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Integral substitution makes both limits zero Watch

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    I'm having a problem with this integral, see the attached pdf, I have written up my problem in latex, if you have any thoughts I would be glad to hear them.

    Cheers.
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  1. File Type: pdfBad Integral Substitution.pdf (87.1 KB, 60 views)
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    Looking at the Wikipedia article on integration by substitution http://en.wikipedia.org/wiki/Integra...y_substitution (particularly the second paragraph) then your 'g' is \sin^{2n}\theta which is not monotonically strictly increasing on [0,2pi] so you can't use it as a substitution.

    You could probably work around this issue by splitting up your integral into several integrals over different intervals in [0,2pi] but I haven't tried this myself.
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    Yes thanks I think that should work.
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    The problem is in the last line for the integral I_1

    The square root part is always taken as positive. But \cos\theta is negative from \frac{\pi}{2}<\theta<\frac{3 \pi}{2}

    You last integral should be:
    I_1=\int_0^{\frac{\pi}{2}}(1-\sin^2\theta)^{1/2}\cos\theta d\theta+\int_\frac{\pi}{2}^{\fra  c{3\pi}{2}}(1-\sin^2\theta)^{1/2}\cos\theta d\theta+\int_\frac{3\pi}{2}^{2\p  i}(1-\sin^2\theta)^{1/2}\cos\theta d\theta
 
 
 
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