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Math Help - Integral Problem

  1. #1
    Senior Member bugatti79's Avatar
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    Jul 2010
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    Integral Problem

    Hi experts,
    How does one solve

    -\int\frac{sinx}{cos^2x}dx

    I have attempted to use the identity cos^2x=1/2(1+cos2x) to give
    -\int\frac{sinx}{\frac{1}{2}(1+cos2x)}
    However I am not sure how to proceed any further because of the double angle....

    In another attempt, i tried tackling it using the method of substitution such as below
    let u=cosx then du = -sinxdx
    then
    \int\frac{du}{u^2} = -\frac{1}{u} + c

    But this answer doesnt look right either....

    Any tips or suggestions on what I am doing wrong will be appreciated. Thanks
    Bugatti
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  2. #2
    Super Member
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    Your second solutions looks fine to me, try differentiating the right side and see if you end up with your original function.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by bugatti79 View Post
    In another attempt, i tried tackling it using the method of substitution such as below
    let u=cosx then du = -sinxdx
    then
    \int\frac{du}{u^2} = -\frac{1}{u} + c

    But this answer doesnt look right either....
    The latter is correct
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  4. #4
    Senior Member bugatti79's Avatar
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    Integral Problem

    Thanks guys,

    I should have been more thorough and checked by differentiating back.
    cheers
    bugatti79
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  5. #5
    MHF Contributor chisigma's Avatar
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    near Piacenza (Italy)
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    Once You realize that is...

    \frac{d}{dx} \cos x = - \sin x (1)

    ... the indefinite integral...

    \displaystyle - \int \frac{\sin x}{\cos^{2} x}\ dx (2)

    ... becomes incredibly easy ...

    Kind regards

    \chi \sigma
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