$\int \frac{\sin \frac{x}{2} . \tan \frac{x}{2}}{\cos x} d x$ is equal to

Moderate

A

$\frac{1}{\sqrt{2}} \log \frac{1 - \sqrt{2} \sin \frac{x}{2}}{1 + \sqrt{2} \sin \frac{x}{2}} - \log \frac{1 + \sin \frac{x}{2}}{1 - \sin \frac{x}{2}} + C$
B

$\frac{1}{\sqrt{2}} \log \frac{1 + \sqrt{2} \sin \frac{x}{2}}{1 - \sqrt{2} \sin \frac{x}{2}} - \log \frac{1 + \sin \frac{x}{2}}{1 - \sin \frac{x}{2}} + C$
C

$\frac{1}{\sqrt{2}} \log \frac{1 + \sqrt{2} \sin \frac{x}{2}}{1 - \sqrt{2} \sin \frac{x}{2}} - \log \frac{1 - \sin \frac{x}{2}}{1 + \sin \frac{x}{2}} + C$
D

None of these

Solution

$\int \frac{\sin \frac{x}{2} . \tan \frac{x}{2}}{\cos x} d x = \int \frac{\sin \frac{x}{2} . (\frac{\sin \frac{π}{2}}{\cos \frac{π}{2}})}{1 - 2 \sin^{2} \frac{x}{2}} d x$
$= \int \frac{\sin^{2} \frac{x}{2} \cos \frac{x}{2} d x}{(1 - \sin^{2} \frac{x}{2}) (1 - 2 \sin^{2} \frac{x}{2})}$

$= 2 \int \frac{z^{2} d z}{(1 - z^{2}) (1 - 2 z^{2})}$

$[P u t t i n g \sin \frac{x}{2} = z \Rightarrow \frac{1}{2} \cos \frac{x}{2} d x = d z]$

$= 2 \int (\frac{1}{1 - 2 z^{2}} - \frac{1}{1 - z^{2}}) d z$

$= \int \frac{d z}{{(\frac{1}{\sqrt{2}})}^{2} - z^{2}} - 2 \int \frac{d z}{1 - z^{2}}$

$= \frac{1}{2. \frac{1}{\sqrt{2}}} \log \frac{\frac{1}{\sqrt{2}} + z}{\frac{1}{\sqrt{2}} - z} - 2. \frac{1}{2} \log \frac{1 + z}{1 - z} + C$

$= \frac{1}{\sqrt{2}} . \log \frac{1 + \sqrt{2} \sin \frac{x}{2}}{1 - \sqrt{2} \sin \frac{x}{2}} - \log \frac{1 + \sin \frac{x}{2}}{1 - \sin \frac{x}{2}} + C .$

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