Question

Evaluate : $\int \frac{\cos x}{\sin (x-\frac{\pi }{6})\sin (x+\frac{\pi }{6})}dx$

• A. $\log \left|\frac{2\sin x-1}{2\sin x+1}\right|+C$
• B. $\log \left|\frac{2\sin x+1}{2\sin x-1}\right|+C$
• C. $\log \left|\frac{\sin x+1}{\sin x-1}\right|+C$
• D. $\log \left|\frac{\sin x-1}{\sin x+1}\right|+C$

Answer 1

The correct option is A $\log \left|\frac{2\sin x-1}{2\sin x+1}\right|+C$

$\int \frac{\cos x}{\sin (x-\frac{\pi }{6})\sin (x+\frac{\pi }{6})}dx$
$\int \frac{\cos x}{{\sin }^{2}x-{\sin }^2\frac{\pi }{6}}dx$

Let $\sin x=t$

or $dt=\cos xdx$

$\therefore I=\int \frac{dt}{t^2-\frac{1}{4}}$
$=\frac{1}{2\frac{1}{2}}\log \left|\frac{t-\frac{1}{2}}{t+\frac{1}{2}}\right|+C$
$=\log \left|\frac{2t-1}{2t+1}\right|+C$
$=\log \left|\frac{2\sin x-1}{2\sin x+1}\right|+C$ where $t=\sin x$

Hence, option 'A' is correct.