dx x + 3 x = - Vidyadip

MCQ

$\smallint \frac{{dx}}{{\cos x + \sqrt 3 \sin x}} = $

A
$\log \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) + c$
B
$\log \tan \left( {\frac{x}{2} - \frac{\pi }{{12}}} \right) + c$
✓
$\frac{1}{2}\log \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) + c$
D
$\frac{1}{2}\log \tan \left( {\frac{x}{2} - \frac{\pi }{{12}}} \right) + c$

✓

Answer

Correct option: C.

$\frac{1}{2}\log \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) + c$

c
$I=\int \frac{d x}{\cos x+\sqrt{3} \sin x}$

$\Rightarrow I=\int \frac{d x}{2\left[\frac{1}{2} \cos x+\frac{\sqrt{3}}{2} \sin x\right]}$

$=\frac{1}{2} \int \frac{d x}{\left[\sin \frac{x}{6} \cos x+\cos \frac{x}{6} \sin x\right]}$

$=\frac{1}{2} \cdot \int \frac{d x}{\sin \left(x+\frac{\pi}{6}\right)}$

$\Rightarrow I=\frac{1}{2} \cdot \int \cos e c\left(x+\frac{\pi}{6}\right) d x$

But we know that

$\int \cos e c x d x=\log |(\tan x / 2)|+C$

$\therefore I=\frac{1}{2} \cdot \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+C$

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