IE18

Solve: $\sin(x) = \int_0^x e^{x-t}u(t)dt\,$

Differentiate both sides with respect to x.

$\cos(x) = \frac{d}{dx} \int_0^x e^{x-t}u(t)dt\,$

$\cos(x) = \int_0^x \frac{\partial}{\partial x} e^{x-t}u(t)dt + u(x)\,$

$\cos(x) = \int_0^x e^{x-t}u(t)dt + u(x)\,$

$\cos(x) = \sin(x) + u(x)\,$

$u(x) = \cos(x) - \sin(x)\,$