How do you evaluate this Integral by hand?

**JohnDoe** · May 5th 2011, 12:32 PM

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.

**TheCoffeeMachine** · May 5th 2011, 04:48 PM

Okay, here is how you do it. Rewrite the integral and put t = cosφ then sub t = √[(1-β²)/β²]tanx.

$\begin{aligned}\displaystyle I & = \int\frac{(1-\beta^2)\sin{\varphi}}{(1-\beta^2\sin^2{\varphi})^{\frac{3}{2}}}\;{d\varphi} = \int\frac{(1-\beta^2)\sin{\varphi}}{(1-\beta^2+\beta^2\cos^2{\varphi})^{\frac{3}{2}}}\;{d \varphi} = (\beta^2-1)\int\frac{1}{(1-\beta^2+\beta^2t^2)^{\frac{3}{2}}}\;{dt} \\& = \frac{(\beta^2-1)}{\beta^3}\int\frac{1}{\bigg(\frac{1-\beta^2}{\beta^2}+t^2\bigg)^{\frac{3}{2}}}\;{dt} = \frac{(\beta^2-1)}{\beta^3}\int\frac{1}{\bigg[\left(\sqrt{\frac{1-\beta^2}{\beta^2}}\right)^2+t^2\bigg]^{\frac{3}{2}}}\;{dt} \\& = \frac{(\beta^2-1)\beta^2}{\beta^3(1-\beta^2)}\int\cos{x}\;{dx} = -\frac{1}{b}\sin{x}+k = -\frac{1}{b}\sin\tan^{-1}\left(\frac{\beta t}{\sqrt{1-\beta^2}}\right)+k \\& = -\frac{1}{b}\sin\tan^{-1}\left(\frac{\beta \cos{\varphi}}{\sqrt{1-\beta^2}}\right)+k = -\frac{\cos{\varphi}}{\sqrt{1-\beta^2\sin^2{\varphi}}}+k. \end{aligned}$

**JohnDoe** · May 7th 2011, 01:28 PM

Thank you. Also Maple evaluates a much more complicated integral with many terms. What may be the reason behind this?

**pickslides** · May 7th 2011, 01:35 PM

There are sometimes many different ways to integrate a particular equation, Maple probably has a default method as a first attempt.

**TheCoffeeMachine** · May 7th 2011, 01:55 PM

Originally Posted by JohnDoe

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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