## anonymous 5 years ago Please evaluate the integral: (sin^2)9xdx

1. anonymous

do you mean $\int\limits_{}^{}\sin^2(9x)dx$?

2. anonymous

yes I do

3. anonymous

I know you have to use a trig indentity, sin^2= 1/2 (1-cos^2) but after that I'm alittle shakey

4. anonymous

that is the right way to start, although it's sin^2(x) = (1-cos^2(2x))/2, if we apply that to our integral, we get: $(1/2)\int\limits_{}^{}(1- \cos^2(18x))$. The integral of 1 is simple, but be sure to watch out for the 18x inside the cosine: $= (x - [\sin(18x)/18])/2 + C$ (don't forget +C )

5. anonymous

Thanks for your help, I really do appreciate it. : ) I see where I went wrong, and I didn't think about taking the 1/2 out.

6. anonymous

Correction here. It's 1/2(1 - cos(2x)), without the square.

7. anonymous

^ thanks. He's right. The identity is without the square on the cosine. pretty sure everything else is good?

8. anonymous

I believe it is.

9. anonymous