Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function such that its derivative $f'$ is continuous and $f(\pi) = -6$. If $F:[0, \pi] \to \mathbb{R}$ is defined by $F(x) = \int_{0}^{x} f(t)\,dt$, and if $\int_{0}^{\pi} (f'(x) + F(x))\cos x\,dx = 2$, then the value of $f(0)$ is ______

$f(\pi) = -6, \quad F(x) = \int_{0}^{x} f(t)\,dt$ $\int_{0}^{\pi} (f'(x) + F(x))\cos x\,dx = 2$ $\text{To find: } f(0)$ $I = \int_0^\pi f'(x)\cos x\,dx + \int_0^\pi F(x)\cos x\,dx$ $\int_0^\pi f'(x)\cos x\,dx = [f(x)\cos x]_0^\pi - \int_0^\pi f(x)(-\sin x)\,dx$ $= f(\pi)\cos\pi - f(0)\cos 0 + \int_0^\pi f\sin x\,dx$ $= -f(\pi) - f(0) + \int_0^\pi f(x)\sin x\,dx = 6 - f(0) + \int_0^\pi f\sin x\,dx$ $\int_0^\pi F(x)\cos x\,dx = [F(x)\sin x]_0^\pi - \int_0^\pi F'(x)\sin x\,dx$ $= 0 - \int_0^\pi f(x)\sin x\,dx = -\int_0^\pi f\sin x\,dx$ $I = 6 - f(0) + \int_0^\pi f\sin x\,dx - \int_0^\pi f\sin x\,dx = 6 - f(0)$ $6 - f(0) = 2 \Rightarrow f(0) = 4$ $\boxed{f(0) = 4}$