U-Substitution

You’ll be able to

Recognize an integrand as f(g(x))·g′(x), the reverse of the chain rule
Carry out u-substitution for indefinite and definite integrals
Change the limits of integration when substituting in a definite integral

Undoing the chain rule

U-substitution is integration’s answer to the chain rule. When an integrand looks like f(g(x))·g′(x) — a composite times the derivative of its inside — set u = g(x) so that du = g′(x) dx. The integral in x becomes a simpler integral in u. The signal to try it: you spot an "inside function" and its derivative (up to a constant) both present in the integrand.

The mechanics

Choose u to be the inside function, compute du = g′(x) dx, and rewrite the whole integrand — integrand and dx — in terms of u. Integrate in u, then substitute back to x for an indefinite integral. A missing constant factor can be adjusted: if you need du but have only part of it, multiply and divide by the constant to balance.

Definite integrals: change the limits

For a definite integral you have a choice. Either substitute back to x at the end and use the original x-limits, or — cleaner — convert the limits to u-values using u = g(x) and evaluate entirely in u. If you change to u-limits you must not substitute back to x; the new limits already carry that information. Mixing x-limits with a u-expression is a frequent error.

U-substitution

∫ f(g(x))·g′(x) dx = ∫ f(u) du, where u = g(x), du = g′(x) dx

For definite integrals, either back-substitute to x, or change the limits: x = a, b become u = g(a), g(b).

Worked example

Evaluate ∫ 2x·(x² + 1)³ dx.

1.Let u = x² + 1, the inside function. Then du = 2x dx — and 2x dx is exactly present in the integrand.
2.Rewrite the integral in u: ∫ u³ du.
3.Integrate: u⁴/4 + C.
4.Substitute back u = x² + 1: (x² + 1)⁴/4 + C.

Answer: ∫ 2x·(x² + 1)³ dx = (x² + 1)⁴/4 + C. Differentiating back gives 4(x² + 1)³·2x/4 = 2x(x² + 1)³, confirming it.

Watch out

When you change the limits of a definite integral to u-values, do not also substitute back to x — the u-limits already account for the substitution. Either finish in u with u-limits, or return to x with the original limits, never a mix.

Checkpoint

For ∫ cos(x²)·2x dx, what is the best choice of u?

Checkpoint

When evaluating ∫[0 to 2] 2x·e^(x²) dx with u = x², what do the limits become?

On the exam

On the exam, converting the limits to u-values is usually faster and less error-prone than back-substituting. Write the new limits explicitly next to the integral so you evaluate the u-antiderivative at the correct endpoints.

Answer the 2 checkpoints as you read.