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14Trig substitution, partial fractions, numerical and improper integrals
$\int x\sqrt{9-x^{2}}\,dx$ takes ten seconds: the $x$ standing in front is exactly what $u=9-x^{2}$ needs. Rub that single $x$ out and the survivor, $\int \sqrt{9-x^{2}}\,dx$, stops every rule we own. It is not that the answer is hidden — the graph of $y=\sqrt{9-x^{2}}$ is a quarter circle of radius $3$ over $[0,3]$, so the number is $9\pi/4$ and we can read it off the picture. We simply have no machinery that produces it.
By the end of this section you can take an integral you have never seen, name in under a minute which of four tools it wants, carry that tool out, and check your own answer without an answer key.
In 60 seconds
Three shapes, three tools: a root of a quadratic wants a triangle, a ratio of polynomials wants a template of simpler fractions, and an integrand with no elementary antiderivative wants either a table entry or a handful of samples.
an endpoint is infinite, or the integrand blows up at one
Three most common mistakes
Stopping at $\theta$. $\tfrac12\theta+\tfrac12\sin\theta\cos\theta$ answers a question nobody asked; the question was in $x$, and the is what finishes the job.
Putting a single constant over an . Above $x^{2}+4$ the numerator is $Bx+C$ — two unknowns, always, no matter how simple the fraction looks.
Writing $\bigl[F(x)\bigr]_{1}^{\infty}$. The upper endpoint of that bracket is a limit to be taken, not a number to be substituted.
Midterm 1, Midterm 2 and the Final carry 28 percent each; quizzes 10 and homework 6. The weights say nothing about which of the four tools a given question wants, and that decision is made in the first ten seconds — so practise the choosing, not only the executing.
How much time do you have?
10 minutes
You leave able to spot which of the three structures you are looking at and to run the two that appear most often.
In 60 seconds card, Trading a root for a triangle, Splitting a ratio of polynomials into pieces you already know, Formula card
45 minutes
Add the two rewrites that make the templates apply at all, plus the ordered strategy — enough for a full exam question of this type.
everything in the 10 minute path, When the quadratic is not a template yet, Choosing the technique before you start writing, Full exam-style question, Practice C
full read
The parts that transfer: reading an entry out of a table without being fooled, what to do when no antiderivative exists, and the interleaved set where the type of the question is hidden from you.
all blocks in order, Reading an answer out of a table, When there is no antiderivative: sampling instead of solving, Integrals that run to infinity, Scaffolding comes off, Practice A to D, Mistake ledger
By the end of this section
Choose the substitution a root of a quadratic asks for, carry it out, and convert the answer back to $x$ with a reference triangle.
Rewrite a quadratic with a linear term by , so that one of the three templates applies to it.
Decompose a into partial fractions and integrate each piece.
Decide, before writing anything, which technique is cheapest for a given integrand and say why the others cost more.
Match an integrand to a table entry by a linear substitution, and check the result by differentiating it.
Approximate a definite integral with the midpoint, trapezoidal or Simpson rule and bound the error you have committed.
Settle whether an improper integral converges, by evaluating a limit or by naming a comparator whose verdict is already known.
Syllabus coverage
7.3
Trigonometric substitution
The three templates, the reference triangle that converts the answer back, and the rewrite that has to happen first when the quadratic carries a linear term.
covered
7.4
Integration of rational functions by partial fractions
Properness check and long division, the four template cases, solving for the unknowns, and the standard antiderivative of every piece.
covered
7.5
Strategy for integration
The ordered list of questions to ask an unfamiliar integrand, and the honest last branch: some integrands have no elementary antiderivative at all.
covered
7.6
Integration using tables and technology, together with the numerical rules that take over when no entry fits
Matching an integrand to a table entry through a linear substitution is the first half; the midpoint, trapezoidal and Simpson rules, taught in the block on sampling, are the second half.
covered
improper integrals
Improper integrals, revisited where the techniques of this week are what make them computable at all
The syllabus lists improper integrals with the previous week's sections; they come back here because a or a completed square is usually the missing antiderivative in the limit.
covered
Recall first
Substitution
$\int f(g(x))\,g'(x)\,dx=\int f(u)\,du$ with $u=g(x)$.
Half of the decisions in this section are the single question: is a $du$ already sitting in the numerator? If it is, none of the heavy machinery below is needed.
One per template. The identity, not the algebra, is what removes the root.
Two standard antiderivatives
$\int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\arctan\frac{u}{a}+C$ and $\int\frac{du}{\sqrt{a^{2}-u^{2}}}=\arcsin\frac{u}{a}+C$.
After a decomposition or a completed square, every surviving piece is one of these two or a logarithm. Note the $\frac{1}{a}$ in the first and its absence in the second.
The tangent substitution lands on it again and again; without it, half of the $\sqrt{a^{2}+x^{2}}$ problems stall one line from the end.
An improper integral is a limit
$\int_{a}^{\infty}f=\lim_{t\to\infty}\int_{a}^{t}f$, and $\int_{a}^{b}f=\lim_{t\to b^{-}}\int_{a}^{t}f$ when $f$ blows up at $b$.
The definition was set up in the previous section. What was missing there was a way to compute the inner integral; that is what this section supplies.
