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Math Help - How do you evaluate this Integral by hand?

  1. #1
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    Exclamation How do you evaluate this Integral by hand?

    Hello everyone,
    I am curious about how to evaluate the following integral without the use of a computer.
    \int\((1-\beta^{2})\sin(\varphi)d\varphi/(1-\beta^{2}\sin(\varphi)^{2})^{3/2}
    Can you show some steps so that I can get the idea of it? Thanks in advance. This integral is from Berkeley Physics Course V2.
    Last edited by Ackbeet; May 5th 2011 at 02:44 PM.
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  2. #2
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    Okay, here is how you do it. Rewrite the integral and put t = cosφ then sub t = √[(1-β)/β]tanx.

    \begin{aligned}\displaystyle I & =  \int\frac{(1-\beta^2)\sin{\varphi}}{(1-\beta^2\sin^2{\varphi})^{\frac{3}{2}}}\;{d\varphi} = \int\frac{(1-\beta^2)\sin{\varphi}}{(1-\beta^2+\beta^2\cos^2{\varphi})^{\frac{3}{2}}}\;{d  \varphi} = (\beta^2-1)\int\frac{1}{(1-\beta^2+\beta^2t^2)^{\frac{3}{2}}}\;{dt} \\& = \frac{(\beta^2-1)}{\beta^3}\int\frac{1}{\bigg(\frac{1-\beta^2}{\beta^2}+t^2\bigg)^{\frac{3}{2}}}\;{dt} = \frac{(\beta^2-1)}{\beta^3}\int\frac{1}{\bigg[\left(\sqrt{\frac{1-\beta^2}{\beta^2}}\right)^2+t^2\bigg]^{\frac{3}{2}}}\;{dt} \\& =  \frac{(\beta^2-1)\beta^2}{\beta^3(1-\beta^2)}\int\cos{x}\;{dx} = -\frac{1}{b}\sin{x}+k = -\frac{1}{b}\sin\tan^{-1}\left(\frac{\beta t}{\sqrt{1-\beta^2}}\right)+k \\&  = -\frac{1}{b}\sin\tan^{-1}\left(\frac{\beta \cos{\varphi}}{\sqrt{1-\beta^2}}\right)+k = -\frac{\cos{\varphi}}{\sqrt{1-\beta^2\sin^2{\varphi}}}+k. \end{aligned}
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  3. #3
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    Thank you. Also Maple evaluates a much more complicated integral with many terms. What may be the reason behind this?
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  4. #4
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    There are sometimes many different ways to integrate a particular equation, Maple probably has a default method as a first attempt.
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  5. #5
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    Quote Originally Posted by JohnDoe View Post
    Thank you. Also Maple evaluates a much more complicated integral with many terms. What may be the reason behind this?
    I'm afraid I've no idea how computer algebra systems work, but I'm glad that my solution was simpler!
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