Reversing the chain rule: choosing the substitution, computing the differential, rewriting every x in terms of the new variable, back-substituting or changing the limits, adjusting constants but never variables, and recognising when substitution cannot help.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 5 — Integration
Substitution
Objectives
Five outcomes. The method is four lines, and three of these are about the ways it goes wrong.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 506-515 — the section these objectives are drawn from
Warm-up
Section 3.6's chain rule differentiates a composition, and it always leaves the inner function's derivative behind as a factor.
Discussion prompt
Differentiate the sixth power of x squared plus one, divided by six. What shape does the answer have?
Hint: The chain rule contributes an extra factor.
Answer:
\[ \frac{d}{dx}\left[\frac{(x^{2}+1)^{6}}{6}\right] = (x^{2}+1)^{5}\cdot 2x \]
The answer is a composition multiplied by the inner function's derivative — and that shape is unmistakable. So any integrand looking like that came from a chain rule, and undoing it recovers the original.
This section is nothing more than recognising that shape and reversing it. The one condition is that the inner derivative must already be present in the integrand, at least up to a numerical factor, and Section 5.4's table stalled on exactly the integrands where it is.
Concept
When an integrand contains a composition multiplied by the inner function's derivative, naming the inner function turns the whole integral into a simple one in the new variable.
substitution — Replacing the inner function of a composition by a new variable, and its derivative times the differential by that variable's differential, so the integral becomes one the basic table covers.
\[ \int f(g(x))g'(x)\,dx = \int f(u)\,du = F(u)+C \]
The method has one requirement and it is strict: the inner function's derivative must already appear. A missing constant factor can be supplied; a missing factor of the variable cannot.
Figure (svg): Substitution as the chain rule read backwards
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 535-543
Section
Section 1
Concept
Look for a composition whose inner function's derivative also appears as a factor. That inner function is the substitution, and its presence is what makes the method available.
choosing the substitution — Take the inner function of the composition — what sits inside a power, a root, an exponential or a trigonometric function — and check whether its derivative appears elsewhere in the integrand.
\[ u = g(x), \qquad du = g'(x)\,dx \]
The usual candidates are what is inside a bracket raised to a power, under a radical, in an exponent, or inside a trigonometric function. Trying one and finding its derivative absent is normal and costs a line.
Figure (svg): Choosing u: the inside of the composition, whose derivative is present
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 535-543 — the substitution method
Picture it
Which choice works, and when none does.
Figure (svg): Choosing u: the inside of the composition, whose derivative is present
The third row is the common case: the derivative is present up to a numerical factor, which is fixable. The fourth has no such factor at all, and no choice rescues it.
Worked example
Example 5.39. The inner derivative is exactly present.
\[ \text{Evaluate } \int (x^{2}+1)^{5}\,2x\,dx. \]
Identify the inner function
Why: Inside the fifth power.
\[ u = x ^{2} + 1 \]
Compute the differential
Why: Differentiate and multiply by dx.
\[ \,du = 2 x \,dx \]
Locate it in the integrand
Why: The remaining factor.
\[ 2 x \,dx\text{ is present exactly} \]
Rewrite the whole integral
Why: Every x is gone.
\[ \int u ^{5} \,du \]
Integrate and substitute back
Why: The power rule.
\[ (x ^{2} + 1) ^{6} / 6 + C \]
Figure (svg): Substitution as the chain rule read backwards
\[ \frac{(x^{2}+1)^{6}}{6}+C \]
Verify: differentiate the answer back
Why: The chain rule gives the fifth power of x squared plus one, times 2x — the original integrand exactly. That check is available for every substitution and settles the matter completely, which makes this one of the few techniques where an error need never survive. Note that the answer is precisely the function from the warm-up, which is where the integrand came from.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 538-539
Sorting
Is the inner derivative present, up to a constant?
Sort into buckets
Sort each integrand.
The fourth is worth knowing about: the sine of x squared has no elementary antiderivative at all, so the failure is not a limitation of substitution but of the elementary functions. Multiplying it by x makes it routine.
