Using residue theorem to calculate integral $\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}$ - where is my mistake?

Question

I am to calculate: $$ \int\limits_0^{2\pi} \frac{dx}{10+6\sin x} $$

We can set $\gamma(t)=e^{it}$ for $t \in [0, 2\pi]$ and then $z = e^{it}$, $\dfrac{dz}{iz}=dt$, $\sin t =\dfrac{1}{2i}(z-\frac{1}{z})$ so that:

$$ \int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = \int\limits_\gamma \frac{dz}{\left(10+\frac{3}{i}(z-\frac{1}{z})\right)iz} = \int\limits_\gamma \frac{dz}{\left(10iz+3z^2-3\right)} $$ Roots of denominator are $-3i$ and $\frac{-i}{3}$ but since the winding number of $-3i$ is equal to 0 we have: $$ \int\limits_\gamma \frac{dz}{10iz+3z^2-3} = 2 \pi\, i\, Res\left(f,\frac{-i}{3}\right)\cdot1 $$ Calculating residue: $$ Res\left(f,\frac{-i}{3}\right) = \lim_{\large z \to \frac{-i}{3}} \frac{1}{(z+3i)}=\frac{3}{8i} $$ summing up: $$ \int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = 2\pi i \cdot \frac{3}{8i} = \frac{3\pi}{4} $$ But wolfram says it is equal to $\dfrac{\pi}{4}$. Could you help me spot my mistake?

Answer 1

The mistake comes from calculating the residue.

We have $Res(f, -\frac{i}{3})=lim_{z\rightarrow\frac{-i}{3}}\frac{z+\frac{i}{3}}{(z+3i)(3z+i)}=\frac{1}{3}lim_{z\rightarrow\frac{-i}{3}}\frac{3z+i}{(z+3i)(3z+i)}=\frac{1}{3}lim_{z\rightarrow\frac{-i}{3}}\frac{1}{(z+3i)}=\frac{1}{3}\frac{1}{\frac{-i}{3}+3i}=\frac{1}{3}\frac{1}{\frac{8i}{3}}=\frac{1}{3}\frac{-3i}{8}=\frac{-i}{8}$.

Answer 2

$$ \operatorname{Res}\left(f,\frac{-i}{3}\right) = \lim_{z \to \frac{-i}{3}} \frac{1}{\color {red}3(z+3i)}=\frac{1}{8i}. $$

Answer 3

You made mistake while finding residue $$ Res\left(f,\frac{-i}{3}\right) = \lim_{z \to \dfrac{-i}{3}} \frac{1}{3(z+3i)}=\frac{1}{3\left(-\frac i3+3i\right)}=\frac{1}{8i} $$ $$ \therefore \int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = 2\pi i \cdot \frac{1}{8i} = \frac{\pi}{4} $$