Question
Evaluate $\int \frac{\cos ^2 x}{1+\sin x} d x$

Answer

Let
$I=\int \frac{\cos ^2 x}{1+\sin x} d x$
$I=\int \frac{1-\sin ^2 x}{1+\sin x} d x$
$=\int \frac{(1-\sin x)(1+\sin x)}{1+\sin x} d x$
$I=\int(1-\sin x) d x$
$=x+\cos x+\text { C }$

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