Worked example
Checkpoint 5.39. Several candidates, one that works.
\[ \text{Evaluate } \int \frac{\cos x}{\sin^{2}x}\,dx. \]
Look for a composition
Why: The square of a sine.
\[ \text{candidate } u = \sin x \]
Compute the differential
Why: Differentiate.
\[ \,du = \cos x \,dx \]
Check the integrand
Why: The numerator.
Rewrite
Why: Every x is gone.
\[ \int \,du / u ^{2} \]
Integrate and substitute back
Why: The power rule with exponent -2.
\[ -1 / \sin x + C \]
Figure (svg): The solution to Worked example recognising the inner function shown as a ladder of expressions, one row per legal move
\[ -\frac{1}{\sin x}+C = -\csc x+C \]
Verify: differentiate back and note the alternative form
Why: Differentiating negative one over sine by the quotient or chain rule gives cosine over sine squared, matching. The answer can be written as minus the cosecant, and a table of integrals would list it that way — so two apparently different answers are the same function. That is common with substitution, and differentiating is the way to confirm two forms agree rather than assuming one is wrong.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 539-541
Trap
\[ \int (x^{2}+1)^{5}dx, \quad u = x^{2}+1 \]
Rewrite as the integral of u to the fifth
Why: The student ignores that du needs a factor of 2x.
\[ du = 2x\,dx \quad \text{but there is no } x \text{ in the integrand} \]
The dx cannot be converted, so the integral cannot be rewritten in u at all.
\[ \text{expand instead: } \int(x^{10}+5x^{8}+\cdots)dx \]
Check that the inner derivative is present BEFORE committing
Why: Substitution is available only when it is.
Expanding a fifth power is tedious but works. The check costs one line and prevents several lines of invalid work, so it is worth making before the substitution is written down.
Fill the middle
A substitution chosen; its differential needed.
Fill in the blanks
u = x^2x+1 \;\Longrightarrow\; du = ___\,dx
Why: The differential is the derivative times dx. Checking whether that expression appears in the integrand is what decides whether the substitution is available.
Matching
The inner function of the composition.
Match the pairs
Why: In every case u is what sits inside — a bracket, a trigonometric function, an exponent, or a squared function. The second needs only a constant adjustment; the third needs a factor of one half.
Prediction
Commit before reasoning.
Predict first
What must be true of an integrand for substitution to work?
Correct: A composition and the inner derivative, up to a constant.
\[ \int(x^{2}+1)^{5}2x\,dx \;\checkmark\; \quad \int(x^{2}+1)^{5}dx \;\times \]
Why: A composition alone is not enough: the fifth power of x squared plus one is a composition and substitution fails on it, because there is no 2x to absorb. The derivative's presence is what allows the dx to be converted into a du, and without that conversion the integral cannot be written in the new variable at all. A missing numerical factor is fixable; a missing variable factor is not.
Section
Section 2
Concept
After substituting, the integral must contain the new variable only. A single leftover x makes the expression impossible to integrate with respect to either variable.
complete substitution — Rewriting every occurrence of the original variable, including inside the differential, so the integral is expressed purely in the new variable.
\[ \int f(g(x))g'(x)\,dx \;\longrightarrow\; \int f(u)\,du \]
When a stray x remains, it can sometimes be eliminated by solving the substitution equation for x and replacing it. When it cannot, the substitution has failed and a different choice or method is needed.
Figure (svg): The substitution procedure, with the step most often skipped marked
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 538-546 — carrying out substitution
Picture it
The procedure, with the danger marked.
Figure (svg): The substitution procedure, with the step most often skipped marked
Step four is the checkpoint. Scanning the rewritten integral for any surviving x costs a second and catches the error that invalidates everything downstream.
