Periodicity of an inhomogeneous system of ODEs 
Consider the linear $2 \times 2$-system: $x'=Ax+b(t),$ where $b \in C(\mathbb{R}, \mathbb{R}^2)$ is a $2 \pi /{\omega}$-periodic and 
  $$A=
\begin{pmatrix} 
0 & \omega \\ -\omega & 0
\end{pmatrix}$$
  for some $\omega >0.$

(a). Find the general solution (depends on two constants $c_1 , c_2 \in \mathbb{R}$).
(b). Show that the general solution is periodic if and only if 
$$\int_{0}^{2 \pi/{\omega}} 
\begin{pmatrix} 
\cos s & -\sin s \\ \cos s & \cos s
\end{pmatrix} \cdot b(s)~ds = \begin{pmatrix} 
0 \\ 0
\end{pmatrix}
$$
My approach. I've done part (a). I need help in solving part (b). Thank you for your time.
Using the solution formula for an inhomogenous linear system
$$x=X(t)c + X(t) \int_{0}^{t} X^{-1}(s)b(s)~ds,$$
we get 
$$x=\begin{pmatrix}
\cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t
\end{pmatrix} \cdot \begin{pmatrix}
c_1 \\ c_2
\end{pmatrix}+\begin{pmatrix}
\cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t
\end{pmatrix}  \int_{0}^{t} \begin{pmatrix}
\cos \omega s & -\sin \omega s \\ \sin \omega s & \cos \omega s \end{pmatrix}
\cdot b(s)~ds,
$$
where $t \in \left[ 0 , 2 \pi /{\omega}  \right].$
 A: Suppose $x(t)$ is periodic. Then $x(0)=x(p)$ for some $p>0$. 
Direct computation shows that $x(0)=\begin{pmatrix}c_1\\c_2\end{pmatrix}$. And
$$x(p)=\begin{pmatrix}
\cos \omega p & \sin \omega p \\ -\sin \omega p & \cos \omega p
\end{pmatrix} \cdot \begin{pmatrix}
c_1 \\ c_2
\end{pmatrix}+\begin{pmatrix}
\cos \omega p & \sin \omega p \\ -\sin \omega p & \cos \omega p
\end{pmatrix}  \int_{0}^{p} \begin{pmatrix}
\cos \omega s & -\sin \omega s \\ \sin \omega s & \cos \omega s \end{pmatrix}
\cdot b(s)~ds$$
For this to be equal to $X(0)$ for arbitrary $c_1, c_2$, the second the term has to be $0$ and $\cos \omega p=1, \sin\omega p=0$. This shows that $p=\frac{2\pi}{\omega}$. (We want this to be the period, so we choose the smallest number that satisfies the conditions.)
Plugging this into the second term, you can get the desired result. 
For the other direction, suppose the integral is zero. We show that $X(t+\frac{2\pi}{\omega})=X(t)$.
Write out $x(t+\frac{2\pi}{\omega})$, noting that $\cos(\omega (t+2\pi/\omega))=\cos \omega t$ and $\sin(\omega (t+2\pi/\omega))=\sin \omega t$, we have
$$x(t+\frac{2\pi}{\omega})=\begin{pmatrix}
\cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t
\end{pmatrix} \cdot \begin{pmatrix}
c_1 \\ c_2
\end{pmatrix}\\
\quad +\begin{pmatrix}
\cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t
\end{pmatrix}  \int_{0}^{t+\frac{2\pi}{\omega}} \begin{pmatrix}
\cos \omega s & -\sin \omega s \\ \sin \omega s & \cos \omega s \end{pmatrix}
\cdot b(s)~ds\\
=x(t)+\int_{t}^{t+\frac{2\pi}{\omega}} \begin{pmatrix}
\cos \omega s & -\sin \omega s \\ \sin \omega s & \cos \omega s \end{pmatrix}
\cdot b(s)~ds$$ 
Now since the integrand is of period $2\pi/\omega$, this integral is equal to 
$$\int_{0}^{2 \pi/{\omega}} 
\begin{pmatrix} 
\cos\omega s & -\sin\omega s \\ \cos\omega s & \cos\omega s
\end{pmatrix} \cdot b(s)~ds = \begin{pmatrix} 
0 \\ 0
\end{pmatrix}.$$
This shows the other direction.
