MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: the substitution rule (MATH 141)

Q: How do I choose u for a substitution in calculus?

Look for a function sitting inside another one, a power, a root, an exponential, whose derivative also appears as a factor of the integrand, possibly up to a constant. That inside function is u. If the derivative is missing by a constant, supply the constant; if it is missing by a factor containing x, this u is the wrong choice.

Q: Do I have to change the limits of integration when I substitute?

Either change them, or go back to x before evaluating, but never half of each. If the integral is written in u, the limits must be values of u: plug the old limits into u equals g of x. If you prefer to keep the old limits, substitute back to x first. An integral in u with limits in x gives a wrong number.

Q: What happens when the new lower limit is bigger than the upper limit?

Nothing goes wrong: keep the limits in the order the substitution gives them. The minus sign that usually comes with du and the reversed order cancel each other. You may also swap the limits and drop the minus sign at the same time, but never do only one of the two, or the sign of the answer flips.

Q: Why is the integral of tan x equal to minus ln of cos x?

Write tan x as sine over cosine and let u be cos x. Then du is minus sine of x dx, so the integral becomes minus the integral of du over u, which is minus ln of the absolute value of cos x, plus C. The same answer is often written as ln of the absolute value of sec x. The minus sign comes from the derivative of cosine.

Q: What do I do if an x is left over after the substitution?

Solve the substitution for x and replace it. With u equals x minus 1, write x as u plus 1; with u equals x squared plus 1, write x squared as u minus 1. The integral in u must contain no x at all. Never pull the leftover x out of the integral as if it were a constant: only numbers can be moved through the integral sign.

This sheet is not a summary of section 5.5 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on the substitution rule in MATH 141 at McGill University, and which precise gesture avoids each loss.

Everything below is done by hand, as on the exam, and every antiderivative quoted has been checked the only way that settles it: by differentiating it back to the integrand.

The thread of the chapter

A substitution is finished only when the integral contains NO $x$ at all: not in the integrand, not in the differential, not in the bounds. Every mark lost on this chapter is an $x$ left somewhere: a variable factor divided out of $du$ , a stray $x$ treated as a constant, a bound that stayed in $x$ .

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Substitution rule
Change of variables
Chain rule
Indefinite integral
Definite integral
Change of limits of integration
Logarithmic integral
Integral of the tangent function
Inverse tangent integral
Inverse sine integral
Odd and even functions
Symmetric interval of integration

The essentials

The rule, and what du has to swallow

• $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$ with $u = g(x)$ and $du = g'(x)\,dx$ . Read backwards, it is the chain rule $\frac{d}{dx}F(g(x)) = F'(g(x))\,g'(x)$ .
• Choose $u$ = the INSIDE function whose derivative is also a factor of the integrand: $x^3 + 1$ in $x^2(x^3 + 1)^5$ , $\ln x$ in $\frac{(\ln x)^3}{x}$ , $\sqrt x$ in $\frac{\cos\sqrt x}{\sqrt x}$ .
• A CONSTANT missing from $du$ is supplied: $u = 1 + x^2$ gives $x\,dx = \frac{1}{2}du$ . A factor containing $x$ is NEVER supplied by dividing: then this $u$ is the wrong one.
• An $x$ still present after $du$ is used is rewritten through $u$ : $u = x - 1$ gives $x = u + 1$ ; $u = x^2 + 1$ gives $x^2 = u - 1$ .
• Indefinite integral: end in $x$ , with $+ C$ , and differentiate the answer. The chain rule must give back the integrand exactly.

Before writing anything, ask: which function, differentiated, gives the other factor up to a constant? If none does, substitution is not the tool, and forcing it is where the marks go.

Definite integrals: the bounds travel through u

• Route 1: antiderivative in $x$ , then the ORIGINAL bounds. Route 2: new bounds $u = g(a)$ and $u = g(b)$ , and never come back to $x$ .
• The mixture, an integrand in $u$ with bounds in $x$ , is always wrong. It is an unfinished substitution: an $x$ is still hiding in the bounds.
• New bounds in decreasing order are normal (as with $u = \cos x$ on $\left[0, \frac{\pi}{2}\right]$ ). Keep them; the minus sign of $du$ and the reversed order cancel together.
• No one-to-one condition: $g'$ continuous on $[a, b]$ is enough. $u = \sin x$ on $[0, \pi]$ is legal.
• $f$ continuous on $[-a, a]$ : odd gives $\int_{-a}^a f = 0$ , even gives $2\int_0^a f$ . Both are proved by $u = -x$ .

