Question

\[\int\limits_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx\]  equals to

Answer 1

 π/4

\[\text{We have}, \]
\[ I = \int_0^\frac{\pi}{2} \frac{\sin x}{\sin x + \cos x} d x . . . . . \left( 1 \right)\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\sin\left( \frac{\pi}{2} - x \right)}{\sin\left( \frac{\pi}{2} - x \right) + \cos\left( \frac{\pi}{2} - x \right)} d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\cos x}{\cos x + \sin x} dx \]
\[ \therefore I = \int_0^\frac{\pi}{2} \frac{\cos x}{\sin x + \cos x} dx . . . . . \left( 2 \right)\]
\[\text{Adding} \left( 1 \right) and \left( 2 \right), \text{we get}\]
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{\sin x}{\sin x + \cos x} + \frac{\cos x}{\cos x + \sin x} \right] d x\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{\sin x + \cos x}{\sin x + \cos x} \right] d x\]
\[ = \int_0^\frac{\pi}{2} dx\]
\[ = \left[ x \right]_0^\frac{\pi}{2} \]
\[ = \frac{\pi}{2}\]
\[Hence\ I = \frac{\pi}{4}\]