Let $\dot x(t)=A(t)x(t)+B(t)$ be the given system of linear differential equations and $\Phi(t,s)$ be the transition matrix satisfying $$\frac{d}{dt}\Phi(t,s)=A(t)\Phi(t,s),\quad \Phi(s,s)=I,\quad \forall t,s$$ According to the textbook I am reading now, if the initial condition is $x(0)=x_0$, then the solution $\hat x$ to the given LDE is given by $$\hat x(t)=\Phi(t,0)x_0+\int_0^t \Phi (t,s)B(s)ds$$ And here goes the textbook's proof:
$$\begin{align} \frac{d}{dt}\hat x(t) &= \frac{d}{dt}\Phi(t,0)x_0+\int_0^t \frac{d}{dt}\Phi(t,s)B(s)ds+\Phi(t,t)B(t)\tag{1} \\ &= A(t)\Phi(t,0)x_0+A(t)\int_0^t\Phi(t,s)B(s)ds+B(t) \tag{2} \\ &= A(t)\hat x(t)+B(t) \tag{3} \end{align} $$
I got the $(1)\rightarrow(2)$ part, however I don't understand how $(2)\rightarrow(3)$ holds. To be specific, I can't get why $$A(t)\int_0^t\Phi(t,s)B(s)ds=0$$ Am I missing something? I would really appreciated your help! Thank you in advance.
The step (2) to (3) comes from substituting $$ \int_0^t\Phi(t,s) B(s)\text d s = \hat x(t) - \Phi(t,0)x_0 $$ in (2).