u-Substitution (Indefinite Integrals)

MATH161

Reference: Stewart 4.5 • Chapter: 4 • Section: 5

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u-Substitution (Indefinite Integrals)

Reversing the Chain Rule

The chain rule tells us how to differentiate compositions: $\frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x)$. But what if we're given an integral that looks like $\int f(g(x)) g'(x)\,dx$? We need to "undo" the chain rule.

u-Substitution is the technique that reverses this process. By introducing a new variable $u = g(x)$, we transform a complicated integral into a simpler one.

This is the foundational integration technique—used more frequently than any other method in first-year calculus.

Prerequisite Map

Quick Reference

Property	Value
Section	Stewart 4.5
Course	MATH162
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Substitution Rule

Theorem: If $u = g(x)$ is a differentiable function whose range is an interval $I$ and $f$ is continuous on $I$, then

$$\boxed{\int f(g(x)) g'(x) \, dx = \int f(u) \, du}$$

Why It Works

The substitution rule is simply the chain rule in reverse:

$$\frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x)$$

Integrating both sides:

$$F(g(x)) = \int F'(g(x)) \cdot g'(x) \, dx$$

If we let $f = F'$ and $u = g(x)$, then $du = g'(x)\,dx$, and:

$$\int f(g(x)) g'(x) \, dx = F(g(x)) = F(u) = \int f(u) \, du$$

The Procedure

Step 1: Choose $u$

Look for a function whose derivative also appears in the integrand
Often, $u$ is the "inside" of a composite function
Common choices: the expression under a radical, inside a trig function, or in an exponent

Step 2: Find $du$

Differentiate: $du = g'(x)\,dx$
Solve for $dx$ if needed: $dx = \frac{du}{g'(x)}$

Step 3: Substitute

Replace all $x$-terms with $u$-terms
The integral should now be entirely in terms of $u$

Step 4: Integrate

Evaluate $\int f(u)\,du$

Step 5: Back-substitute

Replace $u$ with $g(x)$ to return to the original variable

Key Insight: Handling Constants

If the derivative of your $u$ appears with a different constant factor, adjust:

$$\int x^3 \cos(x^4 + 2)\,dx$$

Let $u = x^4 + 2$. Then $du = 4x^3\,dx$, so $x^3\,dx = \frac{1}{4}du$.

$$\int x^3 \cos(x^4 + 2)\,dx = \frac{1}{4}\int \cos u\,du = \frac{1}{4}\sin u + C = \frac{1}{4}\sin(x^4 + 2) + C$$

How to Choose $u$

Pattern	Choose $u$ as...
$\int f(ax + b)\,dx$	$u = ax + b$
$\int g(x) \cdot f(g(x)^n)\,dx$ where $g'$ appears	$u = g(x)$
$\int (\text{inside})^n \cdot (\text{derivative of inside})\,dx$	$u = \text{inside}$
$\int \frac{g'(x)}{g(x)}\,dx$	$u = g(x)$ (gives $\ln\vert u\vert $)

Common Mistake: Leftover $x$-terms

After substitution, all $x$-terms must be eliminated. If you have leftover $x$'s, either:

Express them in terms of $u$ (e.g., if $u = 1 + x^2$, then $x^2 = u - 1$)
Try a different substitution

Worked Examples

Example 1: Basic Substitution

Find $\int 2x\sqrt{1 + x^2}\,dx$.

Choose $u$: Let $u = 1 + x^2$ (the expression under the radical).

Find $du$: $du = 2x\,dx$

Substitute: $$\int 2x\sqrt{1 + x^2}\,dx = \int \sqrt{u}\,du = \int u^{1/2}\,du$$

Integrate: $$= \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C$$

Back-substitute: $$= \frac{2}{3}(1 + x^2)^{3/2} + C$$

Example 2: Adjusting for Constants

Find $\int \cos(5x)\,dx$.

Choose $u$: Let $u = 5x$.

Find $du$: $du = 5\,dx$, so $dx = \frac{1}{5}du$.

Substitute and integrate: $$\int \cos(5x)\,dx = \int \cos u \cdot \frac{1}{5}\,du = \frac{1}{5}\sin u + C = \frac{1}{5}\sin(5x) + C$$

Example 3: Expressing $x$ in Terms of $u$

Find $\int x^5\sqrt{1 + x^2}\,dx$.

Choose $u$: Let $u = 1 + x^2$, so $du = 2x\,dx$.

Handle the extra $x^4$: Since $u = 1 + x^2$, we have $x^2 = u - 1$, so $x^4 = (u-1)^2$.

