Integrating by Substitution — AP® Calculus AB

Part 1 · Hook

Why this matters

\sin(x^2)

into

2x\cos(x^2)

. Going backward, an integral like

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

looks impossible until you spot the inside function

x^2

and its derivative

2x

sitting right there. Substitution is the art of spotting that pair, and it is the most used technique on the exam.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $y = (1 + x^2)^4$ , then $\frac{dy}{dx} =$

$8x(1 + x^2)^3$
$4(1 + x^2)^3$
$4x(1 + x^2)^3$
$8x(1 + x^2)^4$

Show the answer

Chain rule: $4(1 + x^2)^3 \cdot 2x$ .

Correct: $8x(1 + x^2)^3$ :
$4(1 + x^2)^3$ :
$4x(1 + x^2)^3$ :
$8x(1 + x^2)^4$ :

2. $\int u^3\,du =$

$\frac{u^4}{4} + C$
$3u^2 + C$
$u^4 + C$
$\frac{u^3}{3} + C$

Show the answer

Power rule backward.

Correct: $\frac{u^4}{4} + C$ :
$3u^2 + C$ :
$u^4 + C$ :
$\frac{u^3}{3} + C$ :

Part 4 · See it

See it first

Left: the graph of y = 2x cos(x squared) from x = 0 to the square root of pi/2, about 1.25, with the region under it shaded; its area is 1. Right: the graph of y = cos u from u = 0 to pi/2, with the region under it shaded; its area is also 1. With u = x squared, the x-limits become the u-limits. — With u = x², the region under 2x cos(x²) on [0, √(π/2)] and the region under cos u on [0, π/2] have the same area, 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The chain rule gives $\frac{d}{dx}F(g(x))$ $= F'(g(x))\,g'(x)$ so $\int f(g(x))\,g'(x)\,dx$ $= F(g(x)) + C$ when $F' = f$
Let $u = g(x)$ ; then $du = g'(x)\,dx$ the integral becomes $\int f(u)\,du$ , a basic form
A constant factor is missing or extraadjust it: $x\,dx = \frac{1}{2}du$ when $du = 2x\,dx$
The integral is definitechange the limits to $u = g(a)$ and $u = g(b)$ and never go back to $x$

Part 6 · Key ideas

Key ideas

Look for an inside function $u = g(x)$ whose derivative $g'(x)$ is also a factor, up to a constant.
Write $du = g'(x)\,dx$ , rewrite the whole integral in $u$ (no $x$ left), integrate, then substitute back.
Definite integrals: new limits $u = g(a)$ and $u = g(b)$ ; keeping the $x$ -limits is the classic slip.
Linear inside: $\int f(kx)\,dx$ $= \frac{1}{k}F(kx) + C$ , such as $\int e^{3x}\,dx = \frac{1}{3}e^{3x} + C$ .

Part 7 · Misconception

A common mistake

The wrong idea: $\int_0^1 x(1 + x^2)^3\,dx$ $= \frac{1}{2}\int_0^1 u^3\,du$ : the limits stay the same.

What actually happens: The limits belong to the variable. In $u$ they are $u(0) = 1$ and $u(1) = 2$ : $\frac{1}{2}\int_1^2 u^3\,du$ $= \frac{15}{8}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Substitution in a definite integral

Evaluate

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

with the substitution

u = 1 + x^2

.

The integral and the chosen substitution
Piece	Here
integral	$\int_0^1 x(1 + x^2)^3\,dx$
substitution	$u = 1 + x^2$
limits in x	$x = 0$ to $x = 1$

1. With $u = 1 + x^2$ , what does $x\,dx$ become?

$\frac{1}{2}\,du$
$du$
$2\,du$
$\frac{du}{x}$

Show the answer

$u = 1 + x^2$ gives $\frac{du}{dx} = 2x$ , so $du = 2x\,dx$ and $x\,dx = \frac{1}{2}\,du$ .

