Question

\[\frac{dy}{dx} = \left( \cos^2 x - \sin^2 x \right) \cos^2 y\]

Answer 1

We have,
\[\frac{dy}{dx} = \left( \cos^2 x - \sin^2 x \right) \cos^2 y\]
\[ \Rightarrow \frac{dy}{dx} = \cos 2x \cos^2 y\]
\[ \Rightarrow \frac{1}{\cos^2 y}dy = \cos 2x dx\]
\[ \Rightarrow \sec^2 y dy = \cos 2x dx\]
Integrating both sides, we get
\[\int \sec^2 y dy = \int\cos 2x dx\]
\[ \Rightarrow \tan y = \frac{\sin 2x}{2} + C\]
\[\text{ Hence, }\tan y = \frac{\sin 2x}{2} +\text{ C is the required solution }.\]