The value of $\int \frac{1}{\sin \left(x-\frac{\pi}{3}\right) \cos x} d x$, is
Answer & explanation
Correct answer: option 4
We have,
$I=\int \frac{1}{\sin \left(x-\frac{\pi}{3}\right) \cos x} d x=\frac{1}{\cos \frac{\pi}{3}} \int \frac{\cos \left\{x-\left(x-\frac{\pi}{3}\right)\right\}}{\sin \left(x-\frac{\pi}{3}\right) \cos x} d x$
$\Rightarrow I=2 \int \frac{\cos x \cos \left(x-\frac{\pi}{3}\right)+\sin x \sin \left(x-\frac{\pi}{3}\right)}{\sin \left(x-\frac{\pi}{3}\right) \cos x} d x$
$\Rightarrow I=2 \int\left\{\cot \left(x-\frac{\pi}{3}\right)+\tan x\right\} d x$
$\Rightarrow I=2\left\{\log \left|\sin \left(x-\frac{\pi}{3}\right)\right|-\log |\cos x|\right\}+C$
$\Rightarrow I=2 \log \left|\sin \left(x-\frac{\pi}{3}\right) \sec x\right|+C$