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
First of all, when involving two variables, you should clarify the variable which you are differentiating. Following from your question, because of the fundamental theorem of calculus, we have:
$$
\frac{d\left( \displaystyle\int\_{0}^{x} t \cdot f(t)\mathrm{d}t \right)}{dx}=x \cos(x)+x^2
$$
Implies:
$$
xf(x)=x\cos(x)+x^2 \implies f(x)=\cos(x)+x
$$
Let's go back to the beginning. If we substitute the value of $f$, we get:
Implies:

$$
\left( \displaystyle\int\_{c}^{x} t \cdot (\cos(t)+t)\mathrm{d}t \right)=x\sin (x)+\cos (x)+\dfrac{x^3}{3}
$$
After doing all the calculations, we will have:
$$
c\sin (c)+\cos (c)+\dfrac{c^3}{3}=0
$$
I used an external program to solve this, but you can use Bolzano's Theorem to see that the equation has a solution for some $c\in [-1,0]$