When you are computing the definite integral $\int_a^b f(t)dt$ your usual procedure has been:
(i) Find an antiderivative $F(t)$ of $f(t)$. That means, find a function $F(t)$ such that F'(t)=f(t).
(ii) Then your "answer" is $F(b)-F(a)$.
The procedure is exactly the same if $a$ and $b$ are not constants, but are functions of some variable $x$. So (ii) remains unchanged.
In your particular integral, we have $a=a(x)=x^3$ and $b=b(x)=-3$. The "lower" limit of integration is a constant, which you can think of as a constant function, and the upper limit is the function $x^3$.
Let's calculate. We have $f(t)=t^2$. So one antiderivative of $f(t)$ is $F(t)$, where $F(t)=\frac{t^3}{3}$
Now do the familiar substitution process. We get $\int_{-3}^{x^3}t^2dt=F(x^3)-F(-3)=\frac{(x^3)^3 -(-3)^3}{3}$
We might want to simplify this to $\frac{x^9+27}{3}$
Comments: The variable $t$ is called a "dummy variable" roughly because ultimately it plays no role in the answer. You would get exactly the same thing if you had $u$ as the variable, or $w$, instead of $t$.
In principle, you could also have used the letter $x$ as the dummy variable. But don't ever do it when one or both of the limits of integration involves $x$. Although it is technically not wrong, the chances of getting confused are too high.