Trigonobernic calculus

Similar solution with @JD Money 's

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {\sin x}{\sin x + \cos x} dx & \small \color{#3D99F6} \text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac {\sin x}{\sin x + \cos x} + \frac {\sin(\pi -x)}{\sin (\pi -x) + \cos (\pi -x)} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac {\sin x}{\sin x + \cos x} + \frac {\cos x}{\cos x+\sin x} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 dx = \frac x2 \ \bigg|_0^\frac \pi 2 = \frac \pi 4 \end{aligned}$

$\implies a = \boxed{4}$ .

Use integral from a to b of f(x)dx = integral from a to b of f(a+b-x)dx