Let the cubic polynomial be $P(x)=ax^3+bx^2+cx+d$ where $a,b,c,d\in \mathbb{R}$.

Use the local extrema information

Since $P(x)$ has a local maximum at $x=-1$ and a local minimum at $x=1$, we must have $P'(-1)=0,\qquad P'(1)=0.$

Now, $P'(x)=3ax^2+2bx+c.$

So, $P'(1)=3a+2b+c=0$ $P'(-1)=3a-2b+c=0$

Subtracting, $4b=0 \implies b=0.$

Then, $3a+c=0 \implies c=-3a.$

Hence, $P(x)=ax^3-3ax+d.$

Use the condition that $P(-3)=0$

Since the polynomial vanishes at $x=-3$, $P(-3)=a(-27)-3a(-3)+d=0$ $-27a+9a+d=0$ $-18a+d=0 \implies d=18a.$

Therefore, $P(x)=a(x^3-3x+18).$

Use the definite integral condition

Given $\int_{-1}^{1} P(x)\,dx=18.$

So, $\int_{-1}^{1} a(x^3-3x+18)\,dx=18.$

Factor out $a$: $a\int_{-1}^{1}(x^3-3x+18)\,dx=18.$

Now, over $[-1,1]$:

• $x^3$ is odd, so $\int_{-1}^{1}x^3\,dx=0$
• $-3x$ is odd, so $\int_{-1}^{1}(-3x)\,dx=0$
• $18$ is constant, so $\int_{-1}^{1}18\,dx=18(2)=36$

Thus, $a\cdot 36=18 \implies a=\frac12.$

Therefore, $P(x)=\frac12x^3-\frac32x+9.$

Check the nature of extrema

$P'(x)=\frac32x^2-\frac32=\frac32(x^2-1).$ So critical points are at $x=\pm 1$.

$P''(x)=3x.$ Then, $P''(-1)=-3<0$ so $x=-1$ is a local maximum, and $P''(1)=3>0$ so $x=1$ is a local minimum.

Hence all conditions are satisfied.

Find the sum of all coefficients

The sum of coefficients of a polynomial is $P(1)$.

Compute: $P(1)=\frac12-\frac32+9=8.$

So the required integer is $\boxed{8}.$