1.

`ysin2xdx-(1+y^2+cos^2x)dy=0`

Answer» `Pdx+Qdy=0`
`ysin2xdx+(-1-y^2-cos^2x)dy=0`
`(dp)/(dy)=sin2x`
`(dQ)/(dx)=-2cosx*(-sinx)=2sinxcosx`
`intPdx=-1/2ycos2x+g(y)`
`intQdy=-y-y^33/3-ycos^2x+h(x)`
`cos^2x=1/2(1+cos2x)`
`g(y):-y-y^3/3-y/2` and `h(x)=0`
`-1/2ycos2x-3/2y-y^3/3=0`.


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