Substitution rule (u-substitution): Rewrite integrals by changing variables

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Principles | FAQ | How This Fits

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The Principle

The move: Replace a repeated inner expression with a new variable $u$, convert the matching differential, and integrate in the simpler variable.

The invariant: This rewrites the integral as an equivalent antiderivative problem under a valid change of variable.

Pattern: $\int f(g(x))g'(x)\,dx \quad \longrightarrow \quad \int f(u)\,du$

Applies directly ✓	Does not apply directly ✗
$\int 2x\cos(x^2)\,dx \to \int \cos(u)\,du$	$\int \cos(x^2)\,dx \not\to \int \cos(u)\,du$

Left: choosing $u=x^2$ gives $du=2x\,dx$, so the needed differential is present. Right: the inner expression $x^2$ is visible, but its derivative factor is missing, so the direct substitution step is not justified yet.

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Conditions of Applicability

Condition: $u=g(x)$; $\frac{du}{dx}=g'(x)$

Before applying, check: identify a candidate inner expression $g(x)$, then confirm the integrand contains the matching differential factor needed to replace $g'(x)\,dx$ by $du$.

If the condition is violated: a direct change of variables drops or invents a factor, so the rewritten integral is not equivalent to the original one.

• The usable pattern is not just “something complicated inside another function.” The matching derivative factor has to be present, up to a constant multiple you can account for honestly.
• Constant factors are allowed if you keep them visible, as in $\int \cos(5x)\,dx = \frac{1}{5}\int \cos(u)\,du$ after $u=5x$.
• If the derivative factor is missing entirely, as in $\int e^{x^2}\,dx$, substitution does not apply directly to the whole integral even though the inner expression is obvious.
• Compare this move with the derivative chain rule: differentiation multiplies by the inner derivative, while substitution recognizes that same factor during integration.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: switch to $u$ because an inner expression looks convenient, but ignore whether the matching differential is present → the new integral is not equivalent to the original one, so the antiderivative is wrong.

Debug: after choosing $u=g(x)$, point to the exact factor in the integrand that becomes $du$. If you cannot point to it, the substitution is not licensed yet.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• Why does replacing $g(x)$ by $u$ only work when the differential factor $g'(x)\,dx$ is accounted for at the same time?
• In what sense is the rewritten $u$-integral the same antiderivative task rather than a new problem with new meaning?

For the Principle

• When you scan an integrand, what tells you that an inner expression is genuinely usable for substitution instead of merely noticeable?
• How do constant multiples change the bookkeeping in a valid substitution without changing whether the move is available?

Between Principles

• How does this rule reverse the chain structure created by the derivative chain rule, and how does it differ from applying the integral power rule directly?

Generate an Example

• Create one integral where $u$-substitution applies in one step and one near-miss where the inner expression is visible but the differential factor is missing. Explain what condition check separates them.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Choose u = g(x), replace the matching differential g'(x) dx by du, and integrate the simpler u-expression.

Write the canonical equation: _____$\int f(g(x))g'(x)\,dx = \int f(u)\,du$

State the canonical condition: _____$u=g(x); \frac{du}{dx}=g'(x)$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from $\int 2x\cos(x^2+1)\,dx$, reach a finished antiderivative.

Step	Expression	Operation
0	$\int 2x\cos(x^2+1)\,dx$	-
1	$\int \cos(u)\,du$	Let $u=x^2+1$, so $du=2x\,dx$ and rewrite the integral in $u$
2	$\sin(u)+C$	Integrate the simpler $u$-expression
3	$\sin(x^2+1)+C$	Substitute back to the original variable

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Drills

Forward step (Format A)

Apply substitution directly and finish the antiderivative.

$\int 3x^2 e^{x^3}\,dx$

Reveal

Let $u=x^3$, so $du=3x^2\,dx$. Then

$\int 3x^2 e^{x^3}\,dx = \int e^u\,du = e^u + C = e^{x^3}+C$

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Apply substitution directly and finish the antiderivative.

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

Reveal

Let $u=5x$, so $du=5\,dx$ and $dx=\frac{1}{5}du$. Then

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

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Does substitution apply directly to the whole integral? Explain before writing anything else.

$\int e^{x^2}\,dx$

Reveal

No. The obvious inner expression is $x^2$, but choosing $u=x^2$ gives $du=2x\,dx$, and the factor $2x$ is not present.

The near-miss is structural: the inside is visible, but the matching differential is missing, so the direct substitution step is not justified.

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Which integrals are direct substitution matches?

(i) $\int 6x(x^2+1)^4\,dx$ \quad (ii) $\int (x^2+1)^4\,dx$ \quad (iii) $\int \frac{1}{x}\ln(x)\,dx$ \quad (iv) $\int \sin x\,dx$

Reveal

(i) and (iii).

• (i) works with $u=x^2+1$ and $du=2x\,dx$; the constant factor is manageable.
• (iii) works with $u=\ln(x)$ and $du=\frac{1}{x}\,dx$.
• (ii) has a visible inner expression but no matching differential.
• (iv) is integrable, but substitution is not the intended direct move here because there is no inner expression to change variables around.

