Find $\int \frac{\cos x}{(1-\sin x)(2-\sin x)} dx$

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

The correct answer is Option (2) → $\ln \left| \frac{2-\sin x}{1-\sin x} \right| + C$

$\int \frac{\cos x}{(1-\sin x)(2-\sin x)} dx=\int \frac{\cos x}{(\sin x-1)(\sin x-2)} dx$

Put $\sin x = t ⇒\cos x dx = dt$

$ I= \int \frac{dt}{(t-1)(t-2)}$

$=\int\left( \frac{-1}{t-1} + \frac{1}{t-2}\right)$

$ = -\log |t-1| + \log |t-2|$

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

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