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Question
\[\frac{x \sin x}{1 + \cos x}\]
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Solution
\[\text{ Let } u = x \sin x; v = 1 + \cos x\]
\[\text{ Then }, u' = x \cos x + \sin x; v' = - \sin x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{x \sin x}{1 + \cos x} \right) = \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x \right) - x \sin x\left( - \sin x \right)}{\left( 1 + \cos x \right)^2}\]
\[ = \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x \right) + x \sin^2 x}{\left( 1 + \cos x \right)^2}\]
\[ = \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x \right) + x \left( 1 - \cos^2 x \right)}{\left( 1 + \cos x \right)^2}\]
\[ = \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x \right) + x\left( 1 + \cos x \right)\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)^2}\]
\[ = \frac{\left( 1 + \cos x \right)\left( x \cos x + \sin x + x - x\cos x \right)}{\left( 1 + \cos x \right)^2}\]
\[ = \frac{\left( 1 + \cos x \right)\left( x + \sin x \right)}{\left( 1 + \cos x \right)^2}\]
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