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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you mean \[\int\limits_{}^{}\sin^2(9x)dx\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know you have to use a trig indentity, sin^2= 1/2 (1cos^2) but after that I'm alittle shakey

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is the right way to start, although it's sin^2(x) = (1cos^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
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks 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
 5 years ago
Best ResponseYou've already chosen the best response.0Correction here. It's 1/2(1  cos(2x)), without the square.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0^ thanks. He's right. The identity is without the square on the cosine. pretty sure everything else is good?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Opps, that was also my bad. Sorry about that.
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