2(c)

The equation is exact: $M_y = 2 x \cos y = N_x$.

$$F(x,y) = \int 2 x \sin y \, dx = x^2 \sin y + h(y)$$

$$N = 1 + x^2 \cos y = \frac{\partial F}{\partial y} = x^2 \cos y + h'(y)$$

so h'(y) = 1, h(y) = y and the solution is $x^2 \sin y + y =$constant.