The Substitution Rule

The Fundamental Theorem reduces evaluating a definite integral to finding an antiderivative, but the basic table only covers integrands that are themselves derivatives of familiar functions. An integral like $\int 2 x 1 + x^{2} d x$ is not in the table. It becomes tractable once the composite structure is noticed: the factor $2 x$ is the derivative of the inner quantity $1 + x^{2}$ under the root. Substitution is the technique that exploits exactly this pattern, and it is the reverse of the Chain Rule.

Reversing the Chain Rule

The Chain Rule states $\frac{d}{d x} F (g (x)) = F^{'} (g (x)) g^{'} (x)$ . Reading it right to left, $F (g (x))$ is an antiderivative of $F^{'} (g (x)) g^{'} (x)$ , so

\int F^{'} (g (x)) g^{'} (x) d x = F (g (x)) + C .

The integrand has a recognizable form: an outer derivative $F^{'}$ evaluated at an inner function $g (x)$ , multiplied by the inner function's derivative $g^{'} (x)$ . Introduce a new variable for the inner function, $u = g (x)$ , with differential $d u = g^{'} (x) d x$ . Then the whole integral rewrites in $u$ alone.

The mechanical picture: $u = g (x)$ replaces the inner function, and the differential $d u = g^{'} (x) d x$ absorbs the leftover factor $g^{'} (x) d x$ . Everything in $x$ leaves the integral, replaced by an integral in $u$ that is, ideally, in the table.

Substitution as a round trip, shown on

\int 2 x cos (x^{2}) d x

: the composite integral in

x

becomes a table integral in

u = x^{2}

; integrate, then back-substitute to return to

x

The differential notation is what makes the bookkeeping reliable: treat $d u$ and $d x$ as if they were algebraic quantities linked by $d u = g^{'} (x) d x$ , and the rule reduces to substituting one for the other under the integral sign.

Worked substitutions

When the matching derivative carries a constant factor, absorb it into $d u$ .

Anatomy of a substitutable integrand on

\int cos (x^{2}) 2 x d x

: the inner function

x^{2}

names

u

, the outer factor becomes

cos u

, and the leftover

2 x d x

is exactly

d u

, so the product collapses to

\int cos u d u

The procedure is the same every time.

Algorithm:

\textsc{Substitute}

— evaluate

\int f(g(x))\,g'(x)\,\d x

by change of variable

1
choose $u \gets g(x)$ , an inner function whose derivative appears in the integrand
2
compute $\d u \gets g'(x)\,\d x$
3
rewrite the integral entirely in $u$ , replacing $g(x)$ by $u$ and $g'(x)\,\d x$ by $\d u$
4
antidifferentiate the resulting integral in $u$
5
back-substitute $u \gets g(x)$ to return to the original variable

The one judgment call is the first line. Choose $u$ to be a function inside the integrand whose differential also occurs, up to a constant factor. If no such factor is present, try $u$ equal to the inner part of a composition and see whether the leftover simplifies. Guessing wrong is normal; a substitution either collapses the integral or it does not, and a failed attempt is quickly abandoned.

More indefinite integrals

Constant inner derivative. For $\int 2 x + 1 d x$ , take $u = 2 x + 1$ , $d u = 2 d x$ , $d x = \frac{1}{2} d u$ : $\int 2 x + 1 d x = \frac{1}{2} \int u^{1/2} d u = \frac{1}{3} u^{3/2} + C = \frac{1}{3} (2 x + 1)^{3/2} + C .$
Radical in a denominator. For $\int \frac{x}{1 - 4 x ^{2}} d x$ , take $u = 1 - 4 x^{2}$ , $d u = - 8 x d x$ , so $x d x = - \frac{1}{8} d u$ : $\int \frac{x}{1 - 4 x ^{2}} d x = - \frac{1}{8} \int u^{- 1/2} d u = - \frac{1}{4} 1 - 4 x^{2} + C .$
Extra factor of $x$ handled by algebra. For $\int x^{5} 1 + x^{2} d x$ , split $x^{5} = x^{4} \cdot x$ and set $u = 1 + x^{2}$ , $x d x = \frac{1}{2} d u$ , $x^{2} = u - 1$ , so $x^{4} = (u - 1)^{2}$ : $\int x^{5} 1 + x^{2} d x = \frac{1}{2} \int (u - 1)^{2} u d u = \frac{1}{2} \int (u^{5/2} - 2 u^{3/2} + u^{1/2}) d u,$ which integrates term by term to $\frac{1}{7} (1 + x^{2})^{7/2} - \frac{2}{5} (1 + x^{2})^{5/2} + \frac{1}{3} (1 + x^{2})^{3/2} + C$ .

