Question:
Evaluate: $\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x$
Solution:
Given, $\int \frac{1}{\sin ^{2} x \cdot \cos ^{2} x} d x$
$=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} d x\left[\right.$ since, $\left.\sin ^{2} x+\cos ^{2} x=1\right]$
$=\int \frac{\sin ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} d x$
$=\int \frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x} d x$
$=\int\left(\sec ^{2} x+\operatorname{cosec}^{2} x\right) d x$
$=\tan x-\cot x+c$
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