Question
Evaluate $\int\frac{1}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
✓
Answer
Consider the integral
$\text{I}=\int\frac{1}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\big(\sec^2\text{x}+\text{cosec}^2\text{x}\big)\text{dx}$
$=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$
$=\frac{\sin\text{x}}{\cos\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}+\text{C}$
$=\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin\text{x}\cos\text{x}}+\text{C}$
$=-\frac{2\cos2\text{x}}{\sin2\text{x}}+\text{C}$
$=-2\cot2\text{x}+\text{C}$
$\text{I}=\int\frac{1}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\big(\sec^2\text{x}+\text{cosec}^2\text{x}\big)\text{dx}$
$=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$
$=\frac{\sin\text{x}}{\cos\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}+\text{C}$
$=\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin\text{x}\cos\text{x}}+\text{C}$
$=-\frac{2\cos2\text{x}}{\sin2\text{x}}+\text{C}$
$=-2\cot2\text{x}+\text{C}$
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