Substitution works inside definite integrals too. The cleanest way is to change the limits along with the variable, so you never have to switch back to $x$.
Key idea
$$\int_a^b f\big(g(x)\big)g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.$$ When $x$ runs from $a$ to $b$, $u = g(x)$ runs from $g(a)$ to $g(b)$: the new limits are $u$-values.
Why it works
If $F$ is an antiderivative of $f$, then $F(g(x))$ is an antiderivative of the left integrand, so by the Fundamental Theorem the left side is $F(g(b)) - F(g(a))$. That is exactly the right side. In words: the limits of an integral always describe the variable of integration, so once the variable is $u$, the limits must be $u$-values.
Worked example: Change the limits
Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^3\,dx$.
- $u = x^2 + 1$, $du = 2x\,dx$.
- New limits: $x = 0 \Rightarrow u = 1$; $x = 1 \Rightarrow u = 2$.
- $\int_1^2 u^3\,du = \Big[\frac{u^4}{4}\Big]_1^2 = \frac{16 - 1}{4}$.
Answer: $\frac{15}{4}$
Worked example: Trig
Evaluate $\displaystyle\int_0^{\pi/2}\sin^2 x\cos x\,dx$.
- $u = \sin x$, $du = \cos x\,dx$. Limits: $\sin 0 = 0$, $\sin\frac{\pi}{2} = 1$.
- $\int_0^1 u^2\,du = \frac{1}{3}$.
Answer: $\frac{1}{3}$
Worked example: An exponential
Evaluate $\displaystyle\int_0^2 x e^{x^2}\,dx$.
- $u = x^2$, $x\,dx = \frac{1}{2}du$. Limits: $0 \to 0$ and $2 \to 4$.
- $\frac{1}{2}\int_0^4 e^u\,du = \frac{1}{2}(e^4 - 1)$.
Answer: $\frac{e^4 - 1}{2} \approx 26.8$
There is a second, equally correct route: do the indefinite integral first, substitute back to $x$, and then use the original limits. In the first example: $\int 2x(x^2 + 1)^3\,dx = \frac{(x^2 + 1)^4}{4} + C$, and $\Big[\frac{(x^2 + 1)^4}{4}\Big]_0^1 = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$. Changing the limits is usually shorter, because you never need to substitute back.
Common mistake
Changing the variable but keeping the old limits: $\int_0^1 u^3\,du = \frac{1}{4}$ in the first example, which is wrong. Either change the limits to $u$-values, or substitute back to $x$ before using the original limits, but never mix the two. Check: write "$u = \ldots$" next to each limit.