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Math Help - Evalute this DI

  1. #1
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    Evalute this DI

    \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
    \int_{0}^{\pi}\frac{\texttt{sin}^2n\theta}{\texttt  {sin}^2\theta}\texttt{d}\theta
    What's 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 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  I_n = \int_0^{\pi} \frac{ \sin^2{n\theta} }{ \sin^2{\theta}}~d\theta


    We have

     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  \sin^2(A) - \sin^2(B) = \sin(A-B)\sin(A+B)

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

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

    Again  J_{n} - J_{n-1} =

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

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

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

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

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

    Finally  I_n = n\pi
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