Worked example
Example 5.41. Solve the substitution for x.
\[ \text{Evaluate } \int x\sqrt{x-1}\,dx. \]
Choose the inner function
Why: Under the radical.
\[ u = x - 1 \]
Compute the differential
Why: Simple.
\[ \,du = \,dx \]
Rewrite what can be rewritten
Why: The radical.
\[ \int x \sqrt{u} \,du \]
Eliminate the stray x
Why: Solve the substitution.
\[ x = u + 1 \]
Rewrite and integrate
Why: Expand and use the power rule.
\[ (\frac{2}{5}) u ^{\frac{5}{2}} + (\frac{2}{3}) u ^{\frac{3}{2}} \]
Figure (svg): The substitution procedure, with the step most often skipped marked
\[ \frac25(x-1)^{5/2}+\frac23(x-1)^{3/2}+C \]
Verify: differentiate the answer back
Why: Differentiating gives the three-halves power of x minus one, plus the one-half power — and factoring out the square root leaves the square root times x minus one plus one, which is x times the square root. That matches the integrand. Note the essential move: the stray x was eliminated by solving the substitution equation, which works whenever that equation can be inverted, and it extends substitution well beyond the exact-derivative case.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 541-543
Fill the middle
A substitution leaving an x behind.
Fill in the blanks
u = x-1 \;\Longrightarrow\; x = u+1
Why: Solving the substitution equation for x lets the stray factor be rewritten in u. This works whenever the substitution can be inverted, and it extends the method well past the exact-derivative case.
Worked example
Checkpoint 5.41. A constant inner derivative.
\[ \text{Evaluate } \int \cos(5x)\,dx. \]
Choose the inner function
Why: Inside the cosine.
\[ u = 5 x \]
Compute the differential
Why: Differentiate.
\[ \,du = 5 \,dx \]
Solve for dx
Why: Divide.
\[ \,dx = \,du / 5 \]
Rewrite
Why: Every x is gone.
\[ (\frac{1}{5}) \int \cos u \,du \]
Integrate and substitute back
Why: The sine.
\[ (\frac{1}{5}) \sin(5 x) + C \]
Figure (svg): The solution to Worked example a substitution inside a trigonometric function shown as a ladder of expressions, one row per legal move
\[ \frac{\sin(5x)}{5}+C \]
Verify: differentiate back and note the general pattern
Why: The chain rule gives one fifth times cosine of 5x times 5, which is cosine of 5x — correct. The pattern generalises: any integrand with a linear inner function picks up a factor of one over that function's slope. It is worth recognising directly rather than substituting each time, since linear inner functions are extremely common.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 543-544
Error analysis
A student substitutes and integrates.
Annotate
On: \( \int x\sqrt{x-1}\,dx, \; u=x-1 \;\Longrightarrow\; \int x\sqrt{u}\,du = \frac{2x}{3}u^{3/2}+C \)
The rule is absolute: no x may remain after the substitution. Scanning the rewritten integrand for one takes a second and prevents an error that the arithmetic will not reveal.
Ranking
A substitution from start to finish.
Put in order
Why: Step c is the one that fails silently. An expression mixing x and u can be integrated with respect to neither, so anything computed after a leftover x is meaningless however careful the arithmetic.
Two truths and a lie
All three are about carrying out the method.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Pulling a factor outside an integral requires it to be constant with respect to the variable of integration, and x varies as u does — they are related by the substitution. Only genuine numbers may be moved.
Prediction
Commit before reasoning.
Predict first
Why is an integral containing both x and u meaningless?
Correct: Because the expression can be integrated with respect to neither.
\[ \int x\sqrt{u}\,du: \; x \text{ varies with } u \]
Why: The differential du says the integration is with respect to u, so anything else in the integrand must be expressed in u or be a constant. An x is neither: it varies as u varies, so it cannot be treated as a constant, and it is not written in u so the integral cannot be evaluated. The expression is not merely untidy but has no defined value.
Section
Section 3
Concept
When the inner derivative is present up to a numerical factor, the constant can be adjusted freely. When a factor of the variable is missing, nothing can supply it.
adjusting the constant — Multiplying and dividing by a number so that the integrand contains exactly the differential required. Only numerical factors may be moved across an integral sign.
\[ \int x e^{x^{2}}dx = \frac12\int e^{u}\,du \]
The asymmetry comes straight from the constant-multiple rule of Section 5.2, which holds for numbers and fails for anything containing the variable of integration.
