Question
Evaluate $\int\frac{2}{1-\cos2\text{x}}\text{ dx}$
✓
Answer
Let $\text{I}=\int\frac{2}{1-\cos2\text{x}}\text{ dx}$
$=\int\frac{2}{\sin^2\text{x}+\cos^2\text{x}-(\cos^2\text{x}-\sin^2\text{x})}\text{ dx}$
$=\int\frac{2}{2\sin^2\text{x}}\text{ dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{ dx}$
$=\int\text{cosec}^2\text{x}\text{ dx}$
$=-\cot\text{x}+\text{C}$
$=\int\frac{2}{\sin^2\text{x}+\cos^2\text{x}-(\cos^2\text{x}-\sin^2\text{x})}\text{ dx}$
$=\int\frac{2}{2\sin^2\text{x}}\text{ dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{ dx}$
$=\int\text{cosec}^2\text{x}\text{ dx}$
$=-\cot\text{x}+\text{C}$
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