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Apply substitution directly and finish the antiderivative.

$\int \frac{1}{3x+2}\,dx$

Reveal

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

$\int \frac{1}{3x+2}\,dx = \frac{1}{3}\int \frac{1}{u}\,du = \frac{1}{3}\ln|u|+C = \frac{1}{3}\ln|3x+2|+C$

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Action label (Format B)

What was done between these two steps?

$\int 4x\sin(x^2)\,dx \quad \longrightarrow \quad 2\int \sin(u)\,du$

Reveal

Substitution with coefficient bookkeeping. Let $u=x^2$, so $du=2x\,dx$. Then $4x\,dx = 2\,du$, which leaves

$\int 4x\sin(x^2)\,dx = 2\int \sin(u)\,du$

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What tempting move was attempted here, and why does it not apply directly?

$\int \sin(x^2)\,dx \quad \longrightarrow \quad \int \sin(u)\,du$

Reveal

An invalid direct substitution was attempted.

If $u=x^2$, then $du=2x\,dx$. The needed differential factor is missing, so the rewritten integral is not equivalent to the original one.

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What substitution was used in this step, and why is it valid?

$\int \frac{7}{7x-1}\,dx \quad \longrightarrow \quad \int \frac{1}{u}\,du$

Reveal

Let $u=7x-1$, so $du=7\,dx$. The numerator $7\,dx$ matches the new differential exactly, which makes the step valid:

$\int \frac{7}{7x-1}\,dx = \int \frac{1}{u}\,du$

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Transition identification (Format C)

Which transition uses substitution directly?

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

Reveal

Transition (1) uses substitution directly.

• (1) rewrites the integral with $u=x^2+1$ and $du=2x\,dx$.
• (2) is the power rule in the new variable.
• (3) substitutes back.

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Which transition is invalid, and why?

$\int e^{x^2}\,dx \xrightarrow{(1)} \int e^u\,du \xrightarrow{(2)} e^u + C \xrightarrow{(3)} e^{x^2}+C$

Reveal

Transition (1) is invalid.

Choosing $u=x^2$ would require $du=2x\,dx$, but the integrand has only $dx$. Because the matching differential is missing, the first rewrite is not justified and the whole chain collapses.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Compute

$\int 2x\sqrt{x^2+4}\,dx$

and simplify the result.

Full solution

Step	Expression	Move
0	$\int 2x\sqrt{x^2+4}\,dx$	-
1	$\int u^{1/2}\,du$	Let $u=x^2+4$, so $du=2x\,dx$
2	$\frac{u^{3/2}}{3/2}+C$	Power rule in the new variable
3	$\frac{2}{3}u^{3/2}+C$	Simplify the coefficient
4	$\frac{2}{3}(x^2+4)^{3/2}+C$	Substitute back

Check: differentiating $\frac{2}{3}(x^2+4)^{3/2}$ gives $\frac{2}{3}\cdot \frac{3}{2}(x^2+4)^{1/2}\cdot 2x = 2x\sqrt{x^2+4}$, so the antiderivative is consistent.

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Related Principles

Principle	Relationship
Derivative chain rule	Differentiation creates the inner-derivative factor that substitution later recognizes and reverses during integration
Integral power rule	Often supplies the antiderivative after substitution has rewritten the integral into a simpler power of $u$
Indefinite integral as antiderivative	Clarifies the object substitution preserves: a family of functions whose derivative returns the original integrand

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FAQ

What is the substitution rule for integrals?

It rewrites an integral by choosing a new variable $u=g(x)$ and replacing the matching differential $g'(x)\,dx$ with $du$. The goal is to turn the original integral into a simpler antiderivative problem in $u$.

When can I use u-substitution directly?

Use it directly when you can identify an inner expression and the integrand contains the matching differential factor, possibly up to a constant multiple you can handle explicitly. Seeing the inside alone is not enough.

What if the derivative factor is missing?

Then the direct substitution step is not valid for the whole integral. You may need a different technique, a larger strategy, or no elementary antiderivative at all.

Do I always substitute back to x at the end?

For indefinite integrals, yes. The final antiderivative should be written in the original variable unless the context explicitly says otherwise.

How is substitution related to the chain rule?

They are inverse-looking structures. The chain rule differentiates $f(g(x))$ by multiplying by $g'(x)$, while substitution spots that same inner-derivative factor during integration and uses it to change variables.

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How This Fits in Unisium

In Unisium, substitution is trained as move selection, not as a reflexive “let $u$ equal something complicated” slogan. You learn to test whether the matching differential is really present, reject near-misses early, and reinforce that judgment through retrieval practice, self-explanation, and the broader logic of Masterful Learning. After you can spot valid substitutions on sight, practice the same recognition directly in the Unisium app.

Explore further:

• Principle Structures - See where substitution sits in the calculus principle hierarchy
• Substitution rule (definite integral) - Extend the same matching pattern to definite integrals where the bounds change with the variable
• Elaborative Encoding - Build deeper understanding of why the differential match matters
• Retrieval Practice - Make the pattern and condition instantly accessible

Masterful Learning

The study system for physics, math, & programming that works: retrieval, connection, explanation, problem solving, and more.

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