Figure (svg): Adjusting a constant: legitimate, unlike adjusting a variable
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 540-548 — adjusting constants in substitution
Picture it
Two integrands differing by one factor of x.
Figure (svg): Adjusting a constant: legitimate, unlike adjusting a variable
The left works because only a number was needed; the right fails because a variable was. The difference between them is the whole boundary of the method.
Worked example
Example 5.43. The derivative present up to a factor.
\[ \text{Evaluate } \int x e^{x^{2}}dx. \]
Choose the inner function
Why: In the exponent.
\[ u = x ^{2} \]
Compute the differential
Why: Differentiate.
\[ \,du = 2 x \,dx \]
Compare with the integrand
Why: Only x dx is present.
Solve for the available piece
Why: Divide.
\[ x \,dx = \,du / 2 \]
Rewrite and integrate
Why: The constant comes outside.
\[ (\frac{1}{2}) e ^{x ^{2}} + C \]
Figure (svg): Adjusting a constant: legitimate, unlike adjusting a variable
\[ \frac12 e^{x^{2}}+C \]
Verify: differentiate the answer back
Why: The chain rule gives one half times the exponential times 2x, which is x times the exponential — the original integrand. The one half was legitimate because 2 is a number and passes freely across the integral sign. Compare with the exponential alone, where a factor of x would be needed and cannot be created: that integral has no elementary antiderivative at all.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 545-546
Sorting
Only genuine constants.
Sort into buckets
Sort each factor.
The last is the tempting one: pi looks like a letter and is a number. What matters is not whether the symbol is a letter but whether it varies with the variable of integration.
Worked example
Checkpoint 5.43. The boundary of the method.
\[ \text{Why can } \int e^{x^{2}}dx \text{ not be done by substitution?} \]
Attempt the substitution
Why: The obvious choice.
\[ u = x ^{2}, \,du = 2 x \,dx \]
Look for the required factor
Why: The integrand has none.
Consider supplying one
Why: Multiply and divide by x.
\[ \text{the } \frac{1}{x}\text{ cannot come outside} \]
State why
Why: It contains the variable.
Draw the conclusion
Why: No elementary antiderivative exists.
Figure (svg): When substitution will not help, and what to do instead
\[ \int e^{x^{2}}dx \text{ has no elementary antiderivative} \]
Verify: distinguish a limitation of the method from a limitation of the functions
Why: This is not merely a case substitution cannot handle: it was proved in the nineteenth century that no combination of elementary functions differentiates to the exponential of x squared. So no technique will ever succeed, and the integral is instead given a name — it is essentially the error function of statistics. Knowing which failures are method-specific and which are fundamental saves a great deal of wasted effort.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 546-548
Trap
\[ \int e^{x^{2}}dx = \frac{1}{2x}\int e^{x^{2}}2x\,dx \]
Multiply and divide by 2x, then pull the reciprocal outside
Why: The student treats the variable like a constant.
The factor contains x, which varies over the interval, so it cannot leave the integral. The manipulation is invalid.
\[ \text{only numbers may be moved: } \int cf = c\int f \]
Check whether the factor is a genuine constant before moving it
Why: The constant-multiple rule says nothing about variable factors.
This error is tempting because it looks like the legitimate adjustment made one line earlier for a numerical factor. The test is simply whether the factor contains the variable of integration.
Fill the middle
Half the required differential is present.
Fill in the blanks
du = 2x\,dx \;\Longrightarrow\; x\,dx = \frac2___}
Why: Dividing the differential by 2 gives exactly what the integrand contains. The one half then comes outside as a numerical constant, which is legitimate in a way that a factor of x would not be.
Two truths and a lie
All three are about adjusting factors.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Multiplying and dividing by x is algebraically valid, but the resulting reciprocal cannot leave the integral because it varies. The constant-multiple rule applies only to genuine constants, which is exactly the boundary of what substitution can adjust.
Prediction
Commit before reasoning.
Predict first
Why can a constant be moved across an integral sign when a variable cannot?
