Number of possible solution(s)? 
What is the number of possible solution(s)of the equation $\displaystyle\int_{0}^{x} t^2-8t+13\,\mathrm dt=x\sin\left(\dfrac ax\right)$? 

I tried applying Leibniz's rule and differentiated both sides. I got the equation $x^2-8x+13=\sin\left(\dfrac ax\right)-\dfrac ax\cos\left(\dfrac ax\right)$. How should I proceed now? Thanks in advance.
 A: Here is my attempt. First integrate the LHS and conclude:
$$\int_0^x t^2-8t+13\,\mathrm dt= \frac{t^3}{3}-4t^2+13t\bigg\lvert_{t=0}^{t=x}=\frac{x^3}{3}-4x^2+13x$$
Then the equation becomes
$$\frac{x^3}{3}-4x^2+13x=x\sin\left(\frac{a}{x}\right)$$
Since the $x$ cannot be zero, divide both sides by $x$ and we have
$$\frac{x^2}{3}-4x+13=\sin\left(\frac{a}{x}\right)$$
Thus for any $x$ that is a solution to the original solution, it also has to solve the above equation. Now observe the LHS can be rewritten as:
$$\frac{1}{3}(x^2-12x+36)+1=\frac{1}{3}(x-6)^2+1$$
Conclude that the LHS achieves its minimum $1$ at $x=6$.
Now since $\sin\left(\dfrac{a}{x}\right)$ has range $[-1,1]$. The LHS and the RHS can only intersect at most once. Thus we can consider two cases:


*

*For all $a$ such that $\sin\left(\dfrac a6\right)=1$. The solution to the equation is $x=6$.

*For all other values of $a$, no solution exists.
Please let me know if you find any error.
A: Hint: You will get the equation
$$\frac{1}{3}x^2+4x+13=\sin  \left(\frac{a}{x}\right)$$
Now it must be $$\left|\sin  \left(\frac{a}{x}\right)\right|\le 1$$
Can you proceed?
