\begin{align} \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \end{align}
Substitute $t = \pi/2 - x$, $dt = -dx$
\begin{align} I &= \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx\newline &= \int_{\pi/2}^{0} \frac{\sin (\pi/2 - t)}{\sin (\pi/2 - t) + \cos (\pi/2 - t)} (-dt)\newline &= -\int_{\pi/2}^{0} \frac{\cos t}{\cos t + \sin t} dt\newline &= \int_{0}^{\pi/2} \frac{\cos t}{\cos t + \sin t} dt = I\newline \implies 2I &= \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx + \int_{0}^{\pi/2} \frac{\cos t}{\cos t + \sin t} dt\newline 2I &= \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx = \int_{0}^{\pi/2} dx\newline &= \frac{\pi}{2}\newline \implies I &= \frac{\pi}{4} \end{align}