Solve: `(xcosx-sinx)dx=x/ysinxdy`

Explanation

We have, `(xcosx-sinx)dx=x/ysinxdy`
`therefore (xycosx-ysinx)dx=xsinxdy`
`rArr xcosx=(ydx+xdy)sinx`
`rArr xycosx=(ydx+xdy)sinx`
`rArr cotxdx=(d(xy))/(xy)`
On integrating, we get
`log_(e)sinx=log_(e)xy+log_(e)c`
or `sinx=cxy`