The bounds are read on the vertical axis: for

u = 1 + x^2

, the interval

0 \le x \le \sqrt 3

becomes

1 \le u \le 4

. Neither bound stays where it was.

Route 2 is shorter, and the line when $x = a$ , $u = \ldots$ ; when $x = b$ , $u = \ldots$ is exactly what the marker looks for.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Which u, and which antiderivative comes out

Read a line as: for this integrand, this choice of $u$ gives this antiderivative. Each example is a definite integral computed by hand. The red lines are rules students invent.

Integrand	Choice of u	Antiderivative
$x\sqrt{1 + x^2}$	$u = 1 + x^2$ , $x\,dx = \frac{1}{2}du$	$\frac{1}{3}(1 + x^2)^{3/2} + C$
Example: $\int_0^{\sqrt 3}x\sqrt{1 + x^2}\,dx = \frac{1}{3}(8 - 1) = \frac{7}{3}$ .
$\frac{x}{1 + x^2}$	$u = 1 + x^2$	$\frac{1}{2}\ln(1 + x^2) + C$
Example: $\int_0^1\frac{x}{1 + x^2}\,dx = \frac{1}{2}\ln 2$ .
$\frac{x}{1 + x^4}$	$u = x^2$	$\frac{1}{2}\arctan(x^2) + C$
Example: $\int_0^1\frac{x}{1 + x^4}\,dx = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}$ .
$\tan x$	$u = \cos x$	$-\ln\|\cos x\| + C$
Example: $\int_0^{\pi/3}\tan x\,dx = -\ln\frac{1}{2} = \ln 2$ .
$\frac{1}{x^2 + 9}$	$u = \frac{x}{3}$ , $dx = 3\,du$	$\frac{1}{3}\arctan\frac{x}{3} + C$
Example: $\int_0^3\frac{dx}{x^2 + 9} = \frac{1}{3}\cdot\frac{\pi}{4} = \frac{\pi}{12}$ .
$x\sqrt{x - 1}$	$u = x - 1$ , $x = u + 1$	$\frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C$
Example: $\int_1^2 x\sqrt{x - 1}\,dx = \frac{2}{5} + \frac{2}{3} = \frac{16}{15}$ .
$\frac{1}{1 + x^2}$	$u = 1 + x^2$	$\ln(1 + x^2)$ no such rule
Example: The derivative of $\ln(1 + x^2)$ is $\frac{2x}{1 + x^2}$ ; and $\int_0^1\frac{dx}{1 + x^2} = \frac{\pi}{4} \approx 0.785$ , not $\ln 2 \approx 0.693$ . What to do: No $x$ on top, no $du$ : it is the table, $\int\frac{dx}{1 + x^2} = \arctan x + C$ .
$(x^3 + 1)^5$	$u = x^3 + 1$	$\frac{(x^3 + 1)^6}{18x^2}$ no such rule
Example: Its derivative at $x = 1$ is $32 - \frac{64}{9}$ , not the integrand's $2^5 = 32$ . What to do: No $x^2$ in the integrand: expand $(x^3 + 1)^5$ by the binomial theorem and integrate term by term.

Every blue line is proved in one line by differentiating the last column. If the derivative of your answer carries an extra term, the rule you used does not exist.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Dividing by a variable to manufacture du

the whole question

What not to write

“ $u = x^3 + 1$ , $dx = \frac{du}{3x^2}$ , so $\int(x^3 + 1)^5\,dx = \frac{(x^3 + 1)^6}{18x^2} + C$ .”

What to write

“There is no $x^2$ to form $du$ , so this $u$ does not apply: $(x^3 + 1)^5 = x^{15} + 5x^{12} + 10x^9 + 10x^6 + 5x^3 + 1$ , integrated term by term.”

Why: Once $dx = \frac{du}{3x^2}$ is written, the integral in $u$ still contains $x$ , and pulling $\frac{1}{3x^2}$ out treats a variable as a constant. Differentiating the answer produces the extra term $-\frac{(x^3 + 1)^6}{9x^3}$ : that term is the error made visible.

2. Keeping the bounds of x in an integral in u

2 to 3 marks, the whole evaluation

What not to write

“ $u = x^2$ : $\int_0^2 x e^{x^2}\,dx = \frac{1}{2}\int_0^2 e^u\,du = \frac{e^2 - 1}{2}$ .”

