Find an integrating factor and solve the given equation.
$$1+\left({x \over y}−\sin(y)\right)y′=0.\tag{1}$$
Let $M(x,y)=1$ and $N(x,y)={x \over y}−\sin(y)$.
Then, $M_y={\partial \over \partial y}M(x,y)=0$ and $N_x={\partial \over \partial x}N(x,y)={1 \over y}$.
Notice that $M_y =0\ne{1 \over y} =N_x$, which implies that the given equation is not exact.
We are looking for an integrating factor $\mu(x,y)$ such that after multiplying $(1)$ by $\mu$, the equation becomes exact.
That is, $(\mu M)_y=(\mu N)_x$.
$${\partial \mu \over \partial y }={\partial \over \partial x}\left[\mu\left({x \over y}-\sin(y)\right)\right]$$
$${\partial \mu \over \partial y }={\partial \mu \over \partial x }\left({x \over y}-\sin(y)\right)+\mu {1 \over y}$$
Suppose that $\mu$ is a function of only $y$, we get
$${d \mu \over dy}=\mu {1 \over y}$$
$$\int{1 \over \mu}d \mu=\int{1 \over y}dy $$
$$\ln|\mu|=\ln|y|$$
$$\mu=y$$
Multiplying $(1)$ by the integrating factor $\mu=y$, we get an exact equation,
$$y+(x-y \sin(y))y'=0\tag{2}$$
Since $(2)$ is exact, there exists a solution $\phi (x,y)=C$ such that
$\phi_x(x,y)=y\tag{3}$
$\phi_y(x,y)=x-y \sin(y)\tag{4}$
Integrating $(3)$ with respect to $x$, we get
$$\phi(x,y)=yx+g(y)\tag{5}$$
for some function $g$ of $y$.
Differentiating $(5)$ with respect to $y$, we get
$$\phi_y(x,y)=x+g'(y)\tag{6}$$
Equating $(4)$ with $(6)$, we have
$$x-y\sin(y)=x+g'(y)$$
$$-y\sin(y)=g'(y)$$
$$\int -y\sin(y)dy=\int g'(y)dy$$
$$g(y)=-\int y\sin(y)dy$$
Integration by parts gives,
$$g(y)=y\cos(y)-\sin(y)\tag{7}$$
Substituting $(7)$ into $(5)$,
$$\phi(x,y)=yx+y\cos(y)-\sin(y)$$
Thus, the solutions of the differential equation are given implicitly by
$$yx+y\cos(y)-\sin(y)=C$$
