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

Answer

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

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