I found an integral online that I cannot seem to solve. I tried putting in Wolfram Alpha, however, I did not get any output. The integral is: $$\int\frac{dx}{\sqrt{1-\cos^3(x)}}$$ Does anyone have any hints on how to solve this?
How to solve this integral which involves trigonometry?
2
$\begingroup$
calculus
integration
trigonometry
indefinite-integrals
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4the result looks terrible – 2017-01-21
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0@Dr.SonnhardGraubner Did Wolfram Alpha give you an output? Or did you use Mathematica to evaluate the integral? – 2017-01-21
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0one idea is to write $$1-\cos(x)^3=(1-\cos(x))(\cos(x)^2+\cos(x)+1)$$ – 2017-01-21
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0Wolframalpha gave me an output. The result indeed looks horrible, and on top of that it uses elliptic integrals of the first and of the third kind. Where does your problem come from? – 2017-01-21
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0maybe there is a typo in the integral? – 2017-01-21
1 Answers
1
If there is not any typo, then the integral is surely not trivial.
Well, it's better for you. It will introduce you into the world of Elliptic Integrals, which sooner or later you shall study (maybe).
So I will write something for the sake of curiosity!
The result of the integral is
$$\int \frac{1}{\sqrt{1-\cos^3(x)}}\ \text{d}x = \frac{\sin (2 x)-2 \cos ^{\frac{3}{2}}(x) E\left(\left.\frac{x}{2}\right|2\right)}{\sqrt{\cos ^3(x)}}$$
Where the $E$ function is called: Elliptic integral of the second kind.
It's defined as follows (in your case):
$$E\left(\left.\frac{x}{2}\right|2\right) = \int_0^{x/2}\sqrt{1 - 2\sin^2\theta}\ \text{d}\theta$$
Or in general:
$$E\left(\left.\phi\right|m\right) = \int_0^{\phi}\sqrt{1 - m\sin^2\theta}\ \text{d}\theta$$