# Replacement $\int_{0}^{2\pi }\frac{1}{a+\sin(3t)}dt$

I can not understand why a replacement I tried in resolution with the residual method of the goniometric integrals leads to a wrong result.

$$\int_{0}^{2\pi }\frac{1}{a+\sin(3t)}dt$$

$$a\in \mathbb{R} |a|<1$$

I thought

\begin{align}\sin(3t)&=\frac{e^{3\vartheta i}-e^{-3\vartheta i}}{2i}\\ z&=e^{3\vartheta i}\\ d\vartheta&=\frac{dz}{3iz}\\ \frac{1}{3}\oint_{\gamma }\frac{1}{z^2+2aiz-1}dz&=\frac{2\pi }{3\sqrt{a^{2}-1}}\end{align}

\begin{align}\sin(3t)&=\frac{e^{3\vartheta i}-e^{-3\vartheta i}}{2i}\\ z&=e^{\vartheta i}\\ d\vartheta&=\frac{dz}{iz}\\ \oint_{\gamma }\frac{2z^{2}}{z^6+2aiz^3-1}dz&=\frac{2\pi }{\sqrt{a^{2}-1}}\end{align}

the second resolution gives the correct result but is much longer. The two integrals should coincide. The results of residues I think are correct I checked them with Mathematica. I fear there is something even trivial wrong in replacement but do not understand what.

Thank everybody.

• i have $$\frac{2 \pi \sqrt{\frac{a^2}{a^2-1}}}{a}$$ – Dr. Sonnhard Graubner Mar 23 '17 at 14:25
• is $$a$$ assumed to be positive? – Dr. Sonnhard Graubner Mar 23 '17 at 14:26
• why this adjective "goniometric" ? – Jean Marie Mar 23 '17 at 14:36
• @JeanMarie I'm thinking it's intended to be "angular" or "polar". – Joffan Mar 23 '17 at 16:06
• @Joffan, thanks, it looks very likely ; could you confirm Stefano Barone ? It's your question after all... – Jean Marie Mar 23 '17 at 16:13

First of all in both cases the contour on which we have to integrate is same but in first case as t varies from 0 to 2$\pi$ we are integrating around the contour 3 times but in second case only once
Note that $\sin 3t$ goes three times around the circle in the space of $t\in [0,2\pi]$ but you are only integrating once around in your "first thought" option.