Solve the integral thing

Likely to have a simpler solution, but mine is as follows:

$\begin{aligned} I & = \int_{-\frac \pi 2}^\frac \pi 2 \frac {dx}{a+\sin x} & \small \color{#3D99F6} \text{Using }t = \tan \frac x2 \implies dt = \frac 12 \sec^2 \frac x2 \\ & = \int_{-1}^1 \frac {2\ dt}{at^2+2t+a} & \small \color{#3D99F6} \implies \sin x = \frac {2t}{1+t^2} \\ & = \frac 2a \int_{-1}^1 \frac {dt}{t^2+\frac 2a t+1} \\ & = \frac 2a \int_{-1}^1 \frac {dt}{\left(t+\frac 1a\right)^2 -\frac 1{a^2} +1} \\ & = \frac {2a}{a^2-1} \int_{-1}^1 \frac {dt}{\frac {a^2}{a^2-1} \left(t+\frac 1a\right)^2 +1} & \small \color{#3D99F6} \text{Let } u = \frac a{\sqrt{a^2-1}} \left(t+\frac 1a\right) \\ & = \frac 2{\sqrt{a^2-1}} \int_{-\frac {a-1}{\sqrt{a^2-1}}}^{\frac {a+1}{\sqrt{a^2-1}}} \frac {du}{u^2+1} \\ & = \frac 2{\sqrt{a^2-1}} \tan^{-1} u\ \bigg|_{-\frac {a-1}{\sqrt{a^2-1}}}^{\frac {a+1}{\sqrt{a^2-1}}} \\ & = \frac 2{\sqrt{a^2-1}} \left(\tan^{-1} \frac {a+1}{\sqrt{a^2-1}} + \tan^{-1} \frac {a-1}{\sqrt{a^2-1}}\right) \\ & = \frac 2{\sqrt{a^2-1}} \tan^{-1} \left(\frac {\frac {a+1}{\sqrt{a^2-1}} + \frac {a-1}{\sqrt{a^2-1}}}{1- \frac {a+1}{\sqrt{a^2-1}} \times \frac {a-1}{\sqrt{a^2-1}}}\right) \\ & = \frac 2{\sqrt{a^2-1}} \times \frac \pi 2 \\ & = \frac \pi{\sqrt{a^2-1}} \end{aligned}$

Therefore, we have $a^2 - 1 = a$ . $\implies a^2-a-1=0$ , $\implies a = \dfrac {1+\sqrt 5}2 \approx \boxed{1.618} = \varphi$ , the golden ratio .