What to write

“When $x = 0$ , $u = 0$ ; when $x = 2$ , $u = 4$ : $\int_0^2 x e^{x^2}\,dx = \frac{1}{2}\int_0^4 e^u\,du = \frac{e^4 - 1}{2}$ .”

Why: The numbers $0$ and $2$ are values of $x$ , and the new integral is in $u$ . The bound line is part of the substitution, like $du$ : an integral with $u$ inside and $x$ on the bounds is still half in $x$ .

3. Reordering reversed bounds and keeping the minus sign

1 to 2 marks, and a negative value for a positive integrand

What not to write

“ $u = 1 - x^2$ , $du = -2x\,dx$ : $\int_0^1 x\sqrt{1 - x^2}\,dx = -\frac{1}{2}\int_0^1\sqrt u\,du = -\frac{1}{3}$ .”

What to write

“When $x = 0$ , $u = 1$ ; when $x = 1$ , $u = 0$ : $-\frac{1}{2}\int_1^0\sqrt u\,du = \frac{1}{2}\int_0^1\sqrt u\,du = \frac{1}{3}$ .”

The integrand

x\sqrt{1 - x^2}

is positive on

(0, 1)

, so its integral is the shaded area,

\frac{1}{3}

. A negative answer is impossible, whatever the substitution did to the bounds.

Why: The reversed order and the minus of $du$ come as a pair: either keep both, $-\frac{1}{2}\int_1^0$ , or drop both, $\frac{1}{2}\int_0^1$ . Using one without the other flips the sign, and the figure shows it cannot be negative.

4. Treating a stray x as a constant

the whole question

What not to write

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

What to write

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

Why: Only a number passes through the integral sign. The derivative of the wrong answer is $x\sqrt{x - 1} + \frac{2}{3}(x - 1)^{3/2}$ : at $x = 2$ it gives $\frac{8}{3}$ where the integrand is $2$ . Solve $u = g(x)$ for $x$ and replace it.

5. Losing the minus sign in the integral of tan x

1 to 2 marks, and a negative value for every definite integral of tan on the first quarter turn

What not to write

“ $\int\tan x\,dx = \ln|\cos x| + C$ .”

What to write

“ $u = \cos x$ , $du = -\sin x\,dx$ : $\int\tan x\,dx = -\int\frac{du}{u} = -\ln|\cos x| + C = \ln|\sec x| + C$ .”

\tan x

is positive, so its antiderivative must climb:

-\ln(\cos x)

does,

\ln(\cos x)

falls below zero. The sign error is visible without a single computation.

Why: The derivative of $\cos x$ is $-\sin x$ , and the minus sign goes into the answer. Quick test: $\tan x > 0$ on $\left(0, \frac{\pi}{2}\right)$ , so its antiderivative must INCREASE there; $\ln(\cos x)$ decreases.

6. Dropping the absolute value, then calling the integral undefined

2 marks, the whole value

What not to write

“ $\int_{-1}^0\frac{2x}{x^2 - 4}\,dx = \left[\ln(x^2 - 4)\right]_{-1}^0$ , and $\ln(-4)$ does not exist.”

What to write

“ $\left[\ln|x^2 - 4|\right]_{-1}^0 = \ln 4 - \ln 3 = \ln\frac{4}{3}$ .”

Why: $\ln|u|$ is an antiderivative of $\frac{1}{u}$ where $u < 0$ too. The integrand is continuous on $[-1, 0]$ , so the integral exists. Drop the bars only after proving the argument positive, as for $x^2 + 3x + 5$ .

7. Quoting symmetry on an interval that is not symmetric

the whole question

What not to write

“ $x^3$ is odd, so $\int_{-1}^2 x^3\,dx = 0$ .”

What to write

“Only $[-1, 1]$ is symmetric: $\int_{-1}^2 x^3\,dx = 0 + \int_1^2 x^3\,dx = 4 - \frac{1}{4} = \frac{15}{4}$ .”

Why: The theorem has two hypotheses, an interval $[-a, a]$ and continuity on it. On $[-1, 1]$ , $\frac{1}{x^3}$ fails the second: it is not a zero integral but an improper one.

8. Forgetting the factor of a linear substitution

1 to 2 marks

What not to write

“ $\int\frac{dx}{x^2 + 9} = \frac{1}{9}\arctan x + C$ .”

