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Math Help - Integration - Reduction formulas

  1. #1
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    Integration - Reduction formulas

    By writing  \sin^{2n} (x) = \sin^{2n-2}( x ) [1- \cos^2( x )] , use integration by parts (just once, in the second term) to derive the reccurence relation:

    I_{2n} = \left(\frac{2n-1}{2n}\right)I_{2n-2} (n \ge 1), for  I_{2n} = \int_{0}^{\pi} \sin^{2n}( x )dx.

    Hence find the value of I_4 without using your calculator at any stage.
    Last edited by mr fantastic; January 13th 2009 at 04:01 PM. Reason: Fixed the latex
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  2. #2
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    Quote Originally Posted by mitch_nufc View Post
    By writing  sin^{2n}( x ) = [sin^{2n-2}( x )][1-cos^2( x )] , use integration by parts (just once, in the second term) to derive the reccurence relation:
    I_{2n} = (\frac{2n-1}{2n})I_{2n-2} (n \ge 1), for  I_{2n} = \int_{0}^{\pi}sin^{2n}( x )dx.
    Hence find the value of I_4 without using your calculator at any stage.
     I_{2n} = \int_{0}^{\pi}sin^{2n}( x )dx

      = \int_{0}^{\pi}[sin^{2n-2}( x )][1-cos^2( x )]dx

      = \int_{0}^{\pi}sin^{2n-2}( x )-sin^{2n-2}cos^2( x )dx

    Leave the first term as it is, since you're trying to find a solution in terms of \int_{0}^{\pi}sin^{2n-2}(x), since  I_{2n-2} = \int_{0}^{\pi} sin^{2n-2}(x)

    So the integral is now:

      I_{2n} = I_{2n-2}-\int_{0}^{\pi} sin^{2n-2}(x)cos^2( x )dx

    Carry out integration by parts on the 2nd term of the integral and again try to manipulate it so that you get an expression in terms of \int_{0}^{\pi}sin^{2n-2}(x). You should then be able to factor out \int_{0}^{\pi}sin^{2n-2}(x) of the whole expression, and the multiple should be  (\frac{2n-1}{2n}).

    Hint: When it comes to integrating that term by parts, it is difficult to integrate  \int_{0}^{\pi}sin^{2n-2}( x ), hence that is the term you should differentiate!

    Hint #2:  \int cos^2(x) = \frac{x}{2} + \frac{sin(2x)}{4} +C
    Last edited by Mush; January 13th 2009 at 06:18 PM.
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