Correct: Because the rule was proved only for constants.
\[ \sum c\,a_{i} = c\sum a_{i} \quad \text{but} \quad \sum x_{i}a_{i} \ne x\sum a_{i} \]
Why: Section 5.2's constant-multiple rule came from the corresponding fact about sums, where a common factor can be collected from every term — which requires the factor to be the same in every term. A varying factor differs from term to term and cannot be collected. So the asymmetry is not a stylistic rule but a consequence of what the sum can do, and it marks the exact boundary of substitution's reach.
Section
Section 4
Concept
A definite integral can be finished either by converting the limits to the new variable and evaluating there, or by returning to the original variable and using the original limits. Mixing the two gives a wrong answer.
changing the limits — Substituting the original limits into the substitution equation to obtain limits in the new variable, so the integral can be evaluated without returning to the original.
\[ \int_{a}^{b}f(g(x))g'(x)dx = \int_{g(a)}^{g(b)}f(u)\,du \]
Changing the limits is usually shorter and less error-prone, since the answer never has to be rewritten in the original variable. Back-substituting is necessary only when the antiderivative itself is wanted.
Figure (svg): Two ways to finish a definite integral: change the limits, or substitute back
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 546-556 — substitution for definite integrals
Picture it
Both correct, one shorter.
Figure (svg): Two ways to finish a definite integral: change the limits, or substitute back
The left route ends in the new variable and never returns; the right returns and uses the original limits. What must not happen is ending in one variable with the other's limits.
Worked example
Example 5.45. Convert the limits and finish in u.
\[ \text{Evaluate } \int_{0}^{2}x(x^{2}+1)^{3}dx. \]
Choose the substitution
Why: Inside the cube.
\[ u = x ^{2} + 1 \]
Compute the differential
Why: And adjust.
\[ x \,dx = \,du / 2 \]
Convert the lower limit
Why: Substitute x = 0.
\[ u = 1 \]
Convert the upper limit
Why: Substitute x = 2.
\[ u = 5 \]
Integrate and evaluate in u
Why: From 1 to 5.
\[ (\frac{1}{8}) (625 - 1) = 78 \]
Figure (svg): Two ways to finish a definite integral: change the limits, or substitute back
\[ \int_{0}^{2}x(x^{2}+1)^{3}dx = 78 \]
Verify: check by the other route
Why: Back-substituting instead gives one eighth of the fourth power of x squared plus one, evaluated from 0 to 2 — which is one eighth of 625 minus 1, also 78. Both routes agree, as they must. Note that the second route required writing the antiderivative back in x and then evaluating, two extra steps that the first avoided; over many problems that difference in length matters.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 549-550
Fill the middle
An x limit converted into a u limit.
Fill in the blanks
u = x^5+1, \; x=2 \;\Longrightarrow\; u = ___
Why: Substituting the x limit into the substitution equation gives the u limit. Doing this at the same moment as computing the differential keeps the variable and its limits from coming apart.
Worked example
Checkpoint 5.45. The limits collapsing.
\[ \text{Evaluate } \int_{-1}^{1}x(x^{2}+1)^{3}dx \text{ by substitution, and by symmetry.} \]
Substitute
Why: The same choice.
\[ u = x ^{2} + 1 \]
Convert the lower limit
Why: Substitute x = -1.
\[ u = 2 \]
Convert the upper limit
Why: Substitute x = 1.
\[ u = 2 \]
Note the limits coincide
Why: Equal limits.
Confirm by symmetry
Why: The integrand is odd.
Figure (svg): The solution to Worked example a symmetric-looking integral shown as a ladder of expressions, one row per legal move
\[ \int_{-1}^{1}x(x^{2}+1)^{3}dx = 0 \]
Verify: see why the two arguments agree
Why: Section 5.2's rule says an integral with equal limits is zero, and Section 5.4's says an odd integrand over a symmetric interval is zero — and here both apply, giving the same answer by different routes. The coincidence is not accidental: the substitution collapses the limits precisely because the inner function is even, which is what makes the whole integrand odd. Two apparently separate facts turn out to be the same one.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 550-552
Error analysis
A student evaluates a definite integral by substitution.
