\[ y'(x)+y(x) \cos (x)-\frac {1}{2} \sin (2 x)=0 \] ✓ Mathematica : cpu = 0.0282852 (sec), leaf count = 18
\[\left \{\left \{y(x)\to \sin (x)+c_1 e^{-\sin (x)}-1\right \}\right \}\] ✓ Maple : cpu = 0.036 (sec), leaf count = 15
\[y \relax (x ) = \sin \relax (x )-1+{\mathrm e}^{-\sin \relax (x )} c_{1}\]
Hand solution
\begin {equation} \frac {dy}{dx}+y\relax (x) \cos \relax (x) =\frac {1}{2}\sin \left ( 2x\right ) \tag {1} \end {equation}
Integrating factor \(\mu =e^{\int \cos dx}=e^{\sin \relax (x) }\). Therefore (1) becomes2\[ \frac {d}{dx}\left (e^{\sin \relax (x) }y\relax (x) \right ) =\frac {1}{2}e^{\sin \relax (x) }\sin \left (2x\right ) \] Integrating\begin {align*} e^{\sin \relax (x) }y\relax (x) & =\frac {1}{2}\int e^{\sin \relax (x) }\sin \left (2x\right ) +C\\ y\relax (x) & =\frac {e^{-\sin \relax (x) }}{2}\int e^{\sin \relax (x) }\sin \left (2x\right ) +e^{-\sin \relax (x) }C \end {align*}
But \(e^{\sin \relax (x) }\sin \left (2x\right ) \) can be integrated by parts which gives \(e^{\sin \relax (x) }\left (-2+2\sin \relax (x) \right ) \). Hence the above becomes\begin {align*} y\relax (x) & =\frac {e^{-\sin \relax (x) }}{2}\left ( e^{\sin \relax (x) }\left (-2+2\sin \relax (x) \right ) \right ) +e^{-\sin \relax (x) }C\\ & =-1+\sin \relax (x) +e^{-\sin \relax (x) }C \end {align*}