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anonymous

  • 5 years ago

Please evaluate the integral: (sin^2)9xdx

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  1. anonymous
    • 5 years ago
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    do you mean \[\int\limits_{}^{}\sin^2(9x)dx\]?

  2. anonymous
    • 5 years ago
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    yes I do

  3. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Correction here. It's 1/2(1 - cos(2x)), without the square.

  7. anonymous
    • 5 years ago
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    ^ thanks. He's right. The identity is without the square on the cosine. pretty sure everything else is good?

  8. anonymous
    • 5 years ago
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    I believe it is.

  9. anonymous
    • 5 years ago
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    Opps, that was also my bad. Sorry about that.

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