Annotate
On: \( \int_{0}^{2}x(x^{2}+1)^{3}dx = \frac12\int_{0}^{2}u^{3}du = \frac{16}{8} = 2 \)
This is the section's characteristic error because nothing about the arithmetic signals it. The habit worth building is converting the limits at the same moment as the differential, so the two never come apart.
Sorting
Both are valid; one suits each situation.
Sort into buckets
Sort each situation.
The last case is the strongest argument for changing the limits: converting them reveals immediately that they coincide and the integral is zero, which back-substituting would obscure behind a computation.
Ranking
Substitution with limits changed.
Put in order
Why: Placing step b immediately after step a is the practical safeguard: the limits are converted while the substitution is still in mind, rather than being remembered several lines later when the integrand has changed shape.
Prediction
Commit before reasoning.
Predict first
Why do wrong limits after a substitution so often go unnoticed?
Correct: Because the computation runs perfectly and produces a plausible number.
\[ \text{limits } 0,2 \text{ in } x \;\longrightarrow\; 1,5 \text{ in } u \]
Why: Every subsequent step is arithmetically valid, and the answer looks like any other answer — in the worked example, 2 rather than 78, both perfectly reasonable-looking. Unlike an algebraic slip, which usually produces something visibly odd, this fails silently. Converting the limits immediately after computing the differential is the habit that prevents it, because the two are then never separated.
Section
Section 5
Concept
Some integrals resist substitution but yield to another approach; others have no elementary antiderivative at all. Telling the two apart saves a great deal of effort.
non-elementary integrals — Integrals whose antiderivatives cannot be written as any finite combination of elementary functions. Their impossibility is a proved theorem, not a gap in known techniques.
\[ \int e^{x^{2}}dx, \quad \int\frac{\sin x}{x}dx, \quad \int\sqrt{1+x^{4}}\,dx \]
The famous examples are given names and studied as functions in their own right — the error function, the sine integral, the elliptic integrals — and Section 5.3's Part 1 supplies their derivatives despite no formula existing.
Figure (svg): When substitution will not help, and what to do instead
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 543-556 — limitations of substitution
Picture it
What separates the possible from the impossible.
Figure (svg): When substitution will not help, and what to do instead
The last two rows differ by a single factor of x, and that factor is the difference between an integral with no elementary answer and a routine one. Nothing about their appearance suggests such a gap.
Worked example
Example 5.47. A pair of integrands compared.
\[ \text{Compare } \int x\sin(x^{2})dx \text{ with } \int\sin(x^{2})dx. \]
Attempt the first
Why: The inner function is squared.
\[ u = x ^{2}, \,du = 2 x \,dx \]
Check the factor
Why: An x is present.
\[ x \,dx = \,du / 2 \]
Complete it
Why: Integrate the sine.
\[ -\cos(x ^{2}) / 2 + C \]
Attempt the second
Why: The same substitution.
Note the outcome
Why: Nothing can supply one.
Figure (svg): When substitution will not help, and what to do instead
\[ \int x\sin(x^{2})dx = -\tfrac12\cos(x^{2})+C \]
Verify: appreciate how invisible the difference is
Why: The two integrands look almost identical and one is a two-line exercise while the other is a named function of mathematical physics — the Fresnel integral, which describes light diffraction. Nothing in their appearance suggests the gap. That is worth internalising early: whether an integral is elementary depends on fine structural details, not on how complicated it looks.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 552-553
Sorting
Can any technique do this?
Sort into buckets
Sort each integral.
The middle category is worth recognising by shape rather than by attempt. An exponential or trigonometric function of a non-linear expression, with no matching derivative present, is almost always non-elementary.
Worked example
Checkpoint 5.47. The alternatives.
\[ \text{What can be done when substitution does not apply?} \]
Try a different substitution
Why: Another inner function.
Try rewriting
Why: Expand, divide, or use an identity.
\[ \text{Section } 5.4' s\text{ technique} \]
Consider later techniques
Why: Parts, partial fractions, trigonometric substitution.
Consider numerical methods
Why: Section 5.1's rectangles.
