Evaluate the following integrals:


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


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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