Integration by Substitution

Substitution is the working engine behind evaluating integrals in every quantitative science: computing arc length and surface area in geometry relies on substitutions that absorb a square-root Jacobian factor; changing to polar, cylindrical, or spherical coordinates in multivariable calculus is the multivariate generalization of this same chain-rule argument, packaged as the Jacobian determinant in the Change of Variables Theorem. In probability and statistics, the transformation-of-random-variables formula for densities — $f_Y(y) = f_X(g^{-1}(y))\,|(g^{-1})'(y)|$ — is exactly the substitution theorem read as a statement about probability densities. In physics, substitution underlies dimensional analysis and non-dimensionalization of differential equations, where rescaling a variable ($u = x/L$ for some length scale $L$) simplifies an equation's structure. In numerical analysis and scientific computing, substitution is used to map a difficult integration domain (e.g., an infinite or singular interval) to a well-behaved finite domain before applying quadrature rules — a technique that becomes essential once one studies Improper Integrals, where the substitution itself is allowed to send an endpoint to infinity or to a singularity of the integrand. In cryptography and coding theory, generating functions and combinatorial identities are frequently manipulated via substitution in formal power series, a discrete cousin of the same idea.

:::quiz[In Example 2, why was it valid to use the general theorem despite $\sin x$ not being monotonic on $[0,3]$?]

• Because $\sin x$ is bounded, monotonicity is automatically implied
• Because the theorem's proof never assumed monotonicity of $g$ — only $g\in C^1$ and $f$ continuous ✓
• Because $\cos x$ happens to be positive throughout $[0,3]$
• Non-monotonic substitutions require a separate, unproven extension of the theorem

:::faq[What if $f$ is continuous on $g([a,b])$ but not on a whole interval containing it?] The hypothesis is stated with $J$ an interval containing $g([a,b])$ precisely so that Step 2 of the proof — defining $F(u) = \int_c^u f(t)\,dt$ — makes sense for a fixed basepoint $c$ and all $u$ between $c$ and any point of $g([a,b])$. If $g([a,b])$ is itself already an interval (which happens automatically since $g$ is continuous on the connected set $[a,b]$, and continuous images of connected sets are connected), you may simply take $J = g([a,b])$ itself; the extra room in $J$ is a convenience, not a real strengthening, as long as $g([a,b])$ is a nondegenerate interval or a point.

Ask the tutor: How does this substitution formula generalize when the interval $[a,b]$ becomes unbounded, as in improper integrals?

The chain rule tells you how the derivative of a composite function decomposes: if $F$ is an antiderivative of $f$, then

This says that $f(g(x))g'(x)$, which looks like a complicated composite expression, is secretly the derivative of the simpler function $F\circ g$. The Fundamental Theorem of Calculus converts "knowing a derivative" into "knowing a definite integral": integrating a derivative over $[a,b]$ just evaluates the antiderivative at the endpoints. Apply that on both sides of the identity above, and the substitution formula falls out — no independent postulate about $du$ being "$g'(x)\,dx$" is needed; that notation is simply a mnemonic encoding the chain rule. Geometrically, think of $g$ as a reparametrization of the $x$-axis into a $u$-axis. As $x$ sweeps from $a$ to $b$, the point $u=g(x)$ sweeps from $g(a)$ to $g(b)$, possibly backtracking if $g$ is not monotonic. The formula says the signed accumulation of $f(u)$ as $u$ traces this path, weighted by how fast $u$ moves ($g'(x)$), exactly equals the net signed area $\int_{g(a)}^{g(b)} f$. Backtracking is automatically accounted for by the sign of $g'(x)$, which is why no monotonicity hypothesis is required for the exact identity — although monotonicity is exactly what you need if you want to reduce it further to a statement about unsigned areas under a graph.

Example 1 (asymmetric algebraic substitution). Compute

Let $g(x) = 2x^2+5$. Then $g\in C^1[1,2]$ with $g'(x) = 4x$, and $g$ is continuous and strictly increasing on $[1,2]$ (since $g'(x) = 4x>0$ there), with $g(1) = 7$ and $g(2) = 13$. Write the integrand to expose $g'(x) = 4x$:

Here $f$ is continuous on $J = [7,13]\subset (0,\infty)$, so all hypotheses of the theorem hold ($g \in C^1[1,2]$, $f$ continuous on an interval containing $g([1,2]) = [7,13]$). By the theorem,

Compute the right side directly via the Second FTC, using the antiderivative $2u^{1/2}$ of $u^{-1/2}$ on $(0,\infty)$:

Therefore

Numerically $\sqrt{13}\approx 3.6056$, $\sqrt7\approx 2.6458$, so $I \approx \tfrac{0.9598}{2}\approx 0.4799$, matching a numerical estimate of the original integral. Note $g$ was strictly monotonic here, but this was not needed to invoke the theorem — it only makes the geometric picture (a genuine reparametrization without backtracking) especially clean.

Example 2 (non-monotonic substitution, testing the exact-formula robustness). Compute

Notice $\cos x$ changes sign on $[0,3]$ since $3 > \pi/2 \approx 1.5708$: $\cos x > 0$ on $[0,\pi/2)$ and $\cos x<0$ on $(\pi/2, 3]$. So the substitution $u = \sin x$ is not monotonic on $[0,3]$ — $\sin x$ increases from $0$ up to $1$ at $x=\pi/2$, then decreases down to $\sin 3 \approx 0.1411$. This is exactly the situation the general theorem (not the monotonic corollary) is built to handle. Let $g(x) = \sin x$. Then $g \in C^1[0,3]$ (indeed $g\in C^\infty$), with $g'(x) = \cos x$, continuous everywhere. Let $f(u) = u\,e^{u^2}$, continuous on all of $\mathbb R$, in particular on $J = [-1,1] \supset g([0,3])$ (since $|\sin x|\le 1$ always). The integrand is

so all hypotheses of the theorem hold with no monotonicity assumption needed. By the theorem,

since $g(0) = \sin 0 = 0$ and $g(3) = \sin 3$. Using the antiderivative $\tfrac12 e^{u^2}$ of $u e^{u^2}$ (Second FTC):

Numerically, $\sin 3 \approx 0.14112$, so $\sin^2 3 \approx 0.019915$, $e^{\sin^2 3}\approx 1.02012$, giving

The key point: even though $u=\sin x$ retraces part of $[0, 1]$ twice as $x$ ranges over $[0,3]$ (once increasing, once decreasing), the formula required only $g(0)$ and $g(3)$ as endpoints — the "backtracked" portion of the path exactly cancels in the signed integral, which is precisely what the Second FTC computation on the composite $F\circ g$ captures automatically. A naive attempt to instead compute this by splitting into "area swept forward" plus "area swept backward" as unsigned regions would give a different, wrong number; the theorem's signed bookkeeping is what makes the one-line computation valid despite non-monotonicity.