Show that

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

Let $\sin x=t \Rightarrow \cos x d x=d t$

$\therefore \int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}$

Let $\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}$

$1=A(2-t)+B(1-t)$    ...(1)

Substituting t = 2 and then t = 1 in equation (1), we obtain

A = 1 and B = −1

$\therefore \frac{1}{(1-t)(2-t)}=\frac{1}{(1-t)}-\frac{1}{(2-t)}$

$\Rightarrow \int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int\left\{\frac{1}{1-t}-\frac{1}{(2-t)}\right\} d t$

$=-\log |1-t|+\log |2-t|+\mathrm{C}$

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

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