Evaluate the following integrals: 1- 1+ dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$

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Answer

$\int\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\frac{(1-\cos\text{x})^2}{1-\cos^2\text{x}}\text{dx}$
$=\int\frac{1+\cos^2\text{x}-2\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\sin^2\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}-\frac{2\cos\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int(\text{cosec}^2\text{x}+\cot^2\text{x}-2\cot\text{x}\text{ cosec x})\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int(2\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int2\text{cosec}^2\text{x dx}-\int1\text{dx}-\int2\cot\text{x cosec x }\text{dx}$
$=-2\cot\text{x}-\text{x}+2\text{cosec x}+\text{C}$
$=2(\text{cosec x}-\cot\text{x})-\text{x + C}$

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