Given $\int^{x}_{0}tf(t)dt = x\sin(x)+\cos(x)-1$, find $f(\pi), f'(x)$ $$\int^{x}_{0}tf(t)dt = x\sin(x)+\cos(x)-1$$
 find $f(\pi), f'(x)$
This question confuses me because usually, the way I have seen questions like these, they have been in the form:
$$H(x) = \int^{x}_{a} f(t)dt$$
This form is kind of different so I am not sure how I would solve it.
My thinking:
$f(\pi) = x\sin(\pi)+\cos(\pi)-1$
$f'(x) = x\cos(x)$
Is this correct?
 A: I'm afraid your answer is incorrect. 
If you differentiate both sides, like normal, then using the product rule, we get that $$ xf(x)=x\cos x \iff f(x)=\cos x$$ So $f(\pi)=-1, f'(x)=-\sin x$. 
EDIT
As pointed out by @celtschk, we require that $f(x)$ is continous in order for this to work. This is because for us to know for certain $$\frac {d}{dx}\int_{0}^{x} g(t) \mathrm {d}t =g(x)$$
for all $x$ we require something such as the continuity of $xf(x)$ for this solution to be true. Also, we need $f (x) $ to be continuous because of the issues in $f (0) $. 
A: 
Why is my method wrong?

To do correctly, we have to use the concept of differentiation under the integral sign. We thus have, $$\frac {d}{dx} (\int_{0}^{t} tf (t) \mathrm {d}t) = \frac {d}{dx}(x\sin x +\cos x -1) $$ $$\Rightarrow xf (x) -0 = x\cos x +\sin x - \sin x$$ $$\Rightarrow \boxed {f (x) = \cos x} $$
A: Well, strictly speaking the integral equation doesn't fix $f(\pi)$, nor does it guarantee even the existence of $f'(x)$. For example, for arbitrary $\alpha$ be
$$f_\alpha(x) = \cases{
\alpha & for $x\in\pi\mathbb Q$\\
\cos x & otherwise}$$
Then $f_\alpha$ fulfills the integral equality, $f_\alpha(\pi)=\alpha$, and $f_\alpha$ is nowhere differentiable.
However if we add the requirement that $f$ is continuous, it is uniquely defined. By deriving both sides of the integral equation by $x$, we get
that $x f(x) = x\cos x$ must hold almost everywhere. Since the complement of a measure-zero set is dense (if it weren't, the measure zero set would contain an open interval, but open intervals have positive measure), continuity then fixes $f$ to be given by $f(x)=\cos x$ (note that the continuity requirement also fixes the value $f(0)$, which wouldn't even be fixed by $x f(x) = x\cos x$ if that equation needed to hold everywhere).
Obviously, if $f(x)=\cos x$, then $f(\pi)=-1$ and $f'(x)=-\sin x$.
