Question

The value of $\int \frac{\cos x}{\sin (x-\frac{\pi }{6})\sin (x+\frac{\pi }{6})}dx$ is equal to
$(C$ is a constant of integration$)$

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

Answer 1

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

$I=\int \frac{\cos x}{\sin (x-\frac{\pi }{6})\sin (x+\frac{\pi }{6})}dx$
$=\int \frac{\cos x}{{\sin }^{2}x-{\sin }^2\frac{\pi }{6}}dx$
Let $\sin x=t\Rightarrow dt=\cos xdx$
$I=\int \frac{dt}{t^2-{\left(\frac{1}{2}\right)}^2}$

$=\frac{1}{2\times \frac{1}{2}}\log \left|\frac{t-\frac{1}{2}}{t+\frac{1}{2}}\right|+C$

$=\ln \left|\frac{2t-1}{2t+1}\right|+C$

$=\ln \left|\frac{2\sin x-1}{2\sin x+1}\right|+C$