Given $\int_0^x f(t) \, dt = 2\cos x + 3x + 2,$ find $f$ We are told that
$$\displaystyle \int_0^x f(t) \, dt = 2\cos x + 3x + 2$$
and are asked to find $f$. If we apply the Fundamental Theorem of Calculus, then
$$\displaystyle f(x) =  \frac{d}{dx}\int_0^x f(t) \, dt = -2\sin x + 3$$
but
$$\displaystyle \int_0^x -2\sin t + 3 \, dt = 2\cos x+3x - 2$$
It seems to me that the problem stems from using the FTC: Since I don't know anything about $f$, I cannot apply it. My question is: How exactly do I go about finding $f$ ?

As noted, you probably need to have a $-2$ instead of $+2$, as you can see by setting $x=0$.
But the reason I am adding this answer is that, strictly speaking, there are an infinite number of such $f$!
For instance, given one such $f$ (which you found by differentiating) you can pick any point, and change the value of $f$ to be arbitrary without affecting the value of the integrals.
One can in fact show that if $f$ and $g$ are two such function, then they differ only on a set of measure $0$.
This we can show using two theorems:
1) A bounded function $f$ is Riemann integrable iff the set of discontinuities of $f$ is of measure $0$.
2) If $f$ is Lebesgue integrable on $[a,b]$ and $F(x) = \int_{0}^{x} f $ (Lebesgue integral, which is also the Riemann integral in this case), and if $f$ is continuous at $x_0$, then $F$ is differentiable at $x_0$ and $F'(x_0) = f(x)$.
