_0^ / 2 x+ x 1+ x d x= - Vidyadip

MCQ

$\int_0^{\pi / 2} \frac{x+\sin x}{1+\cos x} d x=$

A
$-\log 2$
B
$\log 2$
✓
$\frac{\pi}{2}$
D
$0$

✓

Answer

Correct option: C.

$\frac{\pi}{2}$

(C)
$\int_0^{\frac{\pi}{2}} \frac{x+\sin x}{1+\cos x} d x=\int_0^{\frac{\pi}{2}}\left[\frac{x+2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right] d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{2}}\left[x \sec ^2 \frac{x}{2}+2 \tan \frac{x}{2}\right] d x$
$\begin{array}{l}=\frac{1}{2}\left[2 x \tan \frac{x}{2}\right]_0^{\pi / 2}-\int_0^{\frac{\pi}{2}} \tan \frac{x}{2} d x+\int_0^{\frac{\pi}{2}} \tan \frac{x}{2} d x \\ =\left[x \tan \frac{x}{2}\right]_0^{\pi / 2} \\ =\frac{\pi}{2} \tan \frac{\pi}{4}=\frac{\pi}{2}\end{array}$

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