Chapter 3 Proofs: The Substitution Theorem
This page gives rigorous proofs of three theorems concerning integration by substitution. In each of them, the chain rule for composite functions plays the central role.
1. Substitution for Indefinite Integrals
Theorem (Substitution for Indefinite Integrals)
Let $f(u)$ be a continuous function and $g(x)$ a differentiable function. If $F(u)$ is an antiderivative of $f(u)$ (that is, $F'(u) = f(u)$), then
$$ \displaystyle\int f\bigl(g(x)\bigr)\, g'(x)\, dx = F\bigl(g(x)\bigr) + C $$Alternatively, setting $u = g(x)$, we have formally $du = g'(x)\,dx$, so this can be written as
$$ \displaystyle\int f\bigl(g(x)\bigr)\, g'(x)\, dx = \displaystyle\int f(u)\, du $$Proof
Let $F(u)$ be an antiderivative of $f(u)$; that is,
$$ F'(u) = f(u) $$Differentiate the composite function $F\bigl(g(x)\bigr)$ with respect to $x$. By the chain rule,
$$ \dfrac{d}{dx}\, F\bigl(g(x)\bigr) = F'\bigl(g(x)\bigr) \cdot g'(x) = f\bigl(g(x)\bigr) \cdot g'(x) $$This means that $F\bigl(g(x)\bigr)$ is an antiderivative of $f\bigl(g(x)\bigr)\, g'(x)$. Therefore, by the definition of the indefinite integral,
$$ \displaystyle\int f\bigl(g(x)\bigr)\, g'(x)\, dx = F\bigl(g(x)\bigr) + C $$Since $\displaystyle\int f(u)\, du = F(u) + C$ ($F$ being an antiderivative of $f$), taking $u = g(x)$ gives
$$ \displaystyle\int f\bigl(g(x)\bigr)\, g'(x)\, dx = \displaystyle\int f(u)\, du \bigg|_{u=g(x)} $$as required. $\blacksquare$
2. Substitution for Definite Integrals
Theorem (Substitution for Definite Integrals)
Let $f(u)$ be a continuous function and $g(x)$ be of class $C^1$ (continuously differentiable) on the closed interval $[a,\, b]$. Then
$$ \displaystyle\int_a^b f\bigl(g(x)\bigr)\, g'(x)\, dx = \displaystyle\int_{g(a)}^{g(b)} f(u)\, du $$Proof
Since $f(u)$ is continuous, it has an antiderivative $F(u)$; that is,
$$ F'(u) = f(u) $$Evaluating the left-hand side.
By the result of the previous section, $F\bigl(g(x)\bigr)$ is an antiderivative of $f\bigl(g(x)\bigr)\, g'(x)$. Applying the Fundamental Theorem of Calculus (FTC),
$$ \displaystyle\int_a^b f\bigl(g(x)\bigr)\, g'(x)\, dx = F\bigl(g(x)\bigr) \Big|_a^b = F\bigl(g(b)\bigr) - F\bigl(g(a)\bigr) $$Evaluating the right-hand side.
Similarly, since $F(u)$ is an antiderivative of $f(u)$, by the FTC,
$$ \displaystyle\int_{g(a)}^{g(b)} f(u)\, du = F(u) \Big|_{g(a)}^{g(b)} = F\bigl(g(b)\bigr) - F\bigl(g(a)\bigr) $$Conclusion.
Both the left-hand side and the right-hand side equal $F\bigl(g(b)\bigr) - F\bigl(g(a)\bigr)$. Hence
$$ \displaystyle\int_a^b f\bigl(g(x)\bigr)\, g'(x)\, dx = \displaystyle\int_{g(a)}^{g(b)} f(u)\, du $$as required. $\blacksquare$
Note: Transforming the Interval of Integration
In substitution for definite integrals, note that the change of variables $u = g(x)$ also changes the interval of integration. Since $u = g(a)$ when $x = a$ and $u = g(b)$ when $x = b$, the interval of integration changes from $[a,\, b]$ to $[g(a),\, g(b)]$. It may happen that $g(a) > g(b)$, but the formula still applies as it stands.
3. The Linear Substitution Formula
Theorem (Linear Substitution Formula)
Let $f(x)$ be a continuous function with antiderivative $F(x)$ (that is, $F'(x) = f(x)$). For $a \neq 0$,
$$ \displaystyle\int f(ax + b)\, dx = \dfrac{1}{a}\, F(ax + b) + C $$Proof
Set $g(x) = ax + b$. Then $g'(x) = a$.
Differentiate $\dfrac{1}{a}\, F(ax + b) = \dfrac{1}{a}\, F\bigl(g(x)\bigr)$ with respect to $x$. By the chain rule,
$$ \dfrac{d}{dx}\left[\dfrac{1}{a}\, F\bigl(g(x)\bigr)\right] = \dfrac{1}{a}\, F'\bigl(g(x)\bigr) \cdot g'(x) = \dfrac{1}{a}\, f(ax + b) \cdot a = f(ax + b) $$Therefore $\dfrac{1}{a}\, F(ax + b)$ is an antiderivative of $f(ax + b)$. By the definition of the indefinite integral,
$$ \displaystyle\int f(ax + b)\, dx = \dfrac{1}{a}\, F(ax + b) + C $$as required. $\blacksquare$
Alternative Proof (via Integration by Substitution)
The same result is obtained by applying integration by substitution directly. Setting $u = ax + b$, we have $du = a\, dx$, so $dx = \dfrac{1}{a}\, du$. Therefore
$$ \displaystyle\int f(ax + b)\, dx = \displaystyle\int f(u) \cdot \dfrac{1}{a}\, du = \dfrac{1}{a} \displaystyle\int f(u)\, du = \dfrac{1}{a}\, F(u) + C = \dfrac{1}{a}\, F(ax + b) + C $$This alternative proof is a formal computation, but since the validity of integration by substitution is guaranteed by the proof above, it is logically correct.
References
- Teiji Takagi, Kaiseki Gairon (An Introduction to Analysis), Iwanami Shoten.
- Mitsuo Sugiura, Kaiseki Nyumon I (Introduction to Analysis I), University of Tokyo Press.
Frequently Asked Questions
How is the substitution (change-of-variables) formula proved?
The chain rule for composite functions is the key: setting $G(t)=F(g(t))$ gives $G'(t)=F'(g(t))g'(t)=f(g(t))g'(t)$. By the Fundamental Theorem of Calculus, $\int_\alpha^\beta f(g(t))g'(t)dt=G(\beta)-G(\alpha)=F(g(\beta))-F(g(\alpha))=\int_{g(\alpha)}^{g(\beta)}f(x)dx$. In the multivariable case, $g$ extends to a form involving the Jacobian $|\det J_g|$.
Why do you change the limits of integration when substituting in a definite integral?
Because the integral is then taken with respect to $u = g(x)$, the limits must be expressed in $u$ as well. Since $u = g(a)$ when $x = a$ and $u = g(b)$ when $x = b$, the interval becomes $[g(a),\, g(b)]$.
When is trigonometric substitution useful?
It is useful for integrals containing a radical such as $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, or $\sqrt{x^2-a^2}$. Substituting $x = a\sin\theta$, $x = a\tan\theta$, or $x = a/\cos\theta$ respectively removes the radical via a trigonometric identity.