Integrate the rational function: $\frac{\cos x}{(1-\sin x)(2-\sin x)}$
[Hint: Put $\sin x = t$]

(D) Let $I = \int \frac{\cos x}{(1-\sin x)(2-\sin x)} dx$.
Put $\sin x = t$,then $\cos x dx = dt$.
The integral becomes $I = \int \frac{dt}{(1-t)(2-t)}$.
Using partial fractions: $\frac{1}{(1-t)(2-t)} = \frac{A}{1-t} + \frac{B}{2-t}$.
$1 = A(2-t) + B(1-t)$.
For $t = 1$,$1 = A(2-1) \Rightarrow A = 1$.
For $t = 2$,$1 = B(1-2) \Rightarrow B = -1$.
Thus,$\frac{1}{(1-t)(2-t)} = \frac{1}{1-t} - \frac{1}{2-t}$.
Integrating both sides: $I = \int \left( \frac{1}{1-t} - \frac{1}{2-t} \right) dt$.
$I = -\ln|1-t| - (-\ln|2-t|) + C = \ln|2-t| - \ln|1-t| + C$.
$I = \ln\left|\frac{2-t}{1-t}\right| + C$.
Substituting $t = \sin x$,we get $I = \ln\left|\frac{2-\sin x}{1-\sin x}\right| + C$.