Definite Integral

Question: Compute the following

\[ \begin{aligned} \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \end{aligned} \]

Solution

Substitute \(t = \pi/2 - x\), \(dt = -dx\)

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