Question

$\int \frac{\sin x}{\sin (x-\frac{\pi }{6})\sin (x+\frac{\pi }{6})}dx=$

• A. $\frac{1}{\sqrt{3}}[\log \tan \frac{x}{2}+\log \tan (\frac{x}{2}-\frac{\pi }{12})]$
• B. $\frac{1}{\sqrt{3}}[\log \tan (\frac{x}{2}+\frac{\pi }{12})+\log \tan (\frac{x}{2}-\frac{\pi }{12})]$
• C. $\frac{1}{\sqrt{3}}\left[\log \cot (\frac{x}{2}+\frac{\pi }{12})\right]$
• D. none

Answer 1

Answer: (B)

$\frac{1}{\sqrt{3}}[\log \tan (\frac{x}{2}+\frac{\pi }{12})+\log \tan (\frac{x}{2}-\frac{\pi }{12})]$

$\int \frac{\sin x}{\sin (x-\frac{\pi }{6})\sin (x+\frac{\pi }{6})}dx$
$N^{\prime }=\sin x=\frac{1}{2\cos \left(\pi /6\right)}.2\sin x\cos \frac{\pi }{6}=\frac{1}{\sqrt{3}}[\sin (x+\frac{\pi }{6})+\sin (x-\frac{\pi }{6})]$
$\therefore I=\int \frac{\frac{1}{\sqrt{3}}[\sin (x+\frac{\pi }{6})+\sin (x-\frac{\pi }{6})]}{\sin (x-\frac{\pi }{6})\sin (x+\frac{\pi }{6})}dx$
$\therefore I=\frac{1}{\sqrt{3}}\int [\csc (x-\frac{\pi }{6})+\csc (x+\frac{\pi }{6})]dx$
$=\frac{1}{\sqrt{3}}[\log \tan (\frac{x}{2}-\frac{\pi }{12})+\log \tan (\frac{x}{2}+\frac{\pi }{12})]+C$

$(\because \int \csc xdx=\log \tan (x/2)+C)$