Question

Evaluate $\int \frac{1}{\sqrt{3}\sin x+\cos x}dx$.

• A. $\frac{1}{2}\log \left|\tan (\frac{x}{2}+\frac{\pi }{12})\right|+c$
• B. $\frac{1}{2}\log \left|\tan (\frac{x}{2}+\frac{\pi }{8})\right|+c$
• C. $\frac{1}{3}\log \left|\tan (\frac{x}{2}+\frac{\pi }{12})\right|+c$
• D. none of these

Answer 1

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

Let$\sqrt{3}=r\sin \theta$ and $1=r\cos \theta$
Then, $r=\sqrt{{\left(\sqrt{3}\right)}^2+1^2}=2$ and $\tan \theta =\frac{\sqrt{3}}{1}$ or $\theta =\frac{\pi }{3}$
$\therefore \int \frac{1}{\sqrt{3}\sin x+\cos x}dx$
$=\int \frac{1}{r\sin \theta \sin x+r\cos \theta \cos x}dx$
$=\frac{1}{r}\int \frac{1}{\cos (x-\theta )}dx=\frac{1}{r}\int \sec (x-\theta )dx$
$=\frac{1}{r}\log \left|\tan (\frac{\pi }{4}+\frac{x}{2}-\frac{\theta }{2})\right|+c$
$=\frac{1}{2}\log \left|\tan (\frac{\pi }{4}+\frac{x}{2}-\frac{\pi }{6})\right|+c$
$=\frac{1}{2}\log \left|\tan (\frac{x}{2}+\frac{\pi }{12})\right|+c$