$\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x$ is equal to
Answer & explanation
Correct answer: option 4
Let $I=\int\limits_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x$ ....(i)
Then, $I=\int\limits_{\pi / 5}^{3 \pi / 10} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x$
$\Rightarrow I=\int\limits_{\pi / 5}^{3 \pi / 10} \frac{\cos x}{\cos x+\sin x} d x$ ....(ii)
Adding (i) and (ii), we get
$2 I=\int\limits_{\pi / 5}^{3 \pi / 10} 1 d x=\frac{3 \pi}{10}-\frac{\pi}{5}=\frac{\pi}{10} \Rightarrow I=\frac{\pi}{20}$