By a suitable substitution or differentiation, reduce y'+sin…
By a suitable substitution or differentiation, reduce $y'+\sin y+x\cos y+x=0$ to linear form and find its general integral.
Explanation of Method
Use the tangent half-angle substitution $u=\tan(y/2)$, so $\sin y=\dfrac{2u}{1+u^2}$, $\cos y=\dfrac{1-u^2}{1+u^2}$, and $y'=\dfrac{2u'}{1+u^2}$; this reduces the equation to a linear one in $u(x)$.
Step 1: Apply the half-angle substitution
With $u=\tan(y/2)$: $\sin y=\dfrac{2u}{1+u^2}$, $\cos y=\dfrac{1-u^2}{1+u^2}$, $y'=\dfrac{2u'}{1+u^2}$. Substituting into $y'+\sin y+x\cos y+x=0$:
\[\frac{2u'}{1+u^2}+\frac{2u}{1+u^2}+x\cdot\frac{1-u^2}{1+u^2}+x=0.\]
Step 2: Clear denominators
Multiply through by $1+u^2$:
\[2u'+2u+x(1-u^2)+x(1+u^2)=0.\]
The $\pm xu^2$ terms cancel: $2u'+2u+2x=0$, so
\[u'+u=-x.\]
Step 3: Solve the linear equation in $u$
With integrating factor $e^x$: $\dfrac{d}{dx}(ue^x)=-xe^x$. Since $\displaystyle\int xe^x\,dx=e^x(x-1)$,
\[ue^x=-e^x(x-1)+C=e^x(1-x)+C\quad\Longrightarrow\quad u=(1-x)+Ce^{-x}.\]
Result
Final answer:
\[\boxed{\,\tan\frac{y}{2}=1-x+Ce^{-x},\qquad C\in\mathbb{R}}\]