TITLE: calculate $\int_{0}^{\pi} \frac{dx}{a+\sin^2(x)} $using complex analysis
QUESTION [6 upvotes]: where $a>1$ calculate
$$\int_{0}^{\pi} \dfrac{dx}{a+\sin^2(x)}$$
I tried to use the regular $z=e^{ix}$ in $|z|=1$ contour. ($2\sin(x) = z-\frac1z)$, but it turned out not to work well because $\sin(x)$ in denominator.  could not found a good variables change either.

REPLY [4 votes]: Assume that $a\not\in[-1,0]$ so that the integral converges.
$$
\begin{align}
&\int_{0}^{\pi}\frac{\mathrm{d}x}{a+\sin^2(x)}\\
&=\frac12\int_0^{2\pi}\frac{\mathrm{d}x}{a+\sin^2(x)}\tag{1}\\
&=\frac12\oint\frac1{a-\left(z-1/z\right)^2/4}\frac{\mathrm{d}z}{iz}\tag{2}\\
&=\frac2i\oint\frac{z}{4az^2-(z^2-1)^2}\mathrm{d}z\tag{3}\\
&=\frac2i\oint\frac1{4aw-(w-1)^2}\mathrm{d}w\tag{4}\\
&=\small\frac1{2i\sqrt{a(a+1)}}\oint\left(\frac1{w-\left(1+2a-2\sqrt{a(a+1)}\right)}-\frac1{w-\left(1+2a+2\sqrt{a(a+1)}\right)}\right)\,\mathrm{d}w\tag{5}\\
&=\left\{\begin{array}{}
+\dfrac\pi{\sqrt{a(a+1)}}&\text{if }a\gt0\\
-\dfrac\pi{\sqrt{a(a+1)}}&\text{if }a\lt-1\tag{6}
\end{array}\right.
\end{align}
$$
Explanation:
$(1)$: symmetry using $\sin^2(x+\pi)=\sin^2(x)$
$(2)$: substitute $z=e^{ix}$
$(3)$: algebra
$(4)$: $w=z^2$; note that as $z$ circles the origin once, $w$ circles twice
$(5)$: partial fractions
$(6)$: note which singularity is in the unit circle