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I came across this integral in evaluating the time a particle takes to travel between points on the cycloid. I was able to use substitution to do the integral in the case $b = \pi/2.$ I have not not been able to show the result for other values of $b.$

Is there a way to evaluate this integral analytically?

Sasha
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abel
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  • Mathematica (v. 9) does not evaluate it. – Julián Aguirre Mar 22 '13 at 17:01
  • What you can try is to multiply top and bottom by conjugate: 1-cosx. That will give you a sin²x in the NUM of which you can take the square root.In the DENOM you now have a square root with cosx terms. Then do a u-sub cosx = u Now you have a constant in the NUM and a squareroot of a quadratic term in terms of u (because cosb is just a number) in the DENOM of which the head coefficient is negative. So now we need to complete the square to get it in a form a² - u² on which then another trig sub can be performed. Technically this should work but I am not savvy enough to type this up in LaTeX. – imranfat Mar 22 '13 at 17:17

2 Answers2

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Use the fact that: $$1+\cos(\theta) = 2\cos^2\left(\frac{\theta}{2}\right)$$ $$\cos(\theta)-\cos(b)=\cos ^2\left(\frac{\theta}{2}\right)-\sin ^2\left(\frac{\theta}{2}\right)-\cos(b)$$ Now subsitute $u=2\sin(\theta/2),\ du = \cos(\theta/2)d\theta$ to get: $$\int_0^{u(b)} \frac{2du}{\sqrt{2-u^2-2\cos(b)}}$$ Now you can use an the following integral: $$\int\frac{dx}{a^2-x^2} =\arctan\left(\frac{x\sqrt{a^2-x^2}}{a^2-x^2}\right)$$

Nathaniel Bubis
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Use the fact that

$$1+\cos{\theta} = 2 \cos^2{(\theta/2)}$$ $$\cos{\theta}-\cos{b} = 2 (\sin^2{(b/2)}-\sin^2{(\theta/2)})$$

Then let $u=\cos{(\theta/2)}$ in the integral:

$$2 \int_0^{2 \arcsin{b}} \frac{du}{\sqrt{\sin^2{(b/2)}-u^2}}$$

Mathematica (or you) should be able to handle this one. The stated result follows.

Ron Gordon
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