So I tried to solve $\int_{0}^{2\pi} \frac{d\theta}{a + b\sin\theta}$ using contour integration method, assuming that $a > |b|$. I found the corresponding integral to be $$\oint_{|z|=1} \frac{2}{bz^2 + 2aiz - b} \, dz$$ and the poles $$\frac{-2ai \pm \sqrt{4b^2 - 4a^2}}{2b} $$ I selected the "+" pole. As for the residue at that pole $$\text{Res}(f(z), z_1) = \lim_{z \to z_1} \frac{2}{bz^2 + 2aiz - b}$$ $$= \lim_{z \to z_1} \frac{2}{2bz + 2ai}$$ If I substitute my pole here, I get $$\frac{1}{\sqrt{b^2 - a^2}}$$ But according to Cauchy's Residue theorem, $I = 2\pi i \cdot \sum \text{Res}$. I will be left with a complex answer, which isn't the correct result.
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2Hint: Assuming $|a|> |b|$ from the start might be wise. – Hyperon Apr 14 '25 at 14:04
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@Hyperon Yes a > |b| is included in the question, I forgot to mention, this is why I selected that pole. – Abdullah Omar Nasseef Apr 14 '25 at 14:06
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1OK, so what does this imply for $b^2-a^2$? – Hyperon Apr 14 '25 at 14:11
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2@Hyperon Ohh I see it now, thanks a lot – Abdullah Omar Nasseef Apr 14 '25 at 14:20
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Related: https://math.stackexchange.com/questions/1740458/finding-int-fracdxab-cos-x-without-weierstrass-substitution – Integreek Apr 14 '25 at 15:12
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As a > |b|, $\frac{1}{\sqrt{b^2 - a^2}}$ can be written as $$\frac{1}{i\sqrt{a^2 - b^2}}$$ This cancels out the $i$ in $I = 2\pi i \cdot \sum \text{Res}$ and gives the correct answer, which is $$\frac{2\pi}{\sqrt{a^2 - b^2}}$$ Thanks to @Hyperon for helping me out here