$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2\sqrt{a(a+1)}}$, for $a > 0$ I want to show that
$$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2\sqrt{a(a+1)}}$$
for $a > 0$.
I try several methods


*

*Substitutions to rationalize

*The famous $U = \tan(\frac{\theta}{2})$

*Multiplying by conjugates


and other calculus techniques and still, I can not prove the equality. 
Anyone can give me a hint of how should I begin to work this integral?
I'm very ashamed that I can not solve this problem. 


\begin{align} \int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a +
> \sin^2(\theta)}  &=  \frac{\pi}{2[a(a+1)]^\frac{1}{2}} \end{align}

\begin{align}
I=
\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} 
&= 
\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{(\sqrt{a+1} + \cos\theta)
 (\sqrt{a+1} - \cos\theta} 
\\
&=\frac1{2\sqrt{a+1}}
\left(
\int_0^{\tfrac\pi2}
\frac{d\theta}{\sqrt{a+1} + \cos\theta}
+
\int_0^{\tfrac\pi2}
\frac{d\theta}{\sqrt{a+1} - \cos\theta}
\right)
.
\end{align}
Now we can use  a table integral
\begin{align}
\int \frac{dx}{u+v\cos x}
&=
\frac2{\sqrt{u^2-v^2}}
\arctan\left( \frac{(u-v)\tan\tfrac{x}2}{\sqrt{u^2-v^2}} \right)
,\quad u^2>v^2
\end{align}
for $u=\sqrt{a+1}$, $v=\pm 1$.
\begin{align}
I&=
\frac1{2\,u}
\frac2{\sqrt{u^2-1}}
\left(
\arctan\left(
\sqrt{\frac{u-1}{u+1}}\tan\tfrac\pi4
\right)
+
\arctan\left(
\sqrt{\frac{u+1}{u-1}}\tan\tfrac\pi4
\right)
\right)
,\\
&=
\frac1{u\,\sqrt{u^2-1}}
\left(
\arctan\left(
\sqrt{\frac{u-1}{u+1}}
\right)
+
\arctan\left(
\sqrt{\frac{u+1}{u-1}}
\right)
\right)
.
\end{align}
And since $\arctan x+\arctan \tfrac1x=\tfrac\pi2$ for $x>0$, we have
\begin{align}
I&=
\frac{\pi}{2\,\sqrt{a+1}\,\sqrt{a}}
.
\end{align}
