$\int \frac{1}{\cos x+\sqrt{3} \sin x} d x$ equals

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

We have,

$I =\int \frac{1}{\cos x+\sqrt{3} \sin x} d x $

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

$\Rightarrow I =\frac{1}{2} \int \frac{1}{\cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}} d x$

$\Rightarrow I =\frac{1}{2} \int \frac{1}{\cos \left(x-\frac{\pi}{3}\right)} d x=\frac{1}{2} \int \sec \left(x-\frac{\pi}{3}\right) d x$

$\Rightarrow I=\frac{1}{2} \log \tan \left(\frac{\pi}{4}+\frac{x}{2}-\frac{\pi}{6}\right)+C=\frac{1}{2} \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+C$