Rewrite: $$\int x^5\sqrt{1 + x^2}\,dx = \int x^4 \cdot \sqrt{1 + x^2} \cdot x\,dx = \int (u-1)^2 \sqrt{u} \cdot \frac{1}{2}\,du$$

Expand and integrate: $$= \frac{1}{2}\int (u^2 - 2u + 1)u^{1/2}\,du = \frac{1}{2}\int (u^{5/2} - 2u^{3/2} + u^{1/2})\,du$$

$$= \frac{1}{2}\left[\frac{2}{7}u^{7/2} - \frac{4}{5}u^{5/2} + \frac{2}{3}u^{3/2}\right] + C$$

$$= \frac{1}{7}(1+x^2)^{7/2} - \frac{2}{5}(1+x^2)^{5/2} + \frac{1}{3}(1+x^2)^{3/2} + C$$

Practice Problems

Level 1 Direct Substitution

Evaluate each integral:

$\displaystyle\int \cos(3x)\,dx$
$\displaystyle\int e^{2x}\,dx$
$\displaystyle\int (2x+1)^5\,dx$

Thought Process

For each: identify what to substitute (the inner function), find $du$, adjust for constants.

(a) $u = 3x \Rightarrow du = 3\,dx$ (b) $u = 2x \Rightarrow du = 2\,dx$ (c) $u = 2x + 1 \Rightarrow du = 2\,dx$

Show Answer

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

(b) Let $u = 2x$, $du = 2\,dx$, $dx = \frac{1}{2}du$: $$\int e^{2x}\,dx = \frac{1}{2}\int e^u\,du = \boxed{\frac{1}{2}e^{2x} + C}$$

(c) Let $u = 2x + 1$, $du = 2\,dx$, $dx = \frac{1}{2}du$: $$\int (2x+1)^5\,dx = \frac{1}{2}\int u^5\,du = \frac{1}{2} \cdot \frac{u^6}{6} + C = \boxed{\frac{(2x+1)^6}{12} + C}$$

Level 2 Derivative of Inside Present

Evaluate each integral:

$\displaystyle\int x^2\cos(x^3)\,dx$
$\displaystyle\int \frac{x}{\sqrt{1-4x^2}}\,dx$

Thought Process

(a) Notice $\frac{d}{dx}[x^3] = 3x^2$. The $x^2$ in the integrand is (almost) the derivative of the inside of $\cos(x^3)$.

(b) Notice $\frac{d}{dx}[1 - 4x^2] = -8x$. The $x$ in the numerator relates to the derivative of what's under the radical.

Show Answer

(a) Let $u = x^3$, $du = 3x^2\,dx$, so $x^2\,dx = \frac{1}{3}du$: $$\int x^2\cos(x^3)\,dx = \frac{1}{3}\int \cos u\,du = \frac{1}{3}\sin u + C = \boxed{\frac{1}{3}\sin(x^3) + C}$$

(b) Let $u = 1 - 4x^2$, $du = -8x\,dx$, so $x\,dx = -\frac{1}{8}du$: $$\int \frac{x}{\sqrt{1-4x^2}}\,dx = -\frac{1}{8}\int u^{-1/2}\,du = -\frac{1}{8} \cdot 2u^{1/2} + C = \boxed{-\frac{1}{4}\sqrt{1-4x^2} + C}$$

Level 3 Logarithmic Integrals

Evaluate each integral:

$\displaystyle\int \frac{\ln x}{x}\,dx$
$\displaystyle\int \tan x\,dx$

Thought Process

(a) Notice $\frac{d}{dx}[\ln x] = \frac{1}{x}$. So $\frac{1}{x}$ is the derivative of $\ln x$.

(b) Rewrite $\tan x = \frac{\sin x}{\cos x}$. Notice $\frac{d}{dx}[\cos x] = -\sin x$.

Show Answer

(a) Let $u = \ln x$, $du = \frac{1}{x}\,dx$: $$\int \frac{\ln x}{x}\,dx = \int u\,du = \frac{u^2}{2} + C = \boxed{\frac{(\ln x)^2}{2} + C}$$

(b) Rewrite and substitute: $\tan x = \frac{\sin x}{\cos x}$.

Let $u = \cos x$, $du = -\sin x\,dx$: $$\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = -\int \frac{du}{u} = -\ln\vert u\vert + C = \boxed{-\ln\vert \cos x\vert + C}$$

(Or equivalently: $\ln\vert \sec x\vert + C$)

Level 4 Expressing $x$ in Terms of $u$

Evaluate $\displaystyle\int x\sqrt{2x + 1}\,dx$.

Thought Process

Let $u = 2x + 1$. Then $du = 2\,dx$, so $dx = \frac{1}{2}du$.

But we still have an $x$ that needs to be eliminated. From $u = 2x + 1$: $$x = \frac{u - 1}{2}$$

Now substitute everything.