What to write

“ $x^2 + 9 = 9\left(1 + \left(\frac{x}{3}\right)^2\right)$ ; with $u = \frac{x}{3}$ , $dx = 3\,du$ : $\int\frac{dx}{x^2 + 9} = \frac{1}{3}\arctan\frac{x}{3} + C$ .”

Why: $\frac{1}{9}\cdot\frac{1}{1 + x^2} = \frac{1}{9 + 9x^2}$ , which is not $\frac{1}{x^2 + 9}$ . Factoring the $9$ leaves $\left(\frac{x}{3}\right)^2$ , not $x^2$ , and $dx = 3\,du$ turns $\frac{1}{9}$ into $\frac{1}{3}$ .

Which method to choose

Which u, decided by the FORM of the integrand

Look at the integrand and at the bounds before choosing anything

\frac{\sin x}{1 + x^2}

is odd: on

[-1, 1]

the two shaded lobes are mirror images of opposite sign, so the integral is

0

even though no antiderivative can be written.

If the interval is symmetric, [minus a, a], and the integrand is continuous on it → check parity first: an odd integrand gives 0 with no computation, an even one gives twice the integral over [0, a]

Example: $\int_{-1}^1\frac{\sin x}{1 + x^2}\,dx = 0$
If a function sits inside a power, a root, an exponential or a trig function, and its derivative is a factor up to a constant → u = the inside function; supply the constant

Example: $x^2(x^3 + 1)^5$ : $u = x^3 + 1$ , $x^2\,dx = \frac{1}{3}du$
If a fraction whose numerator is the derivative of its denominator, up to a constant → u = the denominator; the answer is $\ln|u|$

Example: $\frac{2x + 3}{x^2 + 3x + 5}$ gives $\ln(x^2 + 3x + 5) + C$
If $x$ on top, and $x^4$ in $1 + x^4$ or in $\sqrt{a^2 - x^4}$ → $u = x^2$ , then the table: inverse tangent or inverse sine

Example: $\frac{x}{1 + x^4}$ gives $\frac{1}{2}\arctan(x^2) + C$
If the inside function is found, but an $x$ is left over after $du$ → solve $u = g(x)$ for $x$ (or for $x^2$ ) and replace it

Example: $x^3\sqrt{x^2 + 1}$ : $x^2 = u - 1$ , $x\,dx = \frac{1}{2}du$
If a constant $a^2$ added to $x^2$ , or a function of $ax + b$ → linear substitution $u = \frac{x}{a}$ or $u = ax + b$ , and keep its factor

Example: $\frac{1}{x^2 + 9}$ gives $\frac{1}{3}\arctan\frac{x}{3} + C$
If no inside function has its derivative present, and nothing can be rewritten → substitution is not the tool: expand, use the table, or wait for the later techniques

Example: $(x^3 + 1)^5$ : expand into six terms

The test that decides every branch is the same: which function, differentiated, gives the other factor up to a constant? The same denominator $1 + x^2$ gives a logarithm with $x$ on top and an inverse tangent with $1$ on top.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing a definite substitution the marker can follow

When to use it: Any integral where the substitution is not a one-line mental step, and every definite integral

1 Name $u$ and compute $du$ on its own line, with the constant solved for: $u = 1 + x^2$ , $du = 2x\,dx$ , $x\,dx = \frac{1}{2}du$ .
2 Write the bound line: when $x = a$ , $u = g(a)$ ; when $x = b$ , $u = g(b)$ . Keep them in that order, even if decreasing.
3 Rewrite EVERY piece of the integral in $u$ , then check that no $x$ is left: not in the integrand, not in $dx$ , not in the bounds. A leftover $x$ is rewritten through $u$ .
4 Integrate in $u$ and evaluate at the new bounds. Do not return to $x$ .
5 Check: sign of the integrand against the sign of the answer, and a rough size (width times a typical height).

Concluding sentence

“Let $u = 1 + x^2$ , $du = 2x\,dx$ ; when $x = 0$ , $u = 1$ and when $x = \sqrt 3$ , $u = 4$ . Then $\int_0^{\sqrt 3}x\sqrt{1 + x^2}\,dx = \frac{1}{2}\int_1^4 u^{1/2}\,du = \frac{1}{3}\left[u^{3/2}\right]_1^4 = \frac{7}{3}$ .”

