Question

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

• A. $\frac{1}{2}\log \tan (\frac{x}{2}+\frac{\pi }{12})+C$
• B. $\frac{1}{3}\log \tan (\frac{x}{2}-\frac{\pi }{12})+C$
• C. $\log \tan (\frac{x}{2}+\frac{\pi }{6})+C$
• D. $\frac{1}{2}\log \tan (\frac{x}{2}-\frac{\pi }{6})+C$

Answer 1

The correct option is A $\frac{1}{2}\log \tan (\frac{x}{2}+\frac{\pi }{12})+C$

$\int \frac{dx}{\cos x+\sqrt{3}\sin x}$
$=\frac{1}{2}\int \frac{dx}{\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x}$
$=\frac{1}{2}\int \frac{dx}{\cos \frac{\pi }{3}\cos x+\sin \frac{\pi }{3}\sin x}$
$=\frac{1}{2}\int \frac{dx}{\cos (x-\frac{\pi }{3})}$
$=\frac{1}{2}\int \sec (x-\frac{\pi }{3})dx$
$=\frac{1}{2}\log \tan (\frac{x}{2}-\frac{\pi }{6}+\frac{\pi }{4})+C$
$=\frac{1}{2}\log \tan (\frac{x}{2}+\frac{\pi }{12})+C$

Hence, option A is correct.