\begin {align*} y^{\prime } & =\sin y+1\\ y\left ( \pi \right ) & =1 \end {align*}
This has unique solution. Integrating and solving for \(c\) results in the solution\begin {align*} \int \frac {dy}{\sin y+1} & =\int dx\qquad \sin y+1\neq 0\\ y & =-2\arctan \left ( \frac {c_{1}+x+2}{c_{1}+x}\right ) \end {align*}
Applying IC gives\[ 1=-2\arctan \left ( \frac {c_{1}+\pi +2}{c_{1}+\pi }\right ) \] Solving for \(c_{1}\) and substituting in the general solution gives\[ y=-2\arctan \left ( \frac {\left ( x-\pi +2\right ) \tan \left ( \frac {1}{2}\right ) +x-\pi }{-\pi +x-2+\tan \left ( \frac {1}{2}\right ) \left ( x-\pi \right ) }\right ) \]