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Math Help - Integration by Parts

  1. #1
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    Integration by Parts

    Hi How do I do integrate:

    (sinx)^n * (cosx)^2 when it is a definite integral between x = pi/2 and x = 0
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  2. #2
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    Quote Originally Posted by Sunyata View Post
    Hi How do I do integrate:

    (sinx)^n * (cosx)^2 when it is a definite integral between x = pi/2 and x = 0
    Put cos(x) = t
    -sin(x)dx = dt
    dx= - dt/sin(x)
    So Integration I = - intg[(sinx)^n*t^2*dt/sin(x)
    = - intg[(sinx)^(n-1)*t^2*dt.
    Now do the integration by parts.
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  3. #3
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    This is NOT a candidate for "integration by parts". I would be inclined to use the fact that cos^2(x)= 1- sin^2(x) and write
    \int_0^{\pi/2} sin^n(x)(1- sin^2(x))dx= \int_0^{\pi/2} sin^n(x)dx- \int_0^{\pi/2} sin^{n+2}(x) dx.

    Those can be done by the "reduction integrals" given in any Calculus book.
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  4. #4
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    hi
    \int_{0}^{\frac{\pi }{2}}(\sin^n(x)\cos^2(x))dx= \int_{0}^{\frac{\pi }{2}}(\sin^n(x)\cos(x))(\cos(x))dx= \left [ \frac{\sin^{n+1}}{n+1}\cos(x) \right ]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac  {\sin^{n+2}(x)}{n+1}dx= \int_{0}^{\frac{\pi}{2}}\frac{\sin^{n+2}(x)}{n+1}d  x=\frac{1}{n+1}\int_{0}^{\frac{\pi}{2}}\sin^{n+2}(  x)dx
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