1. We need to evaluate

$$\lim_{x\to 0} \frac{ae^x-b\cos x+ce^{-x}}{x\sin x}=2.$$

Since the denominator tends to $0$ as $x\to 0$, the numerator must also vanish to suitable order.

2. Use Taylor expansions near $x=0$:

$$e^x=1+x+\frac{x^2}{2}+\cdots,$$ $$e^{-x}=1-x+\frac{x^2}{2}+\cdots,$$ $$\cos x=1-\frac{x^2}{2}+\cdots,$$ $$\sin x=x+\cdots$$

So

$$x\sin x = x\left(x+\cdots\right)=x^2+\cdots$$

3. Expand the numerator:

$$ae^x-b\cos x+ce^{-x}$$ $$= a\left(1+x+\frac{x^2}{2}+\cdots\right)-b\left(1-\frac{x^2}{2}+\cdots\right)+c\left(1-x+\frac{x^2}{2}+\cdots\right).$$

Collect terms:

$$= (a-b+c) + (a-c)x + \left(\frac a2+\frac b2+\frac c2\right)x^2+\cdots$$

4. For the limit to be finite, the constant and linear terms must vanish:

• Constant term:

$$a-b+c=0$$

• Coefficient of $x$:

$$a-c=0 \implies a=c$$

Substitute $a=c$ into the first equation:

$$a-b+a=0 \implies 2a-b=0 \implies b=2a.$$

5. Then the coefficient of $x^2$ in the numerator is

$$\frac{a+b+c}{2}.$$

Since

$$x\sin x \sim x^2,$$

we get

$$\lim_{x\to 0} \frac{ae^x-b\cos x+ce^{-x}}{x\sin x} = \frac{a+b+c}{2}.$$

Given this limit equals $2$,

$$\frac{a+b+c}{2}=2.$$

Hence,

$$a+b+c=4.$$

6. Therefore, the required integer is

$$\boxed{4}.$$

Comparison with stored answer: the stored correct answer is $4$, which matches our result.