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: 
Let's define:
$$g(t)=tf(t)$$
$$F(x)=\int\_0^xg(t)\,dt$$
so we have:
$$F(x)-F(c)=x\sin(x)+\cos(x)+\frac{x^3}3$$
now if we differentiate both sides it should be clear that:
$$g(x)=xf(x)=\left(x\sin(x)+\cos(x)+\frac{x^3}3\right)'$$
now calculate the derivative and divide through by $x$:
$$xf(x)=\sin(x)+x\cos(x)-\sin(x)+x^2$$
$$f(x)=x+\cos(x)$$

---

Now if we sub back in:
$$\int\_c^x\left[t^2+t\cos(t)\right]dt=x\sin(x)+\cos(x)+\frac{x^3}{3}$$
$$x\sin(x)+\cos(x)+\frac{x^3}{3}-\frac{3c\sin(c)+3\cos(c)+c^3}{3}=x\sin(x)+\cos(x)+\frac{x^3}3$$
and now you are left with:
$$3c\sin(c)+3\cos(c)+c^3=0$$
$$c\sin(c)+\cos(c)+\frac{c^3}{3}=0$$
and now there isn't an analytical way to solve this so a numerical method is the best you can do
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 $-1.673463$.