MCQ
$\int_0^\pi \frac{1}{1+\sin x} dx$ equals
  • $2$
  • B
    $\frac{3}{2}$
  • C
    $\frac{1}{2}$
  • D
    $0$

Answer

Correct option: A.
$2$
$\int_0^\pi \frac{1}{1+\sin x} d x$
$=\int_0^\pi \frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x$
$=\int_0^\pi \frac{1-\sin x}{1-\sin ^2 x} d x$
$=\int_0^\pi \frac{1-\sin x}{\cos ^2 x} d x$
$=\int_0^\pi\left(\sec ^2 x-\sec x \tan x\right) d x$
$=[\tan x-\sec x]_0^\pi$
$=0+1-0+1$
$=2 $

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