_ ^ x + x 1 + x ;dx is equal to - Vidyadip

MCQ

$\int_{}^{} {\frac{{x + \sin x}}{{1 + \cos x}}\;dx} $ is equal to

A
$\frac{1}{2}x\tan \frac{x}{2} + c$
✓
$x\tan \;\frac{x}{2} + c$
C
$x\tan x + c$
D
$\frac{1}{2}x\tan x + c$

✓

Answer

Correct option: B.

$x\tan \;\frac{x}{2} + c$

b
(b)$\int_{}^{} {\frac{{x + \sin x}}{{1 + \cos x}}\,dx = \frac{1}{2}\int_{}^{} {x{{\sec }^2}\frac{x}{2}\,dx + \int_{}^{} {\tan \frac{x}{2}\,dx} } } $
$ = \frac{1}{2}\frac{{x\tan \frac{x}{2}}}{{\frac{1}{2}}} - \int_{}^{} {\tan \frac{x}{2}\,dx} + \int_{}^{} {\tan \frac{x}{2}\,dx} $$ = x\tan \frac{x}{2} + c$.
Trick : By inspection, $\frac{d}{{dx}}\left\{ {x\tan \frac{x}{2} + c} \right\}$
$ = \frac{x}{2}{\sec ^2}\frac{x}{2} + \tan \frac{x}{2} = \frac{1}{2}\left[ {\frac{x}{{{{\cos }^2}\frac{x}{2}}} + \frac{{2\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}} \right] = \frac{{x + \sin x}}{{1 + \cos x}}$.

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