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Math Help - Complex integration

  1. #1
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    Complex integration

    Hi guys! Just working on a puzzle and I realized I have no idea how to integrate this. Help would be much appreciated. Thanks!

      \int _{0} ^{\pi} (2\sin^2\alpha + (\pi-2\alpha)\sin\alpha\cos\alpha)d\alpha
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  2. #2
    MHF Contributor Amer's Avatar
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     \sin ^2 x = \frac{1- \cos 2x }{2}

    \sin x \cos x = \frac{\sin 2x}{2}
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  3. #3
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    Hello, bones!

    If you "break it down", the solution will come to you . . .


    \displaystyle{ \int _{0}^{\pi} (2\sin^2\!x + (\pi-2x)\sin x\cos x)\,dx}

    We have: . \displaystyle{ \int^{\pi}_0\!\bigg(2\sin^2\!x + \pi\sin x\cos x - 2x\sin x\cos x\bigg)\,dx }

    . . . . \displaystyle{=\;\int^{\pi}_0\!\! 2\sin^2\!x\,dx + \pi\!\int^{\pi}_0\!\! \sin x\cos x\,dx - 2\!\int^{\pi}_0 \!\!x\sin x\cos x\,dx }

    . . . . \displaystyle{=\; \int^{\pi}_0\!(1-\cos2x)\,dx + \tfrac{\pi}{2}\!\int^{\pi}_0\!\!\sin2x\,dx - \int^{\pi}_0\!\!x\sin2x\,dx }


    You should be able to integrate the first two.


    The third requires Integration by Parts:

    . . \begin{array}{ccccccc}u &=& x && dv &=& \sin2x\,dx \\ du &=& dx && v &=&\text{-}\frac{1}{2}\cos2x \end{array} \quad \hdots\;\;\text{ etc.}



    Edit: Amer beat me to it . . . *sigh*
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  4. #4
    MHF Contributor chisigma's Avatar
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    With the substitution \alpha= \frac{\theta}{2} is...

    \displaystyle 2\ \sin^{2} \alpha= 1 - \cos \theta

    \displaystyle \sin \alpha\ \cos \alpha = \frac{1}{2}\ \sin \theta (1)

    ... so that is...

    \displaystyle \int_{0}^{\pi} \{2 \sin^{2} \alpha + (\pi -2 \alpha)\ \sin a \cos \alpha \}\ d \alpha =

    \displaystyle = \frac{1}{2} \int_{0}^{2 \pi} (1 - \cos \theta + \frac{\pi}{2}\ \sin \theta - \frac{\theta}{2}\ \sin \theta)\ d \theta =

    \displaystyle = \frac{1}{2} \int_{0}^{2 \pi} (1 - \frac{\theta}{2}\ \sin \theta)\ d \theta =   \frac{1}{2}\ |\theta - \frac{1}{2}\ \sin \theta +\frac{\theta}{2}\ \cos \theta |_{0}^{2 \pi} = \frac{3}{2}\ \pi (2)

    Kind regards

    \chi \sigma
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