Question: > 
> Find a number and a function that satisfies:
> $$\displaystyle\int\_{c}^{x}t\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\dfrac{x^3}{3}$$
> 
> 
> 

I tried taking derivative both sides so I got:

$$\left(\displaystyle\int\_{0}^{x}t\cdot f(t)\,\mathrm{d}t-\displaystyle\int\_{0}^{c}t \cdot f(t)\,\mathrm{d}t\right)^\prime=x \cos(x)+x^2 $$

It's clear that the right part of the integral will be $0$ because it's constant:

$$\left( \displaystyle\int\_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right)^{\prime}=x \cos(x)+x^2$$

But I didn't do rest. Because I don't know how to derive the integral with two functions. I know with one function like $\int\_{0}^{x} f(t)\,\mathrm{d}t$. Thanks!

Answer: 
As said in answers, there is no way to find the zero of function $$f(c)=3c\sin(c)+3\cos(c)+c^3$$ but we can make nice approximations.

By inspection, the solution is "close" to $c=-\frac \pi 2$. So, we can do **one** single iteration of Householder method and obtain, *by hand*, the estimate
$$c \sim -\frac{1872 \pi +552 \pi ^3-5 \pi ^5}{1152+1104 \pi ^2+8 \pi ^4}$$ which is $-1.673493$ while the "exact" solution is $-1.673463$.
You can still think of $t\mapsto tf(t)$ as a function, and differentiate as you normally would using the [fundamental theorem of calculus](https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus):
$$\frac{d}{dx}\left( \int\_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right) = xf(x).$$