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 $u$ , rewrite everything in terms of $u$ , integrate, and then convert back to the original variable.

Steps to apply substitution:

Choose $u$ : 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 $u$ .
Find $du$ : Differentiate $u$ with respect to the original variable. For example, if $u = x^2 + 1$ , then $du = 2x\,dx$ .
Rewrite the integral: Replace every occurrence of the original variable with expressions involving $u$ and $du$ . No leftover $x$ 's should remain.
Integrate in terms of $u$ : The new integral should be one you recognize.
Substitute back: Replace $u$ with the original expression to write your answer in terms of the original variable.

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

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

Notice how the integrand contained both the inner function $x^2$ and its derivative $2x$ . 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:

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

Example: Evaluate $\int_0^1 2x\cos(x^2)\,dx$ .

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

A common mistake is changing the variable to $u$ but forgetting to change the limits. If your limits are still in terms of $x$ , 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, $e^{3x}$ has the inner function $3x$ inside the exponential, and $\cos(x^2)$ has $x^2$ inside cosine.
The derivative of that inner function appearing as a factor. In $\int 2x\cos(x^2)\,dx$ , the factor $2x$ is exactly the derivative of $x^2$ .
A constant multiple is fine. If the derivative is off by a constant factor, you can adjust. For $\int x\cos(x^2)\,dx$ , you'd set $u = x^2$ , get $du = 2x\,dx$ , and write $x\,dx = \frac{1}{2}\,du$ .

Common substitution choices:

Integrand pattern	Typical substitution
$f(ax + b)$	$u = ax + b$
$f(x^n) \cdot x^{n-1}$	$u = x^n$
$f(\sin x)\cdot \cos x$	$u = \sin x$
$f(e^x)\cdot e^x$	$u = e^x$
$\frac{f(\ln x)}{x}$	$u = \ln x$
$f(\sqrt{x}) \cdot \frac{1}{\sqrt{x}}$	$u = \sqrt{x}$

If after substituting you still have the original variable mixed in with $u$ , 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 $du = g'(x)\,dx$ isn't just shorthand. It's what makes the algebra of substitution work cleanly, letting you swap $g'(x)\,dx$ for $du$ 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.