# 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 $$\displaystyle \int_{-a}^a f(x) \; dx = 0$$.