Integration by substitution applies to definite integrals as well
Thanks to the Fundamental Theorem of Calculus, the method of integration by substitution also applies to definite integrals.
We already learned how to use the method of substitution to turn seemingly complicated indefinite integrals into simpler ones we can actually compute.
For example, we learned how to compute integrals like these \[ \int e^{x^2} \, 2x \, dx, \quad\quad\quad \int \frac{3x}{\sqrt{x^2+1}} dx, \]
A natural question to ask is: What about definite integral like \[ \int_0^1 \frac{3x}{\sqrt{x^2+1}} dx \, ? \]
Thanks to the Fundamental Theorem of Calculus, definite integrals can be computed through indefinite integrals, so the same applies, as long as we also perform the same transformation to the limits of integration.
Solution. Applying a substitution, we let $u = x-1$ then $du = dx$.
When $x=0$, $u=-1$.
When $x=2$, $u=1$.
Therefore, \[ \int_0^2 (x-1)^5 dx = \int_{-1}^{1} u^5 du = \left[ \frac{u^6}{6} \right]_{-1}^1 = \frac{1^6}{6} - \frac{(-1)^6}{6} = \frac{1}{6} - \frac{1}{6} = 0. \]
We now state the simple procedure illustrated in the previous example more formally.
As we saw in the previous example, this is nothing but a formal way of saying: When you perform a substitution, you should substitute the limits of integration as well.
In certain situations, we may not want to compute the new limits of integration (e.g., the computation maybe cumbersome or inexact).
Applying a substitution, we let $\color{red}{u = \sin(x)}$, then $\color{blue}{du = \cos(x) dx}$. So $\int [\color{red}{\sin(x)}]^2 \color{blue}{\cos(x) dx} = \int \color{red}{u}^2 \color{blue}{du}$.
So the original integral can be turned into \[ \int_\square^\square u^2 du = \left[\frac{1}{3} u^3 \right]_\square^\square = \left[\frac{1}{3} (\sin x)^3 \right]_1^2 = \frac{\sin^3(2) - \sin^3(1)}{3}. \]
Here, we use "$\square$" to indicate that we should compute the new limits of integration with respect to the new variable $u$, but we are just don't want do that. Instead, we will switch back to $x$ in the next step, thus $\square$'s are placeholders for the limits of integration in the intermediate steps.