Question

\[\frac{dy}{dx} - x \sin^2 x = \frac{1}{x \log x}\]

Answer 1

We have,

\[\frac{dy}{dx} - x \sin^2 x = \frac{1}{x \log x}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x \log x} + x \sin^2 x\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x \log x} + \frac{x}{2}\left( 1 - \cos 2x \right)\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x \log x} + \frac{x}{2} - \frac{x}{2}\cos 2x\]

Integrating both sides, we get

\[\int dy = \int\left( \frac{1}{x \log x} + \frac{x}{2} - \frac{x}{2}\cos 2x \right) dx\]

\[ \Rightarrow \int dy = \int\frac{1}{x \log x}dx + \frac{1}{2}\int x dx - \frac{1}{2}\int\left( x \cos 2x \right)dx\]

\[ \Rightarrow y = \log \left| \log x \right| + \frac{x^2}{4} - \frac{x}{2}\int\left( \cos 2x \right)dx + \frac{1}{2}\int\left[ \frac{d}{dx}\left( x \right)\int\left( \cos 2x \right) dx \right]dx\]

\[ \Rightarrow y = \log \left| \log x \right| + \frac{x^2}{4} - \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} + C\]