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trigonometric integral

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Post Sun Nov 27, 2016 2:23 pm

Posts: 82
Calculation of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx$

Post Sun Nov 27, 2016 6:14 pm

Posts: 8
We know that




So consider the substitution $z=e^{ix}$ and be mindful of where your poles are. This should give you a contour integral of the form


where $C$ is a positively oriented circular contour of unit radius centered over $z=0$. The problem can be finished using Cauchy's integral formula.
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Post Wed Dec 07, 2016 7:27 am

Posts: 27
The following will be assumed:

$\displaystyle \int_0^{2\pi} \frac{\text{d}\theta}{r-\cos{\theta}} = \frac{2\pi}{\sqrt{r^2-1}}$ where $r>1$


Multiply and divide by $b^2$ in order to long divide by inspection.

$\displaystyle \frac{1}{b^2} \int_0^{2\pi} \frac{b^2-a^2 + a^2 - b^2 \cos^2{x}}{a-b\cos{x}} \text{d}x = \frac{1}{b^2} \int_0^{2\pi} a+b\cos{x} \, \text{d}x - \frac{a^2-b^2}{b^3} \int_0^{2\pi} \frac{\text{d}x}{\frac{a}{b}-\cos{x}}$

$\displaystyle = \frac{2\pi a}{b^2} - \frac{a^2-b^2}{b^3} \times \frac{2\pi}{\sqrt{\frac{a^2}{b^2} -1}} = \frac{2\pi}{b^2} \left( a-\sqrt{a^2-b^2} \right) = \frac{2\pi}{a+\sqrt{a^2-b^2}}$

Post Fri Dec 30, 2016 7:28 am

Posts: 82
Thanks Paradoxica

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