Question:
$\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$
Solution:
$\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$
$=\int \frac{1}{\frac{\cos ^{2} x}{1} d x}$
$=\int \frac{\sin ^{2} x}{\cos ^{2} x} d x$
$=\int \tan ^{2} x d x$
$=\int\left(\sec ^{2} x-1\right) d x$
$=\int \sec ^{2} x d x-\int 1 d x$
$=\tan x-x+C$
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