Trigonometric integral $\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}$ The function
$f(x_1,x_2)=\frac{e^{-\sqrt{x^2_1 + x^2_2}}}{\sqrt{x^2_1+x^2_2}}$, $(x_1,x_2)\neq(0,0)$ is integrable in $\mathbb{R}^2\setminus\{(0,0\})$ as $f>0$ and
\begin{align}
\int_{\mathbb{R}^2}f(x_1,x_2)\,dx_1dx_2=2\pi\int^\infty_0e^{-r}\,dr
\end{align}
I am interested in its Fourier transform. Being radial, we have that
\begin{align}
\widehat{f}(t_1,t_2)&=\widehat{f}\big(0,\sqrt{t^2_1+t^2_2}\big)=\int_{(0,\infty)\times(-\pi,\pi)}e^{-r}e^{-2\pi i r\sqrt{t^2_1+t^2_2}\sin\theta}\,drd\theta\\
&=\int^\pi_{-\pi} \frac{d\theta}{1+ 2\pi i \sqrt{t^2_1+t^2_2}\sin\theta}=\int^\pi_{-\pi} \frac{1-2\pi i \sqrt{t^2_1+t^2_2}\sin\theta}{1+ 4\pi^2 (t^2_1+t^2_2)\sin^2\theta}\,d\theta\\
&=2\int^\pi_0\frac{d\theta}{1+ 4\pi^2 (t^2_1+t^2_2)\sin^2\theta}
\end{align}
Question:
At this point, the last integral escapes me. I would like to see if there is either a more direct way to estimate the Fourier transform of $f$, or whether the integral
\begin{align}
\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}\tag{1}\label{one}
\end{align}
can be further evaluated.

Edit: As pointed out by @Koro,  it turns out that \eqref{one} is rather trivial:
\begin{align}
 \int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}&=2\int^{\pi/2}_0\frac{d\theta}{1+a^2-a^2\cos^2\theta}=2\int^{\pi/2}_0\frac{\sec^2\theta}{\sec^2\theta(1 +a^2)-a^2}\,d\theta\\
&=2\int^{\pi/2}_0\frac{\sec^2\theta}{1+(1+a^2)\tan^2\theta}\,d\theta
\stackrel{u=\tan\theta}{=}2\int^{\infty}_0\frac{du}{1+(1+a^2)u^2}\\
&=\frac{\pi}{\sqrt{1+a^2}}
\end{align}
Hence, the Fourier transform of $f$ is
$$\widehat{f}(t_1,t_2)=\frac{2\pi}{\sqrt{1+4\pi^2(t^2_1+t^2_2)}}$$
 A: Let $I=\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}=2\int_0^{\pi/2} \frac{d\theta}{1+a^2\sin^2\theta}$  Noting that $\csc^2\theta=1+\cot^2\theta$, divide numerator and denominator by $\sin^2\theta$ to get: 
$\frac I2=\int_0^{\pi/2} \frac{\csc^2\theta d\theta}{1+\cot^2\theta+a^2}$. Now substitue $t=\cot \theta$ so that $dt=-\csc^2\theta d\theta$ and change the limits accordingly to get 
$\frac I2=\int_0^\infty\frac{dt}{t^2+(1+a^2)}$
Can you take it from here?
A: \begin{aligned}\int_{0}^{\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}&=\frac{1}{2}\int_{0}^{\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}+\frac{1}{2}\int_{\pi}^{2\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\left(2\pi -\theta\right)}}}\\ &=\frac{1}{2}\int_{0}^{2\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}\\ &=\frac{1}{2\,\mathrm{i}}\oint_{\left|z\right|=1}{\frac{\mathrm{d}z}{z\left(1-\frac{a^{2}}{4}\left(z-\frac{1}{z}\right)^{2}\right)}}\\ &=2\,\mathrm{i}\oint_{\left|z\right|=1}{\frac{z}{a^{2}z^{4}-\left(2a^{2}+4\right)z^{2}+a^{2}}\,\mathrm{d}z}\\ &=2\,\mathrm{i}\left(2\pi\,\mathrm{i}\,\mathrm{Res}\left(f,\sqrt{1+\frac{2}{a^{2}}-2\sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}}\right)+2\pi\,\mathrm{i}\,\mathrm{Res}\left(f,-\sqrt{1+\frac{2}{a^{2}}-2\sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}}\right)\right)\\ &=2\,\mathrm{i}\left(2\pi\,\mathrm{i}\times\left(-\frac{1}{8a^{2}\sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}}\right)+2\pi\,\mathrm{i}\times\left(-\frac{1}{8a^{2}\sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}}\right)\right)\\ \int_{0}^{\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}&=\frac{\pi}{\sqrt{1+a^{2}}}\end{aligned}
