# Trigonometric Integrals

1. Jun 13, 2010

### bgut06

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution
I am so lost on this problem... Any suggestions would be great

2. Jun 13, 2010

### phyzguy

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

3. Jun 13, 2010

### bgut06

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!

4. Jun 13, 2010

### phyzguy

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

5. Jun 13, 2010

### bgut06

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??

6. Jun 13, 2010

### phyzguy

It's much simpler if you try u=cos(x), but I think your solution is correct.

Last edited: Jun 13, 2010
7. Jun 13, 2010

### bgut06

Awesome! Thank you so much. I have another question if you don't mind.. . For Integral 3x(cos(2x))^2dx could I use u substitution with the 3x