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Calculus II Unit 1 Review

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1.5 Substitution

1.5 Substitution

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
Calculus II
Unit & Topic Study Guides

Substitution Method for Integration

The substitution method lets you transform a complicated integral into a simpler one by swapping in a new variable. It's the reverse of the chain rule: where the chain rule helps you differentiate composite functions, substitution helps you integrate them. Once you get comfortable spotting the right substitution, a huge number of integrals that look intimidating become straightforward.

Substitution for Indefinite Integrals

The core idea is to replace part of the integrand with a new variable uu, rewrite everything in terms of uu, integrate, and then convert back to the original variable.

Steps to apply substitution:

  1. Choose uu: Look at the integrand and identify an "inner function" whose derivative (or a constant multiple of it) also appears in the integrand. Set that inner function equal to uu.
  2. Find dudu: Differentiate uu with respect to the original variable. For example, if u=x2+1u = x^2 + 1, then du=2xdxdu = 2x\,dx.
  3. Rewrite the integral: Replace every occurrence of the original variable with expressions involving uu and dudu. No leftover xx's should remain.
  4. Integrate in terms of uu: The new integral should be one you recognize.
  5. Substitute back: Replace uu with the original expression to write your answer in terms of the original variable.

Example: Evaluate 2xcos(x2)dx\int 2x\cos(x^2)\,dx.

  • Let u=x2u = x^2, so du=2xdxdu = 2x\,dx.
  • The integral becomes cos(u)du=sin(u)+C\int \cos(u)\,du = \sin(u) + C.
  • Substituting back: sin(x2)+C\sin(x^2) + C.

Notice how the integrand contained both the inner function x2x^2 and its derivative 2x2x. That pairing is exactly what makes substitution work.

Substitution for indefinite integrals, The Chain Rule · Calculus

Substitution in Definite Integrals

The process is the same, with one key difference: you need to handle the limits of integration.

You have two options:

  1. Change the limits (usually faster): When you substitute u=g(x)u = g(x), convert the original limits x=ax = a and x=bx = b into u=g(a)u = g(a) and u=g(b)u = g(b). Then evaluate the integral entirely in terms of uu. No back-substitution needed.
  2. Keep the original limits: Find the antiderivative in terms of uu, substitute back to the original variable, and then evaluate at the original limits.

Example: Evaluate 012xcos(x2)dx\int_0^1 2x\cos(x^2)\,dx.

  • Let u=x2u = x^2, so du=2xdxdu = 2x\,dx.
  • Change limits: when x=0x = 0, u=0u = 0; when x=1x = 1, u=1u = 1.
  • The integral becomes 01cos(u)du=sin(u)01=sin(1)sin(0)=sin(1)\int_0^1 \cos(u)\,du = \sin(u)\Big|_0^1 = \sin(1) - \sin(0) = \sin(1).

A common mistake is changing the variable to uu but forgetting to change the limits. If your limits are still in terms of xx, you must substitute back before evaluating.

Substitution for indefinite integrals, calculus - Steps in evaluating infinite integral - Mathematics Stack Exchange

Recognizing When to Use Substitution

Substitution works best when the integrand contains a composite function and something resembling the derivative of the inner function. Here's what to look for:

  • A function nested inside another function. For instance, e3xe^{3x} has the inner function 3x3x inside the exponential, and cos(x2)\cos(x^2) has x2x^2 inside cosine.
  • The derivative of that inner function appearing as a factor. In 2xcos(x2)dx\int 2x\cos(x^2)\,dx, the factor 2x2x is exactly the derivative of x2x^2.
  • A constant multiple is fine. If the derivative is off by a constant factor, you can adjust. For xcos(x2)dx\int x\cos(x^2)\,dx, you'd set u=x2u = x^2, get du=2xdxdu = 2x\,dx, and write xdx=12dux\,dx = \frac{1}{2}\,du.

Common substitution choices:

Integrand patternTypical substitution
f(ax+b)f(ax + b)u=ax+bu = ax + b
f(xn)xn1f(x^n) \cdot x^{n-1}u=xnu = x^n
f(sinx)cosxf(\sin x)\cdot \cos xu=sinxu = \sin x
f(ex)exf(e^x)\cdot e^xu=exu = e^x
f(lnx)x\frac{f(\ln x)}{x}u=lnxu = \ln x
f(x)1xf(\sqrt{x}) \cdot \frac{1}{\sqrt{x}}u=xu = \sqrt{x}

If after substituting you still have the original variable mixed in with uu, your substitution probably isn't the right one. A good substitution eliminates the original variable completely.

Connection to Other Techniques

  • Chain rule in reverse: Substitution undoes the chain rule. If you can spot the "outer function" and "inner function" structure, substitution will likely work.
  • Differential notation: Writing du=g(x)dxdu = g'(x)\,dx isn't just shorthand. It's what makes the algebra of substitution work cleanly, letting you swap g(x)dxg'(x)\,dx for dudu directly.
  • When substitution isn't enough: Some integrals need other methods like integration by parts (for products of unrelated functions) or partial fractions (for rational functions). If no substitution simplifies the integral, try a different technique.