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Question
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Solution
We have,
\[ \sin^4 x\frac{dy}{dx} = \cos x\]
\[ \Rightarrow dy = \frac{\cos x}{\sin^4 x}dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\frac{\cos x}{\sin^4 x}dx\]
\[ \Rightarrow y = \int\frac{\cos x}{\sin^4 x}dx\]
\[\text{ Putting }\sin x = t\]
\[ \Rightarrow \cos x dx = dt\]
\[ \therefore y = \int\frac{1}{t^4}dt\]
\[ = \frac{t^{- 3}}{- 3} + C\]
\[ = \frac{- \sin^{- 3} x}{3} + C\]
\[ = - \frac{1}{3} {cosec}^3 x + C \]
\[\text{ Hence, }y = - \frac{1}{3} {cosec}^3 x +\text{C is the solution to the given differential equation.}\]
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