Answer
An integrating factor is $ \mu(y)=y$
The differential equation
$$
d x+(x / y-\sin y) d y=0
$$
is not exact but becomes exact when multiplied by the integrating factor $ \mu(y)=y$ and its solutions are given implicitly by
$$
yx + y \cos (y)-\sin (y)=c
$$
where $ c $ is an arbitrary constant
Work Step by Step
$$
d x+(x / y-\sin y) d y=0
\quad \quad (i)
$$
Comparing this Equation with the differential form:
$$
M(x,y) d x+N(x,y) d y=0
$$
we observe that
$$
\begin{aligned}M(x,y) &=1
\\ N(x,y) &=(x / y-\sin y) \end{aligned}
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
\begin{aligned} M_{y}(x, y) &= 0
\\ N_{x}(x, y) & = \frac{1}{y} \end{aligned}
$$
we find that
$$
\frac{N_{x}(x, y)-M_{y}(x, y)}{M(x, y)}=\frac{ \frac{1}{y}-0} {1 }= \frac{1}{y}.
$$
Thus there is an integrating factor $μ $ that is a function of $y$ only, and it satisfies the differential equation
$$
\frac{d \mu}{d y}=\frac{1}{y} \mu .
$$
Hence
$$
\frac{d \mu}{ \mu}=\frac{d y}{ y}
$$
Integrating, we obtain
$$
\mu(y)=y
$$
Multiplying Eq. (i) by this integrating factor, we obtain
$$
y d x+(x-y\sin y) d y=0
$$
The latter equation is exact, since
$$
M_{y}(x, y)=1=N_{x}(x, y).
$$
Thus there is a $\psi (x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=M(x,y) =y ,
\\
\psi_{y}(x, y) &=N(x,y) =(x-y\sin y)
\end{aligned} \quad \quad (ii)
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\begin{split}
\psi(x, y) & =\int {y } dx \\
& = yx +h(y)
\quad\quad\quad (iii)
\end{split}
$$
where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (ii), we compute $\psi_{y}(x, y) $ from Eq. (iii) and set it equal to $N$, obtaining
$$
\psi_{y}(x, y)= x +h^{\prime}(y)=(x-y\sin y)
$$
Thus $h^{\prime}(y)=y\sin y $ and integrating by parts we obtain
$$
h(y)=-y \cos (y)+\sin (y)
$$
the constant of integration can be omitted since any solution of the preceding differential equation is satisfactory.
Now substituting for $h(y)$ in Eq. (iii) gives
$$
\psi(x, y)= yx - y \cos (y)+\sin (y)
$$
Hence solutions of Eq. (i) are given implicitly by
$$
\psi(x, y) =yx + y \cos (y)-\sin (y)=c
$$
where $ c $ is an arbitrary constant