How do I choose what to set equal to u?

Look for a function whose derivative (or a constant multiple of it) also appears in the integrand. Common candidates are expressions inside a power, under a radical, or inside a trigonometric, exponential, or logarithmic function. If your first choice doesn't simplify the integral, try a different piece.

Integration by Substitution — Definition, Formula & Examples

Integration by substitution is a technique for evaluating integrals by replacing a complicated expression with a single new variable, simplifying the integrand into a form you can integrate directly. It is essentially the chain rule of differentiation applied in reverse.

If $g$ is a differentiable function whose range is contained in the domain of a continuous function $f$ , and $F$ is an antiderivative of $f$ , then $\int f(g(x))\,g'(x)\,dx = F(g(x)) + C$ . Equivalently, setting $u = g(x)$ so that $du = g'(x)\,dx$ , the integral becomes $\int f(u)\,du = F(u) + C$ .

Key Formula

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

Where:

$u$ = The substitution variable, set equal to the inner function g(x)
$g(x)$ = The inner function identified inside the integrand
$g'(x)\,dx$ = The differential of the inner function, replaced by du
$f$ = The outer function applied to g(x)

How It Works

You look for a composite structure inside the integrand — an "inner function"

g(x)

whose derivative

g'(x)

also appears (possibly up to a constant factor). Introduce a new variable

u = g(x)

, compute

du = g'(x)\,dx

, and rewrite the entire integral in terms of

u

. After integrating with respect to

u

, substitute back to express the result in terms of

x

. For definite integrals, you can either convert the limits of integration to

u

-values or substitute back and use the original

x

-limits.

Worked Example

Problem: Evaluate

\int 2x\cos(x^2)\,dx

Identify the inner function: The argument of cosine is

x^2

, and its derivative

2x

appears as a factor. Set

u = x^2

u = x^2

Compute du: Differentiate both sides to find the relationship between

du

and

dx

du = 2x\,dx

Rewrite the integral in terms of u: Replace

x^2

with

u

and

2x\,dx

with

du

\int 2x\cos(x^2)\,dx = \int \cos(u)\,du

Integrate and substitute back: Integrate

\cos(u)

and then replace

u

with

x^2

\int \cos(u)\,du = \sin(u) + C = \sin(x^2) + C

Answer:

\sin(x^2) + C

Another Example

Problem: Evaluate

\displaystyle\int_0^2 3x^2(x^3 + 1)^4\,dx

Choose u and find du: Set

u = x^3 + 1

, so

du = 3x^2\,dx

. The factor

3x^2\,dx

in the integrand matches

du

exactly.

u = x^3 + 1,\quad du = 3x^2\,dx

Convert the limits: When

x = 0

u = 0^3 + 1 = 1

. When

x = 2

u = 2^3 + 1 = 9

x=0 \Rightarrow u=1,\quad x=2 \Rightarrow u=9

Rewrite and integrate: The integral becomes a simple power rule problem in

u

\int_1^9 u^4\,du = \left[\frac{u^5}{5}\right]_1^9 = \frac{9^5}{5} - \frac{1^5}{5} = \frac{59049 - 1}{5} = \frac{59048}{5}

Answer:

\dfrac{59048}{5}

Why It Matters

Substitution is the first integration technique taught in Calculus I and II, and it appears in the vast majority of integral problems on AP Calculus and university exams. Physicists and engineers rely on it constantly — for example, when changing a position-based integral into a time-based one in kinematics. Mastering substitution also builds the intuition needed for more advanced methods like integration by parts and trigonometric substitution.

Common Mistakes

Mistake: Forgetting to convert

dx

du

, leaving a mix of

u

and

dx

in the integral.

Correction: After choosing

u

, always solve

du = g'(x)\,dx

for

dx

(or match the entire

g'(x)\,dx

with

du

) and replace every occurrence of

x

before integrating.

Mistake: Using the original

x

-limits after substituting to

u

in a definite integral.

Correction: Either convert the limits by plugging each

x

-value into

u = g(x)

, or substitute back to

x

after integrating and then apply the original limits. Mixing the two produces wrong answers.