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