Recognise a non-elementary integral
Why: No technique will succeed.
\[ \text{name it and use Part } 1 \]
Figure (svg): The solution to Worked example what to do when substitution fails shown as a ladder of expressions, one row per legal move
\[ \text{rewrite} \to \text{other methods} \to \text{numerical} \]
Verify: note which option is always available
Why: Numerical integration never fails: Section 5.1's rectangles work for any integrand that can be evaluated, elementary or not, and give as many digits as wanted. So a non-elementary integral is not an unanswerable question but one whose answer must be a number rather than a formula. That is why the error function is tabulated in every statistics text despite having no closed form.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 553-555
Trap
\[ \int e^{x^{2}}dx: \; \text{try } u=x^{2}, \text{ then } u=e^{x^{2}}, \text{ then }\ldots \]
Keep trying substitutions
Why: The student assumes a technique must exist.
No elementary antiderivative exists — this was proved, not merely unfound. No amount of searching will produce one.
\[ \text{name it: } \tfrac{\sqrt\pi}{2}\,\text{erf}(x)+C, \text{ or integrate numerically} \]
Recognise a non-elementary integral and switch approach
Why: Knowing which failures are fundamental saves the effort.
The practical signals are an exponential of a non-linear expression, a trigonometric function of one, or a ratio like sine over x. Recognising the shapes is quicker than rediscovering their impossibility each time.
Fill the middle
Two integrands, one solvable.
Fill in the blanks
\int x\sin(x^___)dx \text___; \; \int\sin(x^___)dx \text___
Why: A single factor of x supplies the derivative the substitution needs. Without it the integral is the Fresnel integral, a named non-elementary function used in optics.
Matching
The famous examples are studied as functions.
Match the pairs
Why: Each is defined as an integral because no other definition exists, and each is genuinely useful — in statistics, signal processing, optics and number theory respectively. Part 1 of the Fundamental Theorem gives their derivatives immediately, which is how they are analysed.
Prediction
Commit before reasoning.
Predict first
The exponential of negative x squared has no elementary antiderivative. What does that mean?
Correct: It was proved that none exists.
\[ \int_{-\infty}^{\infty}e^{-x^{2}}dx = \sqrt\pi \quad \text{exists, with no elementary antiderivative} \]
Why: This is a theorem, established in the nineteenth century, not a gap in current knowledge — so further searching is guaranteed to fail. The integral itself exists perfectly well: the function is continuous and the area under it is a definite number for any interval, computable numerically to any precision. What does not exist is a closed-form formula, which is a statement about notation rather than about the area.
Comparison
Fill the blanks. The middle is identical; the ends differ.
Comparison matrix
| Indefinite | Definite | |
|---|---|---|
| After integrating | substitute back to x | change the limits, or substitute back |
| The limits | none | must be converted if staying in u |
| The answer | a function of x, plus C | a number |
| The constant | required | unnecessary: it cancels |
The second row is where the definite case goes wrong. Converting the limits at the same moment as the differential is the habit that prevents the section's characteristic error.
Pattern
Given an integral that the basic table does not cover.
Steps three and four are where the method fails silently. Converted limits and a fully rewritten integrand are the two things to confirm before integrating, since neither error shows up in the arithmetic.
Stewart, Calculus: Early Transcendentals 8e, §5.5 The Substitution Rule §5.5, pp. 412-420
Check
Recognising the shape.
Check your understanding
Which of these can be done by substitution?
Answer: A
Why: The factor of x supplies half the inner derivative, and the missing 2 is a constant.
Check
Definite integrals.
Check your understanding
For the integral from 0 to 2 with u = x^2 + 1, what are the limits in u?
Answer: A
Why: Substituting x = 0 gives u = 1, and x = 2 gives u = 5.
Check
The boundary of the method.
Check your understanding
Why can a factor of x not be supplied by multiplying and dividing?
Answer: A
Why: Only genuine constants may be moved across an integral sign.
Real world
A drug's concentration in the bloodstream falls exponentially with a rate constant that itself depends on the amount present, giving a model where the elimination rate is proportional to the concentration times an exponential of it.
Discussion prompt
Explain why substitution is the natural tool here, what quantity the integral gives, and what happens if the model is changed slightly.
