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Math Help - Indefinite trig integral

  1. #1
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    Indefinite trig integral

    \int\frac{cos x}{sin^2x}dx

    I'm confused about the sin^2x I tried the substitution u = sinx \Rightarrow du = cosx dx which is in the integral.
    but what does the new integral become?

    I have \int\frac{du}{u} although I feel it should be \int\frac{du}{u^2} neither seem correct.

    Is the sub rule necessary for this integral?

    Are there any trig identities which would help me out?

    Also what is \frac{d}{dx}sin^2x?
    Last edited by Foxlion; April 26th 2012 at 10:30 AM.
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  2. #2
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    Re: Indefinite trig integral

    You are correct with your substitution u=sinx so sin^2x=(sinx)^2=u^2
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  3. #3
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    Re: Indefinite trig integral

    ok, then the integral is \int\frac{du}{u^2} which is \frac{3}{u^3}+C correct?
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  4. #4
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    Re: Indefinite trig integral

    nvmnd did it with \int u^{-1}du
    Last edited by Foxlion; April 26th 2012 at 11:06 AM.
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  5. #5
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    Re: Indefinite trig integral

    Quote Originally Posted by Foxlion View Post
    ok, then the integral is \int\frac{du}{u^2} which is \frac{3}{u^3}+C correct?
    \int\frac{du}{u^2}=\frac{-1}{u}+C
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  6. #6
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    Re: Indefinite trig integral

    thank you both
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