x (1+ x) d x=? - Vidyadip

MCQ

$\int \frac{\sin x}{(1+\sin x)} d x=?$

A
$x + \tan x - \sec(x) + C$
✓
$x+\frac{2}{\tan \frac{x}{2}+1}+c$
C
$x - \tan x - \sec(x) + C$
D
$x - \tan x + \sec(x) + C$

✓

Answer

Correct option: B.

$x+\frac{2}{\tan \frac{x}{2}+1}+c$

Given
$\int \frac{\sin x}{1+\sin x} d x$
$=\int d x-\int \frac{d x}{1+\sin x}$
$=x-\int \frac{d x}{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}$
$=x-\int \frac{d x}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}$
$=x-\int \frac{\sec ^2 \frac{x}{2} d x}{\left(\tan \frac{x}{2}+1\right)^2}$
Let, $\tan \frac{x}{2}+1=z$
$\Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d z$
So,
$x-\int \frac{2 d z}{z^2}$
$=x+\frac{2}{z}+c$
$=x+\frac{2}{\tan \frac{x}{2}+1}+c$
where $c$ is the integrating constant.

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