The Substitution Method, calculus lesson

Differentiate $\sin(x^2)$ and you get $2x \cos(x^2)$ , an inner function and its derivative side by side. Many integrands carry exactly that fingerprint of the chain rule. Substitution is the method for spotting it and running the chain rule backward.

Key terms

Reverse chain rule

Definition Let $g$ be differentiable on an interval $J$ , and let $F$ be differentiable on an interval containing $g(J)$ with $F' = f$ . By the chain rule

\frac{d}{dx} F\big(g(x)\big) = f\big(g(x)\big) \, g'(x) \qquad \text{for } x \in J,

\int f\big(g(x)\big) \, g'(x) \, dx = F\big(g(x)\big) + C.

In plain words If the integrand is a function of an inner expression times the derivative of that inner expression, the answer is an antiderivative of the outer function evaluated at the inner expression.

Example In $\int 2x \cos(x^2) \, dx$ the inner function is $x^2$ with derivative $2x$ , and the outer function is $\cos$ , with antiderivative $\sin$ . So

\int 2x \cos(x^2) \, dx = \sin(x^2) + C.

Check by differentiating, which gives $\cos(x^2) \cdot 2x$ .

Substitution for indefinite integrals

Definition Let $u = g(x)$ be differentiable on an interval $J$ , with values in an interval $I$ , and let $f$ be continuous on $I$ . Then

\int f\big(g(x)\big) \, g'(x) \, dx = \int f(u) \, du, \qquad \text{where } u = g(x).

The equation means that if $F$ is an antiderivative of $f$ on $I$ , then $F(g(x))$ is an antiderivative of the left integrand on $J$ . In practice you write $du = g'(x) \, dx$ and replace every piece of the integrand. The new integral must contain only $u$ and $du$ .

In plain words Name the inner expression $u$ , find $du$ , and rewrite the integral so $x$ disappears. If some $x$ is left over that you cannot express through $u$ , the substitution has not worked.

Example For $\int x \sqrt{1 + x^2} \, dx$ let $u = 1 + x^2$ , so $du = 2x \, dx$ and $x \, dx = \frac{1}{2} du$ .

\int x \sqrt{1 + x^2} \, dx = \frac{1}{2} \int u^{1/2} \, du = \frac{1}{3} u^{3/2} + C.

Back-substitution

Definition After integrating in $u$ , the result is a function of $u$ . Back-substitution replaces $u$ by $g(x)$ , giving an antiderivative of the original integrand as a function of $x$ . The answer can always be checked by differentiating it with respect to $x$ .

In plain words The question was asked in $x$ , so the answer must be given in $x$ .

Example Continuing the previous example, put $u = 1 + x^2$ back in to get

\int x \sqrt{1 + x^2} \, dx = \frac{1}{3}(1 + x^2)^{3/2} + C.

Its derivative is

\frac{1}{3} \cdot \frac{3}{2} (1 + x^2)^{1/2} \cdot 2x = x \sqrt{1 + x^2},

as it should be.

Change of variable

Definition Sometimes the leftover factors of $x$ must also be written through $u$ . If $u = g(x)$ and $g$ has an inverse on the interval of interest, solve for $x = g^{-1}(u)$ and replace every $x$ , including those outside the inner function. More generally, let $f$ be continuous on an interval and let $x = h(u)$ , where $h$ is differentiable on an interval with $h'(u) \neq 0$ there and takes values where $f$ is continuous. Then $h$ is one-to-one with a differentiable inverse, and

\int f(x) \, dx = \int f\big(h(u)\big) \, h'(u) \, du,

followed by the back-substitution $u = h^{-1}(x)$ .

In plain words You are free to rename the variable in any reversible way, as long as you translate the whole integrand, $dx$ included, into the new variable.

Example For $\int x \sqrt{x - 1} \, dx$ on $x \ge 1$ , let $u = x - 1$ . Then $x = u + 1$ and $dx = du$ .

\int x \sqrt{x - 1} \, dx = \int (u + 1) u^{1/2} \, du = \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C.

Back in $x$ this is

\frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C.

Its derivative is

(x - 1)^{3/2} + (x - 1)^{1/2} = (x - 1)^{1/2} \big( (x - 1) + 1 \big) = x\sqrt{x - 1}.

Definite-integral bounds

Definition If $g'$ is continuous on $[a, b]$ and $f$ is continuous on the range of $u = g(x)$ for $x$ in $[a, b]$ , then

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

In plain words Change the bounds along with the variable. The new bounds are the values of $u$ at the old bounds, and then there is no need to go back to $x$ .

Example For $\int_0^1 x e^{x^2} \, dx$ let $u = x^2$ , so $du = 2x \, dx$ . When $x = 0$ , $u = 0$ , and when $x = 1$ , $u = 1$ .

\int_0^1 x e^{x^2} \, dx = \frac{1}{2} \int_0^1 e^u \, du = \frac{1}{2}(e - 1) \approx 0.859.

Interval and regularity conditions

Definition The definite substitution rule requires $g$ to be continuously differentiable on the whole closed interval $[a, b]$ and $f$ to be continuous on an interval containing every value $g(x)$ for $x$ in $[a, b]$ . It does not require $g$ to be one-to-one. The indefinite rule holds on each interval where these conditions hold.

In plain words The inner function must be smooth across the entire interval, with no breaks or vertical asymptotes. If it fails somewhere inside, the bounds you plug in can lie.

Example The expression $\frac{\sec^2 x}{1 + \tan^2 x}$ is not defined at $x = \frac{\pi}{2}$ , but at every other point of $[0, \pi]$ it equals $1$ , since $1 + \tan^2 x = \sec^2 x$ . So it extends continuously to the constant function $1$ on $[0, \pi]$ , and with that value at $\frac{\pi}{2}$

\int_0^{\pi} \frac{\sec^2 x}{1 + \tan^2 x} \, dx = \int_0^{\pi} 1 \, dx = \pi.

Substituting $u = \tan x$ blindly gives bounds $\tan 0 = 0$ and $\tan \pi = 0$ , so $\int_0^0 \frac{du}{1 + u^2} = 0$ , which is wrong. The inner function $g(x) = \tan x$ is not defined at $x = \frac{\pi}{2}$ , which lies inside $[0, \pi]$ , so it is not continuously differentiable on the whole interval and the rule does not apply.

Check yourself

Find $\int \cos^3 x \sin x \, dx$ .

Show the answer

Let $u = \cos x$ , so $du = -\sin x \, dx$ . The integral is

-\int u^3 \, du = -\frac{u^4}{4} + C = -\frac{\cos^4 x}{4} + C.

Evaluate $\int_1^e \frac{\ln x}{x} \, dx$ .

Show the answer

Let $u = \ln x$ , $du = \frac{dx}{x}$ , with new bounds $0$ and $1$ . The value is

\int_0^1 u \, du = \frac{1}{2}.

Find $\int \frac{dx}{3x + 2}$ and say where the answer is valid.

Show the answer

Let $u = 3x + 2$ , $du = 3 \, dx$ . The integral is

\frac{1}{3} \int \frac{du}{u} = \frac{1}{3} \ln \lvert 3x + 2 \rvert + C,

valid on each of the intervals $x < -\frac{2}{3}$ and $x > -\frac{2}{3}$ .

The Substitution Method

Theory, part 107

Practice, part 108

Key terms

Reverse chain rule

Substitution for indefinite integrals

Back-substitution

Change of variable

Definite-integral bounds

Interval and regularity conditions

Check yourself

How this lesson connects

Builds on

Used later in