Evaluate the following integrals:


Evaluate: $\int \frac{1-\sin x}{\cos ^{2} x} d x$


Given, $\int \frac{1-\sin x}{\cos ^{2} x} d x$

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

$=\int \sec ^{2} x-\tan x \cdot \sec x d x\left[\operatorname{since}, \cos x=\frac{1}{\sec x}\right]$

$=\tan x-\sec x+c$

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