it is really shame on me that I cannot find the derivative of the following integral :(

$$X(t)=e^{-bt}X(0)+\sigma\int_{0}^{b}e^{-b(t-s)}dW(t)$$

Find $dX(t)$ ? where $0 < b, \sigma\in\mathbb{R} $, $X(0)$ is initial distribution of $X(t)$, independent of the Brownian motion $W(t)$. I want so show that $dX(t)= -bX(t)dt+\sigma dW(t)$, but I am getting stuck on computing the derivative of $$\sigma\int_{0}^{b}e^{-b(t-s)}dW(t)$$

could some one please give me some ideas ? thanks so much for your time

PS. the above equation is one of type of the Langevin's equation, more detail could be found here http://en.wikipedia.org/wiki/Langevin_equation