1-sin^2x)du... 31 integral 2u(1-u^2)du... Eventually leading to 31u^2-31/2u^4... And then 31sin^2x-31/2sin^4x.. How does that look??

phyzguy