Hint: The exponent contains the same quantity that multiplies it.
Answer:
The elimination model produces an integrand of the form concentration times the exponential of a multiple of concentration — a composition multiplied by something proportional to the inner function's derivative. That is exactly the shape substitution reverses, and the total drug eliminated over an interval is the integral of that rate.
\[ \int k\,C e^{-\alpha C}\,dC, \qquad u = -\alpha C \]
The integral gives total elimination over the concentration range, which determines how long the drug remains active — the quantity a dosing schedule is built on.
A slight change to the model destroys this. If the elimination rate were proportional to the exponential alone, with no factor of concentration, substitution would fail and no elementary antiderivative would exist — the integral would be the error function's relative. The pharmacologist would then have to integrate numerically, which is entirely workable but gives tables of numbers rather than a formula.
That fragility is worth noticing. Whether a model yields a closed-form answer or only numerical ones can turn on a single factor that has no particular biological significance, so a modeller sometimes chooses between two nearly equivalent models on the grounds that one is integrable and the other is not.
Commit first
Answer, then rate your confidence honestly.
Predict first
After substituting in a definite integral, what must happen to the limits?
Correct: They must be converted, or the answer back-substituted first.
\[ \int_{0}^{2}\ldots dx = \int_{1}^{5}\ldots du, \quad \text{not } \int_{0}^{2}\ldots du \]
Why: The limits belong to whichever variable is being used, so evaluating a u integral between x limits gives a wrong number — 2 rather than 78 in the worked example. Either convert them, which is shorter, or return to the original variable and keep the original limits. What fails is mixing the two, and it fails silently: the arithmetic runs perfectly and the answer looks reasonable.
Explain it
They tried substituting the inside of the bracket for the fifth power of x squared plus one, with no x factor present, and got stuck.
Discussion prompt
In four sentences or fewer, explain why it failed.
Hint: Ask them what du equals.
Answer:
Ask them to write down du: it is 2x dx, and there is no x anywhere in the integrand to supply it. Without that factor the dx cannot be converted into a du, so the integral cannot be written in the new variable at all.
Substitution needs a function AND its derivative both present, not just a composition. For this one, expanding the fifth power is the available route — tedious, but it works, and checking for the derivative first would have saved the attempt.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For spotting, look inside brackets, roots, exponents and trigonometric functions, then check for the derivative. For rewriting, scan for surviving x's before integrating and solve the substitution for x if one remains. For limits, convert them immediately after computing the differential. For recognising failure, check that the inner derivative is present up to a constant before committing. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the chain rule and beneath it the substitution formula, with an arrow between them and a note saying which direction each is read. Below, list the six steps of the method, boxing the one about rewriting every x and writing beside it what goes wrong if a variable survives. In the middle of the page, make a two-column table of what may and may not be pulled outside an integral, with three examples in each and one line on why. Beside it, work one definite integral twice — once by changing the limits and once by back-substituting — and confirm the answers agree. In the lower half, write four integrands: one where substitution works, one where it fails but expanding works, and two that are non-elementary, with the obstacle named beside each. At the bottom, write the pair that differs by a single factor of x and one sentence on what that factor decides.
If your two routes through the definite integral gave different numbers, the limits were mixed with the wrong variable — that is the section's characteristic error, and catching it once on paper is worth more than being warned about it.
Recap
Five things, and three of them are about the ways the method fails silently.
| If you see | Then |
|---|---|
| A composition with its derivative present | Substitute the inner function |
| A composition with no such factor | Expand or rewrite instead |
| A missing numerical factor | Adjust it: constants may move |
| A missing factor of the variable | Nothing can supply it |
| A stray x after substituting | Solve the substitution for x |
| A definite integral | Convert the limits immediately |
| An exponential of a non-linear expression | Suspect a non-elementary integral |
Section 5.6 applies substitution to exponential and logarithmic integrands, where it supplies the antiderivatives that Section 4.10's table could only gesture at — including the general exponential and the integral that defines the logarithm.
OpenStax Calculus Volume 1, §5.5 Substitution §5.5, pp. 506-515 — everything on these slides traces back here
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