The trap: Writing $\int x\sqrt{1 + x^2}\,dx = \int\sqrt u$ with no $du$ and no bound line: the chain of equalities is false as written, and markers treat it as such.

Marking: Typically 1 mark for u and du, 1 for the new bounds, 1 for the antiderivative, 1 for the value. The bound line is the one most often missing.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Differentiate the answer

For an indefinite integral, the chain rule applied to your answer must return the integrand exactly. An extra term means a variable was treated as a constant.

$\frac{d}{dx}\frac{(x^3 + 1)^6}{18} = \frac{6 \cdot 3x^2(x^3 + 1)^5}{18} = x^2(x^3 + 1)^5$ : consistent.
The sign of the integrand fixes the sign of the answer

A positive integrand on $[a, b]$ with $a < b$ gives a positive integral. This catches the reversed-bounds error and the sign of $\int\tan x\,dx$ in five seconds.

$\sin x\,e^{\cos x} > 0$ on $\left(0, \frac{\pi}{2}\right)$ , so the answer is $e - 1$ , never $1 - e$ .
Width times height brackets the value

The integral lies between (b minus a) times the smallest and times the largest value of the integrand. It catches a bound left in $x$ .

$x\,e^{x^2}$ on $[0, 2]$ reaches $2e^4$ ; on $[1, 2]$ alone it is at least $e$ . The value $\frac{e^2 - 1}{2} \approx 3.2$ is far too small; $\frac{e^4 - 1}{2} \approx 26.8$ is plausible.
Both routes give the same number

When time allows, redo the definite integral by substituting back to $x$ and using the original bounds.

$\left[\frac{1}{3}(1 + x^2)^{3/2}\right]_0^{\sqrt 3} = \frac{7}{3}$ , as with the bounds $1$ and $4$ in $u$ .

The typical problem, taken apart

A definite integral where x must be rewritten and the bounds move

Compute $\int_0^3\frac{x^3}{\sqrt{x^2 + 16}}\,dx$ exactly.

No calculator. Every step must be justified as on a MATH 141 final.

Step 1

The inside function is $x^2 + 16$ , with derivative $2x$ . Split $x^3\,dx = x^2\cdot x\,dx$ .

Why

One factor $x$ goes with $dx$ to build $du$ ; the other two form $x^2$ , which is a function of $u$ . Naming the split is the first method mark.

Step 2

$u = x^2 + 16$ , $du = 2x\,dx$ , $x\,dx = \frac{1}{2}du$ , $x^2 = u - 16$ . When $x = 0$ , $u = 16$ ; when $x = 3$ , $u = 25$ .

Why

Every piece in $x$ now has its translation: the leftover $x^2$ , the differential, and both bounds. Nothing in $x$ may survive.

Step 3

$\int_0^3\frac{x^3}{\sqrt{x^2 + 16}}\,dx = \frac{1}{2}\int_{16}^{25}\frac{u - 16}{\sqrt u}\,du = \frac{1}{2}\int_{16}^{25}\left(u^{1/2} - 16u^{-1/2}\right)du$ .

Why

Divide term by term before integrating: $\frac{u}{\sqrt u} = u^{1/2}$ and $\frac{16}{\sqrt u} = 16u^{-1/2}$ . The power rule applies to each.

Step 4

$= \frac{1}{2}\left[\frac{2}{3}u^{3/2} - 32u^{1/2}\right]_{16}^{25} = \frac{1}{2}\left[\left(\frac{250}{3} - 160\right) - \left(\frac{128}{3} - 128\right)\right] = \frac{1}{2}\cdot\frac{26}{3} = \frac{13}{3}$ .

Why

$25^{3/2} = 125$ and $16^{3/2} = 64$ : the numbers $16$ and $25$ were chosen as perfect squares, so the evaluation is done by hand. Evaluated at the new bounds, never at $0$ and $3$ .

Step 5

Check by route 1: $F(x) = \frac{1}{3}(x^2 + 16)^{3/2} - 16(x^2 + 16)^{1/2}$ gives $F(3) - F(0) = \left(\frac{125}{3} - 80\right) - \left(\frac{64}{3} - 64\right) = \frac{13}{3}$ . Size: the integrand rises from $0$ to $\frac{27}{5}$ , so $0 < \frac{13}{3} < 3 \cdot \frac{27}{5}$ .

Why

Two independent routes and a size bracket confirm the value. On an exam, the size check alone takes ten seconds and catches a bound left in $x$ .

