What is U-Substitution? — BUders Math Concepts

Simple Definition

U-substitution is the most fundamental integration technique after the basic antiderivative rules. The idea: replace a complicated expression inside an integral with a single variable $u$, solve the simpler integral, then substitute back.

It works because integration and differentiation are inverses — and u-sub is precisely the chain rule read backwards. When you see an integral of the form $\int f(g(x)) \cdot g'(x)\,dx$, that's a signal: let $u = g(x)$, find $du$, and the integral simplifies to $\int f(u)\,du$.

The key skill is choosing the right u — usually the inner function of a composition, or the expression whose derivative also appears in the integrand.

Formal Statement

Substitution Rule

$$\int f(g(x))\cdot g'(x)\,dx = \int f(u)\,du \quad \text{where } u = g(x)$$

For Definite Integrals — Change the Bounds

$$\int_a^b f(g(x))\cdot g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$$

Intuition

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The chain rule, reversed

The chain rule says $\frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x)$. U-substitution reads this backwards: if you see $F'(g(x)) \cdot g'(x)$ inside an integral, you know its antiderivative is $F(g(x))$. The $g'(x)\,dx$ piece becomes $du$, and the whole integral collapses.

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What to look for

You need two things: a function $g(x)$ and its derivative $g'(x)$ both appearing in the integral (sometimes with a constant factor). The inner function $g(x)$ becomes $u$. Its derivative $g'(x)dx$ becomes $du$. If $g'$ is missing, u-sub might not work — try a different method.

Step by Step — $\int 2x\,e^{x^2}\,dx$

1

Choose u — the inner function

Let $u = x^2$. This is the exponent of $e$, and its derivative $2x$ appears in the integrand.

2

Find du

$\dfrac{du}{dx} = 2x \implies du = 2x\,dx$

3

Substitute

$\displaystyle\int 2x\,e^{x^2}\,dx = \int e^u\,du$

4

Integrate in u

$\displaystyle\int e^u\,du = e^u + C$

5

Substitute back

$= e^{x^2} + C$ ✓

Tip — When du Has a Missing Constant

Adjust with a constant factor

For $\displaystyle\int x\,e^{x^2}\,dx$: let $u = x^2$, so $du = 2x\,dx$, meaning $x\,dx = \frac{du}{2}$.
Then $\displaystyle\int x\,e^{x^2}\,dx = \int e^u \cdot \frac{du}{2} = \frac{1}{2}e^u + C = \frac{1}{2}e^{x^2} + C$.
You can always factor out a constant — just never a variable.

Quick Examples

Example 1

$\displaystyle\int \cos(3x)\,dx$

$u=3x,\ du=3\,dx$
$= \frac{1}{3}\displaystyle\int \cos u\,du$
$= \frac{1}{3}\sin(3x)+C$

Example 2

$\displaystyle\int \frac{2x}{x^2+1}\,dx$

$u=x^2+1,\ du=2x\,dx$
$= \displaystyle\int \frac{du}{u} = \ln|u|+C$
$= \ln(x^2+1)+C$

Example 3 — Definite

$\displaystyle\int_0^1 2x(x^2+1)^3\,dx$

$u=x^2+1;\ x=0\to u=1,\ x=1\to u=2$
$=\displaystyle\int_1^2 u^3\,du = \left[\frac{u^4}{4}\right]_1^2 = \frac{15}{4}$

Example 4

$\displaystyle\int \sin x\,\cos x\,dx$

$u=\sin x,\ du=\cos x\,dx$
$= \displaystyle\int u\,du = \frac{u^2}{2}+C$
$= \frac{\sin^2 x}{2}+C$

Common Mistakes

✗ Forgetting to change dx to du

After choosing $u$, you must replace every $x$ in the integral — including the $dx$. Writing $\int e^u\,dx$ (with $dx$ unchanged) is wrong. The $dx$ must be fully expressed in terms of $du$ before integrating.

✗ Forgetting to substitute back for indefinite integrals

After integrating in $u$, you must replace $u$ back with the original $x$ expression. Leaving the answer in terms of $u$ is incomplete — $u$ is just a temporary variable, not part of the final answer.

✗ Not changing the bounds in definite integrals

For definite integrals, either change the bounds to $u$-values (recommended), or substitute back to $x$ before evaluating. Mixing $u$-bounds with $x$-expressions gives the wrong answer.

Related Concepts

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