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

Answer

Correct option: C.
$\frac{\pi }{4}$
c
(c) $I = \int_0^{\pi /2} {\frac{{\sin x.dx}}{{\sin x + \cos x}}} = \int_0^{\pi /2} {\frac{{\cos x.dx}}{{\cos x + \sin x}}} $,

$\,\,\left( \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right)$

$2I = \int_0^{\pi /2} {dx} \Rightarrow I = \frac{\pi }{4}$.

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