The conclusion, written out

“ $\int_0^3\frac{x^3}{\sqrt{x^2 + 16}}\,dx = \frac{13}{3}$ , with $u = x^2 + 16$ , $x^2 = u - 16$ and bounds $16$ and $25$ .”

The classic mistake on this problem: Pulling the leftover $x^2$ out of the integral as a constant, or evaluating $\frac{1}{3}u^{3/2} - 16u^{1/2}$ at $u = 0$ and $u = 3$ : both leave an $x$ in the computation, the first in the integrand, the second in the bounds.

Learn by heart

• $u$ = the inside function whose derivative is a factor, up to a constant. Constants are supplied; variables never are.
• No $x$ left: not in the integrand, not in $dx$ , not in the bounds. A stray $x$ is rewritten through $u$ .
• Definite: new bounds $g(a)$ , $g(b)$ , kept in their order; or antiderivative in $x$ with the old bounds. Never half of each.
• $\int\frac{g'}{g}\,dx = \ln|g| + C$ . $\int\tan x\,dx = -\ln|\cos x| + C$ , $\int\cot x\,dx = \ln|\sin x| + C$ .
• $\int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\frac{x}{a} + C$ ; $\int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C$ .
• Symmetric $[-a, a]$ , continuous: odd gives $0$ , even gives $2\int_0^a$ . Even times odd is odd.
• $\int_0^a f(a - x)\,dx = \int_0^a f(x)\,dx$ .
• Always differentiate an indefinite answer, and check the sign of a definite one.

Frequently asked questions

How do I choose u for a substitution in calculus?

Look for a function sitting inside another one, a power, a root, an exponential, whose derivative also appears as a factor of the integrand, possibly up to a constant. That inside function is u. If the derivative is missing by a constant, supply the constant; if it is missing by a factor containing x, this u is the wrong choice.

Do I have to change the limits of integration when I substitute?

Either change them, or go back to x before evaluating, but never half of each. If the integral is written in u, the limits must be values of u: plug the old limits into u equals g of x. If you prefer to keep the old limits, substitute back to x first. An integral in u with limits in x gives a wrong number.

What happens when the new lower limit is bigger than the upper limit?

Nothing goes wrong: keep the limits in the order the substitution gives them. The minus sign that usually comes with du and the reversed order cancel each other. You may also swap the limits and drop the minus sign at the same time, but never do only one of the two, or the sign of the answer flips.

Why is the integral of tan x equal to minus ln of cos x?

Write tan x as sine over cosine and let u be cos x. Then du is minus sine of x dx, so the integral becomes minus the integral of du over u, which is minus ln of the absolute value of cos x, plus C. The same answer is often written as ln of the absolute value of sec x. The minus sign comes from the derivative of cosine.

What do I do if an x is left over after the substitution?

Solve the substitution for x and replace it. With u equals x minus 1, write x as u plus 1; with u equals x squared plus 1, write x squared as u minus 1. The integral in u must contain no x at all. Never pull the leftover x out of the integral as if it were a constant: only numbers can be moved through the integral sign.

Practise it

Corrected exercises: The substitution rule, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet The Fundamental Theorem of Calculus Next sheet Integration by parts

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Revision sheet: the substitution rule (MATH 141)

The essentials

The rule, and what du has to swallow

Definite integrals: the bounds travel through u

The rules in table form

Which u, and which antiderivative comes out

The mistakes that cost marks

1. Dividing by a variable to manufacture du

2. Keeping the bounds of x in an integral in u

3. Reordering reversed bounds and keeping the minus sign

4. Treating a stray x as a constant

5. Losing the minus sign in the integral of tan x

6. Dropping the absolute value, then calling the integral undefined

7. Quoting symmetry on an interval that is not symmetric

8. Forgetting the factor of a linear substitution

Which method to choose

Which u, decided by the FORM of the integrand

How the answer is expected to be written

Writing a definite substitution the marker can follow

Check before you hand in

The typical problem, taken apart

A definite integral where x must be rewritten and the bounds move

Learn by heart

Frequently asked questions

How do I choose u for a substitution in calculus?

Do I have to change the limits of integration when I substitute?

What happens when the new lower limit is bigger than the upper limit?

Why is the integral of tan x equal to minus ln of cos x?

What do I do if an x is left over after the substitution?

Corrected exercises: The substitution rule, MATH 141 at McGill

See also

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