Please evaluate the integral:

(sin^2)9xdx

Question

Please evaluate the integral:

(sin^2)9xdx

anonymous · Answer

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

anonymous · Answer

yes I do

anonymous · Answer

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

anonymous · Answer

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 )

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

I believe it is.

anonymous · Answer

Opps, that was also my bad.  Sorry about that.