Question
Evaluate the following integrals:
$\int\frac{1}{1-\cos\text{x}}\text{dx}$
$\int\frac{1}{1-\cos\text{x}}\text{dx}$
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Answer
$\int\frac{1}{1-\cos\text{x}}\text{dx}$
$=\int\frac{1}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{1-\cos^2\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{\sin^2\text{x}}\times\text{dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{dx}+\int\frac{\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\cot\text{x}\times\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{1}{1-\cos\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
$=\int\frac{1}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{1-\cos^2\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{\sin^2\text{x}}\times\text{dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{dx}+\int\frac{\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\cot\text{x}\times\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{1}{1-\cos\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
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