Show Answer

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

$$\int x\sqrt{2x + 1}\,dx = \int \frac{u-1}{2} \cdot \sqrt{u} \cdot \frac{1}{2}\,du$$

$$= \frac{1}{4}\int (u-1)u^{1/2}\,du = \frac{1}{4}\int (u^{3/2} - u^{1/2})\,du$$

$$= \frac{1}{4}\left[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right] + C$$

$$= \frac{1}{10}(2x+1)^{5/2} - \frac{1}{6}(2x+1)^{3/2} + C$$

$$= \boxed{\frac{(2x+1)^{3/2}}{30}(3(2x+1) - 5) + C = \frac{(2x+1)^{3/2}(6x - 2)}{30} + C}$$

Or simplified: $\boxed{\frac{(2x+1)^{3/2}(3x-1)}{15} + C}$

Level 5 Multiple Valid Substitutions

Evaluate $\displaystyle\int \sqrt{2x+1}\,dx$ using two different substitutions:

$u = 2x + 1$
$u = \sqrt{2x + 1}$

Verify both give the same answer.

Thought Process

(a) Standard approach: substitute what's inside the radical.

(b) Alternative: substitute the entire radical. Then $u^2 = 2x + 1$, so $2u\,du = 2\,dx$.

Both methods should yield the same final answer (up to constant).

Show Answer

(a) Using $u = 2x + 1$:

$du = 2\,dx$, so $dx = \frac{1}{2}du$: $$\int \sqrt{2x+1}\,dx = \frac{1}{2}\int u^{1/2}\,du = \frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C = \frac{1}{3}(2x+1)^{3/2} + C$$

(b) Using $u = \sqrt{2x + 1}$:

Then $u^2 = 2x + 1$, so $2u\,du = 2\,dx$, meaning $dx = u\,du$: $$\int \sqrt{2x+1}\,dx = \int u \cdot u\,du = \int u^2\,du = \frac{u^3}{3} + C = \frac{(\sqrt{2x+1})^3}{3} + C$$ $$= \frac{(2x+1)^{3/2}}{3} + C$$

Both methods give: $\boxed{\frac{1}{3}(2x+1)^{3/2} + C}$ ✓

Conceptual Questions (CCI-Style)

Level 2 Understanding the Method

A student tries to evaluate $\int e^{x^2}\,dx$ using substitution with $u = x^2$:

$$u = x^2, \quad du = 2x\,dx$$

$$\int e^{x^2}\,dx = \frac{1}{2x}\int e^u\,du = \frac{e^{x^2}}{2x} + C$$

What is wrong with this calculation?

Thought Process

When we write $dx = \frac{du}{2x}$, the $\frac{1}{2x}$ cannot be pulled outside the integral because it contains $x$, which is being integrated over.

Show Answer

The error: You cannot pull $\frac{1}{2x}$ outside the integral because $x$ is the variable of integration.

When using substitution, any factors involving $x$ must be:

Converted to $u$ (using $x = \sqrt{u}$ in this case), OR
Already present as part of $du$

Since $du = 2x\,dx$, we need an $x$ factor in the integrand to pair with $dx$. The integral $\int e^{x^2}\,dx$ does not have this factor.

Conclusion: $\int e^{x^2}\,dx$ cannot be evaluated using elementary substitution. (In fact, it has no elementary antiderivative—its antiderivative is the error function, $\text{erf}(x)$.)

Level 3 Connection to Chain Rule

Explain why $\int f(g(x))g'(x)\,dx = F(g(x)) + C$ where $F' = f$.

Use the chain rule to verify your answer by differentiating the right-hand side.

Thought Process

If we differentiate $F(g(x))$ using the chain rule, we should get back the integrand.

Show Answer

Verification by differentiation:

Let $H(x) = F(g(x)) + C$.

By the chain rule: $$H'(x) = \frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x)$$

Since $F' = f$: $$H'(x) = f(g(x)) \cdot g'(x)$$

This is exactly the integrand! ✓

Interpretation: The substitution rule works because integration and differentiation are inverse operations, and the chain rule tells us how to differentiate compositions. Substitution "undoes" the chain rule.

Mastery Checklist

[ ] I can identify when substitution is appropriate (derivative of inside appears)
[ ] I can choose an effective $u$ for a given integral
[ ] I can correctly compute $du$ and solve for $dx$
[ ] I can handle constant factors that don't match exactly
[ ] I can express leftover $x$-terms in terms of $u$ when needed
[ ] I always back-substitute to return to the original variable
[ ] I can verify my answer by differentiation

Mental Model

The Disguised Package Analogy:

Imagine you receive a package wrapped in multiple layers. To see what's inside, you unwrap layer by layer.

Similarly, $\int 2x\sqrt{1+x^2}\,dx$ has a "package" $(1+x^2)$ wrapped in a square root, multiplied by packaging tape $2x\,dx$. The substitution $u = 1 + x^2$ unwraps the package, revealing the simple $\int \sqrt{u}\,du$ inside.

Connections

Looking back:

Chain Rule is the differentiation rule that substitution reverses
Antiderivative Rules provide the formulas used after substitution

Looking ahead:

u-Substitution (Definite) handles limits of integration
Integration by Parts handles products where substitution fails
Trig Substitution uses specific substitutions for radical expressions

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Last updated: 2026-01-22