Question
Find the integral: $\int \frac{1-\sin x}{\cos ^{2} x} d x$

Answer

We have
$\int \frac{1-\sin x}{\cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x} d x-\int \frac{\sin x}{\cos ^{2} x} d x$ 
= $\int \sec ^{2} x d x-\int \tan x \sec x d x$ 
= tan x - sec x + C

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