Solve for f in the integral equation $$f(t) = \sin t + \int_{0}^{t} f(s)ds$$ using $ (V^nx)(t) =\int_{0}^{t} \frac{(t-s)^{n-1}}{(n-1)!}x(s)ds $ to to where V is the Volterra operator $V$ on $L^2(0,1)$ is an operator of integration such as the one shown below $$(Vx)(t) = \int_{0}^{t} x(s)ds, 0\leq t\leq 1$$
I tried integrating both sides from $0$ to $t$: $$\int_{0}^{t}f(s)ds = \int_{0}^{t} \sin(s) ds + \int_{0}^{t} \int_{0}^{s} f(s) ds = -\cos(t) + \cos(0) + (V^2f)(t) = 1 - \cos(0) + \int_{0}^{t} (t-s)f(s)ds $$
I'm not sure where to go from here.
I'm not sure of the solution using the Volterra operator, but it should be fairly straightforward to just take the derivative. If you take there derivative, then you'll have:
$$ f'(t) = \cos(t) + f(t) $$
which is the same as: $$ f'(t) - f(t) = \cos(t)$$
Use the integrating factor $e^{-t}$, and you'll find
$$e^{-t} f'(t) - e^{-t}f(t) = (e^{-t} f(t))'$$
So, you have:
$$ (e^{-t} f(t))' = e^{-t} \cos(t)$$ If you integrate both sides, you'll be able to get the solution easily. For the Volterra approach, the right thing is probably not to integrate per se, but to use the fact $\int_0^t f(s) ds = f(t) - \sin(t)$ so simplify the expressions involving the Volterra operator.