5.5 | Integration by Substitution


Antidifferentiation of a Composite Function
Let \(g\) be a function whose range is an interval \(I\) and let \(f\) be a function that is continuous on \(I\). If \(g\) is differentiable on its domain and \(F\) is an antiderivative of \(f\) on \(I\), then
$$
\int f(g(x)) g'(x) \; dx = F(g(x)) + C.
$$
Letting \(u = g(x)\) gives \(du = g'(x) \; dx\) and
$$
\int f(u) \; du = F(u) + C.
$$
Change of Variables for Definite Integrals
If the function \(u = g(x)\) has a continuous derivative on the closed interval \([a,b]\) and \(f\) is continuous on the range of \(g\), then
$$
\int_a^b f(g(x))g'(x) \; dx = \int_{g(a)}^{g(b)} f(u) \; du.
$$
Integration of Even and Odd Functions
Let \(f\) be integrable on the closed interval \([-a,a]\).

  1. If \(f\) is an even function, then \(\displaystyle \int_{-a}^a f(x) \; dx = 2 \int_0^a f(x) \; dx\).
  2. If \(f\) is an odd function, then \(\displaystyle \int_{-a}^a f(x) \; dx = 0\).

E 5.5 Exercises