_ /2 ^ , 1 - x 1 - x dx =

MCQ

$\int\limits_{\pi /2}^\pi {\,\frac{{1 - \sin x}}{{1 - \cos x}}} $ $dx =$

✓
$1 - ln 2$
B
$ln 2$
C
$1 + ln 2$
D
$none$

✓

Answer

Correct option: A.

$1 - ln 2$

a
$I\,\, = \,\int {\frac{{1 - \sin x}}{{2{{\sin }^2}\frac{x}{2}}}\,\,\,\,dx} $= $\int {\left( {\frac{1}{2}\,\cos e{c^2}\frac{x}{2} - \cot \frac{x}{2}} \right)\,dx} $ $= -x cot$$\left. {\frac{x}{2}} \right|_{\,\frac{\pi }{2}}^{\,\pi }$

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