Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{\sin x}{\sin x+\cos x}dx$ equals to

(a) π
(b) π/2
(c) π/3
(d) π/4

Answer 1

(d) π/4

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x}{\sin x+\cos x}\mathit{d}x\mathit{}.....\left(1\right) \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}-x}\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)+\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}\mathit{d}x \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos x}{\cos x+\sin x}dx \\ \therefore I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos x}{\sin x+\cos x}dx.....\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{\sin x}{\sin x+\cos x}+\frac{\cos x}{\cos x+\sin x}\right]\mathit{d}x \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{\sin x+\cos x}{\sin x+\cos x}\right]\mathit{d}x \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence}I=\frac{\mathrm{\pi }}{4} \end{gathered}$$