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Math Help - Integral of (x + sin(x))/(1 + cos(x))

  1. #1
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    Integral of (x + sin(x))/(1 + cos(x))

    \int{\frac{x+sin(x)}{1+cos(x)}}dx


    My attempt so far

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

    \int{\frac{x-x*cos(x)+sin(x)-cos(x)sin(x)}{sin^2(x)}}dx =

    -x*cotan(x)+ln(sin(x))-x*cosec(x)-\frac{cos(x)}{sin^2(x)} + \int{\frac{1}{sin(x)}}dx-ln(sin(x)) =

    -x*cotan(x)-x*cosec(x)-\frac{cos(x)}{sin^2(x)} +  \int{\frac{1}{sin(x)}}dx

    I should get \frac{x(1-cos(x))}{sin(x)} = \frac{x}{sin(x)}-x*cotan(x) pretending that I don't know how to solve \int\frac{1}{sin(x)}dx so apparantly somehow I should be able to paritally integrate \int\frac{x*cos(x)}{sin^2(x)}dx in such a way that I get \int\frac{-1}{sin(x)}dx and so get rid of that integral.
    Last edited by Bart; May 8th 2010 at 06:06 PM.
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  2. #2
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  3. #3
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    = \int{\frac{x}{1+cosx}}dx + \int{\frac{sinx}{1+cosx}}dx

    = \int{\frac{x}{2cos^2{x/2}}}dx +  \int{\frac{2sin(x/2)cos(x/2)}{2cos^2(x/2)}}dx

    = \int{\frac{(x)(sec^2(x/2)}{2}}dx + \int{\tan(x/2)}dx

    Now try to integrate.
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  4. #4
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    Try to only integrate the first one , there is a trick : we don't need to integrate the second one , it will be deleted finally !!


     \frac{1}{2} \int x \sec^2(\frac{x}{2})~dx + \int \tan(\frac{x}{2})~dx

    Using integration by parts ,

     = x\tan(\frac{x}{2}) - \int \tan(\frac{x}{2})~dx  + \int \tan(\frac{x}{2})~dx

    we can see the second integral can be deleted but note that before writing this

     = x\tan(\frac{x}{2}) + C

    we should add one more line , it is

     =  x\tan(\frac{x}{2}) + \int 0~dx

    so followed by

     = x\tan(\frac{x}{2}) + C
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