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Math Help - The beta function integration

  1. #1
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    The beta function integration

    the integral from 0 to pi/4 of (cot2theta)^1/3

    This is suppose to somehow transform into the form of cos^(a)theta.sin^(b)theta
    where a and b are constants

    but i've tried so many times to put it in this form but the 1/3 is just making things difficult for me
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  2. #2
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    Quote Originally Posted by CookieC View Post
    the integral from 0 to pi/4 of (cot2theta)^1/3

    This is suppose to somehow transform into the form of cos^(a)theta.sin^(b)theta
    where a and b are constants

    but i've tried so many times to put it in this form but the 1/3 is just making things difficult for me
    The definition of B(x,y) is \displaystyle B(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}dt, the limits being from t=0 to t=1. The given integral goes from \theta=0 to \theta=\pi/4. So you need a change of variable which achieves that change of limits. I suggest making the substitution t = \sin^2(2\theta), which has that effect. Then dt = 4\sin(2\theta)\cos(2\theta)\,d\theta = 4t^{1/2}(1-t)^{1/2}d\theta, and the integral becomes

    \begin{aligned}\displaystyle\int_0^{\pi/4}\frac{(\cos(2\theta))^{1/3}}{(\sin(2\theta))^{1/3}}d\theta &= \int_0^1\frac{(1-t)^{1/6}}{t^{1/6}}\,\frac{dt}{4t^{1/2}(1-t)^{1/2}} \\ &= \frac14\int_0^1t^{-2/3}(1-t)^{-1/3}\,dt \\ &=\tfrac14 B\bigl(\tfrac13,\tfrac23\bigr).\end{aligned}
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