Integral thing ... 1

$\begin{aligned} I & = \int_{-\frac \pi 2}^\frac \pi 2 \frac {dx}{3+\sin x} & \small \color{#3D99F6} \text{Using }t = \tan \frac x2 \implies dt = \frac 12 \sec^2 \frac x2 \\ & = \int_{-1}^1 \frac {2\ dt}{3t^2+2t+3} & \small \color{#3D99F6} \implies \sin x = \frac {2t}{1+t^2} \\ & = \frac 34 \int_{-1}^1 \frac {dt}{\frac 98\left(t+\frac 13\right)^2 +1} & \small \color{#3D99F6} \text{Let } u = \frac 3{\sqrt 8} \left(t+\frac 13\right) \\ & = \frac 1{\sqrt 2} \int_{-\frac 1{\sqrt 2}}^{\sqrt 2} \frac {du}{u^2+1} \\ & = \frac 1{\sqrt 2} \tan^{-1} u\ \bigg|_{-\frac 1{\sqrt 2}}^{\sqrt 2} \\ & = \frac 1{\sqrt 2} \left(\tan^{-1} \sqrt 2 + \tan^{-1} \frac 1{\sqrt 2}\right) \\ & = \frac 1{\sqrt 2} \tan^{-1} \left(\frac {\sqrt 2 + \frac 1{\sqrt 2}}{1- \sqrt 2 \times \frac 1{\sqrt 2}}\right) \\ & = \frac 1{\sqrt 2} \times \frac \pi 2 = \frac \pi {\sqrt 8} \end{aligned}$

$\implies a = \boxed{8}$