Integration by Substitution

Differentiation has a workhorse for composite functions — the chain rule. Run it backwards and you get the most-used integration technique of all: substitution. The idea is to spot that an integrand is really f(g(x)) multiplied by the derivative g'(x) of the inner function, and to fold that inner function into a single new variable u = g(x).

\int f\big(g(x)\big)\, g'(x)\, dx = \int f(u)\, du, \qquad u = g(x).

The right-hand integral has no composition left — it is in the clean variable u. Substitution is, quite literally, the chain rule read from right to left.

Deriving the rule from the chain rule

Let F be an antiderivative of f, so F' = f. We differentiate the composite F(g(x)) and then integrate the result — the two operations cancel, and what survives is exactly the substitution rule.

Step 1 — differentiate the composite with the chain rule. The outer derivative F' is evaluated at the inner function, times the inner derivative:

\frac{d}{dx}\, F\big(g(x)\big) = F'\big(g(x)\big)\, g'(x).

Step 2 — rename F' = f. By assumption F antidifferentiates f, so F'(g(x)) = f(g(x)):

\frac{d}{dx}\, F\big(g(x)\big) = f\big(g(x)\big)\, g'(x).

Step 3 — integrate both sides in x. Integration undoes the \tfrac{d}{dx} on the left, leaving the composite itself (up to a constant):

\int f\big(g(x)\big)\, g'(x)\, dx = F\big(g(x)\big) + C.

Step 4 — recognise the right-hand side as a u-integral. Set u = g(x). Then F(g(x)) = F(u), and since F' = f that is precisely an antiderivative of f(u):

F\big(g(x)\big) + C = F(u) + C = \int f(u)\, du.

Step 5 — read off the rule. Comparing Steps 3 and 4,

\int f\big(g(x)\big)\, g'(x)\, dx = \int f(u)\, du.

In differential shorthand, u = g(x) gives \dfrac{du}{dx} = g'(x), i.e. du = g'(x)\, dx — so the slogan "let u = g(x), then du = g'(x)\,dx" is not an abuse of notation but a packaged chain rule.

Worked example: \displaystyle\int 2x\cos(x^2)\, dx

The integrand hides a composite — \cos(x^2) — and, helpfully, the derivative of its inner part x^2 is 2x, which is sitting right there as a factor.

Step 1 — choose the inner function. Let

u = x^2 \qquad\Longrightarrow\qquad \frac{du}{dx} = 2x \qquad\Longrightarrow\qquad du = 2x\, dx.

Step 2 — substitute, swapping 2x\,dx for du. The factor 2x\,dx is exactly du, and \cos(x^2) = \cos u:

\int 2x\cos(x^2)\, dx = \int \cos u\, du.

Step 3 — integrate in u. A standard antiderivative:

\int \cos u\, du = \sin u + C.

Step 4 — back-substitute u = x^2. Always return to the original variable:

\int 2x\cos(x^2)\, dx = \sin(x^2) + C.

A quick sanity check: differentiate \sin(x^2) by the chain rule and you recover 2x\cos(x^2) — the integrand we started with.

A definite version: changing the limits

For a definite integral, the cleanest route is to transform the limits along with the variable, so you never have to back-substitute at all. Consider

\int_{0}^{\sqrt{\pi}} 2x\cos(x^2)\, dx.

Step 1 — same substitution. u = x^2, du = 2x\,dx.

Step 2 — push the limits through u = x^2. The limits are values of x; convert them to values of u:

x = 0 \Rightarrow u = 0^2 = 0, \qquad x = \sqrt{\pi} \Rightarrow u = (\sqrt{\pi})^2 = \pi.

Step 3 — rewrite entirely in u. No x remains, limits included:

\int_{0}^{\sqrt{\pi}} 2x\cos(x^2)\, dx = \int_{0}^{\pi} \cos u\, du.

Step 4 — evaluate. Apply the Fundamental Theorem to \sin u:

\Big[\sin u\Big]_{0}^{\pi} = \sin\pi - \sin 0 = 0 - 0 = 0.

Let g be differentiable on an interval and let f be continuous on the range of g. With u = g(x) and du = g'(x)\, dx,

\int f\big(g(x)\big)\, g'(x)\, dx = \int f(u)\, du = F(u) + C, \qquad F' = f,

and for a definite integral, with the limits carried through u = g(x),

\int_{a}^{b} f\big(g(x)\big)\, g'(x)\, dx = \int_{g(a)}^{g(b)} f(u)\, du.

The art is picking u. Reliable instincts:

Let u be the inner function of a composite — what is "inside" a power, root, exponential, sine, or logarithm.
Better still, pick the u whose derivative g'(x) already appears (up to a constant) elsewhere in the integrand. A stray constant is harmless: from du = g'(x)\,dx you can solve g'(x)\,dx = du and absorb the factor.
In \int x\, e^{x^2} dx, take u = x^2, so du = 2x\,dx and x\,dx = \tfrac12 du, giving \tfrac12\int e^u du = \tfrac12 e^{x^2} + C.

For a definite integral you have two equivalent choices. Either change the limits (convert x-limits to u-limits, as above, and never look back), or back-substitute (find the antiderivative in u, replace u = g(x), then plug in the original x-limits). The first is usually tidier — fewer chances to forget to undo the substitution.

See it: an integrand and its antiderivative

The faint curve is the integrand 2bx\cos(bx^2); the bold curve is its antiderivative \sin(bx^2), found by the very substitution u = bx^2. Slide b — the inner stretch — and notice the bold curve's slope always matches the faint curve's height: that is the Fundamental Theorem made visible.