Question

Solve it :-

$\frac{xdx+ydy}{ydx-xdy}=\frac{x\sin (x^2+y^2}{y^3}$

Answer 1

$\frac{xdx+ydy}{\sin (x^2+y^2)}=\frac{x}{y}\frac{ydx-xdy}{y^2}$
Let $x^2+y^2=u$, then $xdx+ydy=du/2$
Let $\frac{x}{y}=v$, then $\frac{ydx-xdy}{y^2}=dv$
$\therefore \frac{1}{2}\frac{du}{\sin u}=vdv$
$⟹\int \text{cosec}udu=\int 2vdv$
$⟹\ln (|\text{cosec}u-\cot u|)=v^2+C$
$⟹\ln \left(|\text{cosec}\right(x^2+y^2)-\cot (x^2+y^2\left)|\right)=(\frac{x}{y}{)}^2+C$