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Math Help - cosx/sin^2x - bit stuck

  1. #1
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    cosx/sin^2x - bit stuck

    I can do these now but I am stuck on this indefinite integration.

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

    So I thought try letting u = sinx. du/dx = cosx. And then.

    \int \frac{1}{u^2} du

    = 2 ln |sinx| + c

    But that is not correct answer. Can anyone please guide me.

    Angus
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    Re: cosx/sin^2x - bit stuck

    Quote Originally Posted by angypangy View Post
    I can do these now but I am stuck on this indefinite integration.
    \int \frac{cos x}{sin^2 x} dx
    So I thought try letting u = sinx. du/dx = cosx. And then.
    \int \frac{1}{u^2} du
    = 2 ln |sinx| + c
    What is the derivative of \frac{-1}{u}~?
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    Re: cosx/sin^2x - bit stuck

    Quote Originally Posted by Plato View Post
    What is the derivative of \frac{-1}{u}~?
    \frac{-1}{u} = -u^{-1} so it would be u^{-2} or \frac{1}{u^2} Is that correct?
    Last edited by angypangy; January 2nd 2012 at 01:19 PM. Reason: tex editing
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    Re: cosx/sin^2x - bit stuck

    Quote Originally Posted by angypangy View Post
    \frac{-1}{u} = -u^{-1} so it would be u^{-2} or \frac{1}{u^2} Is that correct?
    Yes it is. Now look at what you said the antiderivative was.
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    Re: cosx/sin^2x - bit stuck

    Quote Originally Posted by Plato View Post
    Yes it is. Now look at what you said the antiderivative was.
    So, just because the integral of 1/x is lnx, the integral of 1/u^2 is NOT ln x^2?
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    Re: cosx/sin^2x - bit stuck

    Quote Originally Posted by angypangy View Post
    So, just because the integral of 1/x is lnx, the integral of 1/u^2 is NOT ln x^2?
    \int \frac{\cos x}{\sin^2 x} dx=\frac{-1}{\sin(x)}+C
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    Re: cosx/sin^2x - bit stuck

    knowing your trig identities and basic trig derivatives may help ...

    \int \frac{\cos{x}}{\sin^2{x}} \, dx = \int \frac{1}{\sin{x}} \cdot \frac{\cos{x}}{\sin{x}} \, dx = \int\csc{x} \cdot \cot{x} \, dx = -\csc{x} + C
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