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Math Help - Evaluating integral!

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    Evaluating integral!

    evaluate the integral of 1/cosx(sinx)^5
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  2. #2
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    Quote Originally Posted by twilightstr View Post
    evaluate the integral of 1/cosx(sinx)^5
    Make the substitution u = \sin x. Then the integral becomes \int \frac{du}{(1 - u^2) u^5}.
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  3. #3
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    however,
    cosx is in the denominator
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    Quote Originally Posted by twilightstr View Post
    however,
    cosx is in the denominator
    Yes but there is also another cos(x)

    sin(x) =u

    cos(x)dx = du
    dx= du/cos(x)

    Put this in integral
    <br />
\int{\frac{du}{cos(x) \cdot cos(x) \cdot sin^5(x) }}

    \int{\frac{du}{(1-u^2) \cdot u^5}}
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  5. #5
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    1/(1-u^2)(u^5)
    = (1/u)[ 1/(1-u^2)(u^4) ]
    = (1/u)[ (1+u^2)/(u^4) + 1/(1-u^2) ]
    = (1+u^2)/(u^5) + 1/(1-u^2)(u)
    = 1/u^5 + 1/u^3 + 1/u + u/(1-u^2)
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  6. #6
    MHF Contributor chisigma's Avatar
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    Under the assumption that the function we have to integrate is…

    f(x)= \frac {1}{\cos x \cdot sin^{5} x}

    … the ‘standard approach’ is the change of variable…

    t=\tan \frac {x}{2}

    … so that...

    \sin x = \frac {2t}{1+t^{2}}, \cos x= \frac {1-t^{2}}{1+t^{2}}, dx= 2 \cdot \frac {dt}{1+t^{2}}

    ... and we arrive to an integral that contains rational functions and that can be solved without great difficulty. Other substitutions doesn’t work so well … at least in my opinion…

    Regards
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