Correct: $\frac{1}{2}\,du$ : Correct: $du = 2x\,dx$ , so $x\,dx = \frac{1}{2}du$ .
$du$ : $du = 2x\,dx$ , which is twice $x\,dx$ . Divide by 2.
$2\,du$ : $du = 2x\,dx$ gives $x\,dx = \frac{du}{2}$ , not $2\,du$ .
$\frac{du}{x}$ : $du = 2x\,dx$ ; solving for $x\,dx$ (not $dx$ ) gives $\frac{1}{2}du$ , with no $x$ left over.

2. What are the limits of integration in terms of $u$ ?

$u = 0$ to $u = 1$
$u = 1$ to $u = 4$
$u = 0$ to $u = 2$
$u = 1$ to $u = 2$

Show the answer

Substitute each $x$ -limit into $u = 1 + x^2$ : $x = 0 \to u = 1$ , $x = 1 \to u = 2$ .

$u = 0$ to $u = 1$ : Those are the $x$ -limits. Put each into $u = 1 + x^2$ .
$u = 1$ to $u = 4$ : At $x = 1$ , $u = 1 + 1^2 = 2$ , not 4.
$u = 0$ to $u = 2$ : At $x = 0$ , $u = 1 + 0^2 = 1$ ; the constant 1 in $u$ matters.
Correct: $u = 1$ to $u = 2$ : Correct: $x = 0$ gives $u = 1 + 0 = 1$ and $x = 1$ gives $u = 2$ .

3. Which integral in $u$ equals $\int_0^1 x(1 + x^2)^3\,dx$ ?

$\int_1^2 u^3\,du$
$\frac{1}{2}\int_1^2 x\,u^3\,du$
$\frac{1}{2}\int_1^2 u^3\,du$
$\frac{1}{2}\int_0^1 u^3\,du$

Show the answer

$\int_0^1 (1 + x^2)^3 \cdot x\,dx$ $= \int_1^2 u^3 \cdot \frac{1}{2}\,du$ $= \frac{1}{2}\int_1^2 u^3\,du$ .

$\int_1^2 u^3\,du$ : The factor $\frac{1}{2}$ from $x\,dx = \frac{1}{2}du$ is missing.
$\frac{1}{2}\int_1^2 x\,u^3\,du$ : Everything must be in $u$ : the $x$ was absorbed into $du = 2x\,dx$ .
Correct: $\frac{1}{2}\int_1^2 u^3\,du$ : Correct: $(1 + x^2)^3 = u^3$ , $x\,dx = \frac{1}{2}du$ , limits 1 to 2.
$\frac{1}{2}\int_0^1 u^3\,du$ : The limits are still the $x$ -limits. In terms of $u$ they are 1 and 2.

4. Find the value of $\int_0^1 x(1 + x^2)^3\,dx$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{1}{2}\int_1^2 u^3\,du$ $= \frac{1}{2}\cdot\frac{u^4}{4}\Big|_1^2$ $= \frac{1}{8}(16 - 1)$ $= \frac{15}{8}$ .

Answer: $\frac{15}{8}$ ≈ 1.875

5. Find $\int 2x\cos(x^2)\,dx$ . Include $+ C$ .

Type an expression for $\int 2x\cos(x^2)\,dx =$ .

Show the answer

$u = x^2$ , $du = 2x\,dx$ : $\int \cos u\,du$ $= \sin u$ $+ C$ $= \sin(x^2)$ $+ C$ .

Answer: $\int 2x\cos(x^2)\,dx =$ $\sin(x^{2}) + C$

6. Find $\int e^{3x}\,dx$ . Include $+ C$ .

Type an expression for $\int e^{3x}\,dx =$ .

Show the answer

$u = 3x$ , $du = 3\,dx$ : $\int e^{3x}\,dx$ $= \frac{1}{3}\int e^u\,du$ $= \frac{1}{3}e^{3x}$ $+ C$ .

Answer: $\int e^{3x}\,dx =$ $\frac{1}{3}e^{3x} + C$

Part 9 · Summary

Summary

Substitution undoes the chain rule. Choose

u

as an inside function whose derivative is a factor of the integrand, replace

g'(x)\,dx

by

du

, fix any constant factor, and integrate in

u

. In a definite integral, change the limits to values of

u

.

Part 10 · Up next

What comes next

Part 11 · Connections