With practice the explicit $u$ can be skipped for simple cases — $\int cos 5 x d x = \frac{1}{5} sin 5 x + C$ by inspection — but for anything with an algebraic tangle, writing the substitution out prevents errors.

Choosing the substitution

The first line of the procedure is the only real decision. A few recurring shapes point straight at $u$ :

Integrand shape	Take $u =$
a composite $f (g (x))$ with $g^{'} (x)$ present as a factor	the inner function $g (x)$
a power of a linear expression, $(a x + b)^{n}$	the linear expression $a x + b$
a trigonometric power times the derivative of its argument	the argument, or the trig factor whose derivative appears
a radical $q (x)$ with $q^{'} (x)$ nearby	the quantity $q (x)$ under the radical

Three more, chosen so the leftover factor is exactly $d u$ :

Substitution in definite integrals

For a definite integral there are two ways to carry the substitution through, and the second is usually cleaner.

Method 1 — evaluate the indefinite integral first, then apply the limits in $x$ . Find the antiderivative in $x$ , then use the Fundamental Theorem with the original $x$ -limits. Using the result above for $2 x + 1$ ,

0 \int 4 2 x + 1 d x = [\frac{1}{3} (2 x + 1)^{3/2}]_{0}^{4} = \frac{1}{3} (9^{3/2} - 1^{3/2}) = \frac{1}{3} (27 - 1) = \frac{26}{3} .

Method 2 — change the limits when the variable changes. The alternative transforms the bounds along with the integrand, so there is never a need to return to $x$ .

The new limits are the $u$ -values matching the old $x$ -values: when $x = a$ , $u = g (a)$ ; when $x = b$ , $u = g (b)$ . The integral is then finished entirely in $u$ . For the same integral, $u = 2 x + 1$ sends $x = 0 \mapsto u = 1$ and $x = 4 \mapsto u = 9$ :

0 \int 4 2 x + 1 d x = 1 \int 9 \frac{1}{2} u d u = [\frac{1}{3} u^{3/2}]_{1}^{9} = \frac{1}{3} (27 - 1) = \frac{26}{3} .

Changing the variable also changes the limits: the region under

2 x + 1

[0, 4]

maps to the region under

\frac{1}{2} u

[1, 9]

, with the same area.

Changing the limits amounts to remapping the interval of integration: $u = 2 x + 1$ carries $[0, 4]$ in $x$ to $[1, 9]$ in $u$ .

The substitution

u = 2 x + 1

remaps the interval of integration, sending each

x

[0, 4]

u

[1, 9]

; the limits transform with the variable.

The $u$ -limits may run backwards when $g$ decreases.

Symmetry shortcuts

Substitution applied to a symmetric integrand over a symmetric interval $[- a, a]$ produces two shortcuts worth memorizing. Recall that $f$ is even if $f (- x) = f (x)$ (graph symmetric about the $y$ -axis) and odd if $f (- x) = - f (x)$ (graph symmetric about the origin).

The proof splits the integral at $0$ and substitutes $u = - x$ in the piece over $[- a, 0]$ . That substitution flips the interval to $[0, a]$ and turns $f (x)$ into $f (- u)$ ; even symmetry makes the two halves equal and they add, odd symmetry makes them opposite and they cancel. Geometrically, an even function has matching areas on both sides of the axis, and an odd function has a positive region on one side exactly cancelled by a negative region on the other.

Left, an even integrand — the two halves have equal area, so the whole is twice the right half. Right, an odd integrand — the halves are equal in size and opposite in sign, so they cancel to zero.

Symmetry is worth a glance whenever the interval is centered at the origin: it can turn a hard integral into a trivial one.