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Thread: Evalute this DI

  1. #1
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    Evalute this DI

    $\displaystyle \int_{0}^{\pi}\frac{\texttt{sin}^2n\theta}{\texttt {sin}^2\theta}\texttt{d}\theta$
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  2. #2
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    Quote Originally Posted by banku12 View Post
    $\displaystyle \int_{0}^{\pi}\frac{\texttt{sin}^2n\theta}{\texttt {sin}^2\theta}\texttt{d}\theta$
    What's $\displaystyle n$? Any real number or an integer?
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  3. #3
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    Quote Originally Posted by Danny View Post
    What's $\displaystyle n$? Any real number or an integer?
    ya forgot to mention n is an integer
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  4. #4
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    a hint will also help...if not full soln
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  5. #5
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    Let $\displaystyle I_n = \int_0^{\pi} \frac{ \sin^2{n\theta} }{ \sin^2{\theta}}~d\theta $


    We have

    $\displaystyle I_{n+1} - I_n = \int_0^{\pi} \frac{ \sin^2{(n+1) \theta} - \sin^2{n\theta} }{ \sin^2{\theta}}~d\theta $

    Note that $\displaystyle \sin^2(A) - \sin^2(B) = \sin(A-B)\sin(A+B)$

    $\displaystyle I_{n+1} - I_n = \int_0^{\pi} \frac{ \sin{\theta} \sin{ (2n+1)\theta}}{ \sin^2{\theta}}~d\theta $

    Consider $\displaystyle J_n = \int_0^{\pi} \frac{ \sin{ (2n+1)\theta}}{ \sin{\theta}}~d\theta $

    Again $\displaystyle J_{n} - J_{n-1} = $

    $\displaystyle \int_0^{\pi} \frac{ \sin{ (2n+1)\theta} - \sin{ (2n-1)\theta} }{ \sin{\theta}}~d\theta $

    $\displaystyle = 2\int_0^{\pi} \cos{2n\theta}~d\theta = 0 $

    Therefore , $\displaystyle J_n = J_{n-1} = ... = J_1 = J_0 $

    and $\displaystyle J_0 = \int_0^{\pi} d\theta = \pi $

    so $\displaystyle I_{n+1} - I_n = \pi $ with $\displaystyle I_0 = 0 $

    Finally $\displaystyle I_n = n\pi $
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