Question

Integrate w.r.t x: $\frac{{\cos }^{2}x}{1-\sin x}$

Answer 1

$\int \frac{{\cos }^2}{1-\sin x}dx$

$=\int \frac{1-{\sin }^{2}x}{1-\sin x}dx$

$=\int \frac{\left(1-\sin x\right)\left(1+\sin x\right)}{1-\sin x}dx$

$=x-\cos x+c$