MCQ
$\int_0^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x=$
  • A
    $\frac{\pi}{2}$
  • B
    $\frac{\pi}{3}$
  • $\frac{\pi}{4}$
  • D
    $\frac{\pi}{6}$

Answer

Correct option: C.
$\frac{\pi}{4}$
(C)
$\text { Let } I=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x$ ....(i)
$=\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=\int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x$...(ii)
Adding (i) and (ii), we get
$2 I =\int_0^{\frac{\pi}{2}} d x=[x]_0^{\pi / 2}$
$\therefore 2 I =\frac{\pi}{2} \Rightarrow I =\frac{\pi}{4}$
Alternate Method:
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\int_0^{\frac{\pi}{2}} \frac{\cos ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\frac{\pi}{4}$
$\int_0^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x=\frac{\pi}{4}$ 

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