Let $I=\int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx$ Put $\sin x = t \Rightarrow \cos x dx = dt$ ...(i) $\therefore \int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx = \int \frac{1}{(1-t)(2-t)}dt$ Let $\frac{1}{(1-t)(2-t)} = \frac{A}{1-t} + \frac{B}{2-t}$ ⇒1 = A(2 – t) + B(1 – t) ...(ii) Putting t = 1 in (ii), we get 1 = A(2 – 1) ⇒ A = 1 Putting t = 2 in (ii), we get B = –1 $\int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx$ $= \int \frac{1}{1-t} dt - \frac{dt}{2-t}$ $= -\log|1-t| + \log|2-t| + C$ $= \log\left|\frac{2-t}{1-t}\right| + C$ $= \log\left|\frac{2-\sin x}{1-\sin x}\right| + C$ [using (i)]