Question

PLZ HELP
USing an appropriate substitution, solve
\[(x^2 tan y)y'=x^6 sin (3x)+3x ln (sec y)\]

Answer 1

-.-

Answer 2

I need help

Answer 3

Try \(t=\ln \sec y\), then \(t'=\dfrac{\sec y\tan y}{\sec y}y'\), or \(t'=\tan y~y'\).
\[\begin{align*}
x^2\tan y~y'&=x^6\sin3x+3x\ln\sec y\\\\
x^2t'&=x^6\sin3x+3xt\\\\
t'-\frac{3}{x}t&=x^4\sin3x
\end{align*}\]
which is linear in \(t\).