Question

$\int \frac{dx}{\cos x+\sqrt{3}\sin x}$ is equal to

• A. $\log \tan (\frac{x}{2}+\frac{\pi }{3})+k$
• B. $\log \tan (\frac{x}{2}-\frac{\pi }{3})+k$
• C. $\frac{1}{2}\log \tan (\frac{x}{2}+\frac{\pi }{12})+k$
• D. None of these

Answer 1

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

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

Put, $1=r\sin \theta ,\sqrt{3}=r\cos \theta$

$\Rightarrow r=\sqrt{(x^2+y^2)}=\sqrt{{\sqrt{3}}^2+1^2}=2$

and $\tan \theta ={\tan }^{-1}\frac{y}{x}=\frac{\pi }{6}$

$=\int \frac{dx}{r\sin \theta \cos x+r\cos \theta \sin x}$

$=\int \frac{dx}{r\sin (\theta +x)}$

$=\frac{1}{r}\int \csc (\theta +x)dx$

$=\frac{1}{r}\log \left|\tan (\frac{\theta }{2}+\frac{x}{2})\right|+C$

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