dx x + 2x = dx x (1 + 2 x) = . dx (1 - x)(1 + ) (1 + 2 x) =- dt (1 - t)( 1 + t) ( 1 +2t) where x = t = ( -1 6 1 - t + 1 2 1 + t - 4 3 1 + 2t )dt = + 1 6 |1 -t| + 1 2 |1 + t| -2 3 |1 + 2t| + c = 1 6 |1 - | + 1 2 | 1 + x| - 2 3 |1 + 2 | + c

Find: dx x + 2x

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Question

$\text{Find}:\int \frac{dx}{\sin x + \sin 2x}$

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Answer

$\int \frac{\text{dx}}{\sin \text{x} + \sin \text{2x}} = \int \frac{\text{dx}}{\sin \text{x} (1 + 2 \cos \text{x)}} = \int \frac{\sin\text{x}. \text{dx}}{(1 - \cos{\text{x})(1 + \cos\text{x}) (1 + 2\cos \text{x)}}}$
$=-\int \frac{\text{dt}}{(1 - \text{t})( 1 + \text{t}) ( 1 +\text{2t})} \text{where} \cos \text{x = t}$
$=\int \Bigg(\frac{\frac{-1}{6}}{1 - t}+\frac{\frac{1}{2}}{1 + t} - \frac{\frac{4}{3}}{1 + 2t}\Bigg)\text{dt}$
$= + \frac{1}{6} \log|1 -\text{t}| + \frac{1}{2}\log|1 + \text{t}| -\frac{2}{3}\log|1 + \text{2t}| + \text{c}$
$= \frac{1}{6}\log|1 - \cos\text{x}| + \frac{1}{2}\log| 1 + \cos \text{x}| - \frac{2}{3}\log|1 + 2\cos\text{x}| + \text{c}$

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