U-Substitution

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U-Substitution: meaning, method and solved example (AP Calculus AB) – Maths PDF

In short: U-substitution simplifies an integral by naming an inner expression u, rewriting everything in terms of u and du, integrating, and then switching back to x.

AP Calculus ABCalculusAP Calculus AB Unit 6

What is U-Substitution?

U-substitution is the chain rule run backward. When an integrand contains a function together with a multiple of that function’s derivative, renaming the inner piece \(u\) reveals a much simpler integral.

If \(u = g(x)\), then \(du = g^{\prime}(x)\,dx\), and \(\int f(g(x))\,g^{\prime}(x)\,dx\) becomes \(\int f(u)\,du\). For a definite integral you can also convert the limits: replace \(x = a\) and \(x = b\) by \(u = g(a)\) and \(u = g(b)\), and never return to \(x\).

How to spot it

  • The integrand has a bracket raised to a power and, outside it, something resembling the bracket’s derivative.
  • A function like cos(5x) or the square root of an expression appears with a linear or polynomial inside.
  • The integral is a product, but one factor is (up to a constant) the derivative of the other factor’s inside.
  • Expanding the bracket would be long, such as (3x^2 + 1)^4, which is when u-substitution pays off.

The method

  1. For u-substitution, choose \(u\) as the inner expression whose derivative also appears (up to a constant factor).
  2. Compute \(du = u^{\prime}\,dx\) and solve for the leftover \(x\)-part, adjusting constants as needed.
  3. Rewrite the whole integral in \(u\) only, integrate, and for a definite integral convert the limits.
  4. For an indefinite integral, substitute back \(u = g(x)\) and add \(+C\); then differentiate to check.
1. Integral of 6x (3x^2 + 1)^4 dx2. Let u = 3x^2 + 1, so du = 6x dx3. Integral of u^4 du = u^5 / 54. Back to x: (3x^2 + 1)^5 / 5 + C
The messy integral turns into one you can do with the power rule in a single step.

U-Substitution: solved example

Problem

Evaluate \(\int_0^2 x(x^2 + 1)^3\,dx\).

Solution

Let \(u = x^2 + 1\), so \(du = 2x\,dx\) and \(x\,dx = \frac{1}{2}du\). Limits: \(x = 0\) gives \(u = 1\); \(x = 2\) gives \(u = 5\).

The integral becomes \(\frac{1}{2}\int_1^5 u^3\,du = \frac{1}{2}\cdot\frac{u^4}{4}\Big|_1^5 = \frac{1}{8}(625 – 1)\).

That equals \(\frac{624}{8} = \) 78.

Try it

Question 1

Find \(\int \cos(5x)\,dx\).

Show the answer

With \(u = 5x\), \(dx = \frac{1}{5}du\), giving \(\frac{1}{5}\sin(5x) + C\).

Question 2

Find \(\int 8x\sqrt{4x^2 + 9}\,dx\).

Show the answer

Let \(u = 4x^2 + 9\), \(du = 8x\,dx\). Then \(\int u^{1/2}du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(4x^2 + 9)^{3/2} + C\).

U-Substitution vs. Chain Rule

The chain rule differentiates a function inside another function. U-substitution does the reverse job when integrating: it spots the inside and the leftover derivative factor, and strips them away. A quick check is to differentiate your answer; the chain rule should return the original integrand.

Chain Rule

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