Polynomial long division
For polynomials $P$ and $Q$ there are unique $S$ and $R$ with $\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)}$ and $\deg R<\deg Q$.
A partial fraction template is only valid on the remainder part, so the division has to happen first whenever the numerator degree is not lower.
Try it yourself first (3 questions)
1§14.1 — a root is not automatically the variable●○○○○
Before any new machinery, one habit has to be checked. Decide true or false and be ready to defend it with a number, not with a rule.
Given
Claim: $\sqrt{x^{2}}=x$ for every real number $x$.
Find
True or false, with a reason or a counterexample.
Hint 1/4
Do not reason about it in general. Try to break it with one number.
Hint 2/4
$\sqrt{\;\cdot\;}$ always returns the non-negative root, so the output can never be negative.
Hint 3/4
Take $x=-3$: the claim says $\sqrt{(-3)^{2}}=-3$, while $\sqrt{9}=3$.
Hint 4/4
One number breaks it, so the claim is false and the correct identity is $\sqrt{x^{2}}=\lvert x\rvert$.
Show solutionTest the claim
$\sqrt{(-3)^{2}}=\sqrt{9}=3$
the root symbol returns the non-negative root
$3\neq -3$
so the claimed equality fails at this one number
$\sqrt{x^{2}}=\lvert x\rvert$
the identity that is true for every real $x$
Answer $$\text{false}$$
Check
Check the corrected identity on both signs: at $x=-3$ it gives $\lvert -3\rvert=3$ and at $x=3$ it gives $3$, matching $\sqrt{9}=3$ in both cases.
Every time a root of a square appears in this section, the bars are the default and dropping them has to be earned.
2§14.3 — the two standard forms that keep appearing●●○○○
The pieces produced by every decomposition in this section are logarithms, arctangents and powers. This one checks that the arctangent form is stored with its constant attached.
Given
$\displaystyle\int\frac{dx}{x^{2}+9}$
Find
Which antiderivative is correct?
Hint 1/4
You are not being asked to invent anything: this is one of the two standard forms, and the only question is what the constant out front is.
Differentiate: $\frac{1}{3}\cdot\frac{1/3}{1+x^{2}/9}=\frac{1}{3}\cdot\frac{3}{9+x^{2}}=\frac{1}{x^{2}+9}$, which is the integrand.
3§14.3 — what has to happen before a template is written●●○○○
A trap that costs marks every year: the template for partial fractions is only valid for certain rational functions, and this one is not among them yet.
Given
$\displaystyle\int\frac{x^{3}+1}{x^{2}-4}\,dx$
Find
What is the correct first move?
Hint 1/4
Compare the two degrees before you look at anything else.
Hint 2/4
A partial fraction template requires the numerator degree to be strictly lower than the denominator degree.
Hint 3/4
Here the numerator has degree $3$ and the denominator degree $2$, so the requirement fails.
Hint 4/4
Divide first: $\frac{x^{3}+1}{x^{2}-4}=x+\frac{4x+1}{x^{2}-4}$, and the template goes on the remainder.
Show solutionCompare degrees
$\deg(x^{3}+1)=3\;\ge\;\deg(x^{2}-4)=2$
the template is not licensed while this holds
Divide
$x^{3}+1=(x^{2}-4)\cdot x+(4x+1)$
one step of long division
$\frac{x^{3}+1}{x^{2}-4}=x+\frac{4x+1}{x^{2}-4}$
the remainder now has degree $1$, lower than $2$
Answer $$x+\frac{4x+1}{x^{2}-4}$$
Check
Multiply back: $(x^{2}-4)x+4x+1=x^{3}-4x+4x+1=x^{3}+1$, the original numerator.
If the degrees do not fall the right way, nothing else you do to the fraction is legal yet.
Notation
symbol
reads as
means
watch out
$\theta$
theta
the new variable a trigonometric substitution introduces; $x$ is written in terms of it, not the other way round
the final answer may not contain $\theta$
$a$
a
the positive constant inside the root, so $a=\sqrt{9}=3$ in $\sqrt{9-x^{2}}$
$a$ is the square root of the constant, not the constant
$u$
u
the shifted variable after completing the square, usually $u=x+\tfrac{b}{2}$
a definite integral in $u$ needs shifted limits as well
$\Delta x=\frac{b-a}{n}$
delta x equals b minus a over n
the common width of the $n$ subintervals of $[a,b]$
$n$ counts subintervals, so there are $n+1$ sample points
$x_{i}=a+i\,\Delta x$
x sub i
the $i$-th sample point, $i=0,1,\ldots,n$
the midpoint rule does not use these points but the midpoints between them
$M_{n},\;T_{n},\;S_{n}$
M sub n, T sub n, S sub n
the midpoint, trapezoidal and Simpson approximations with $n$ subintervals
$S_{n}$ exists only for even $n$
$K,\;L$
K and L
any numbers with $\lvert f''\rvert\le K$ and $\lvert f^{(4)}\rvert\le L$ on $[a,b]$
they are bounds, not maxima; a larger safe number is allowed and only weakens the estimate
$t$
t
the temporary endpoint that replaces the bad one in an improper integral
the answer is the limit as $t$ moves, not the value at any $t$
Conventions used here
Roots and signs
$\sqrt{u^{2}}=\lvert u\rvert$, never $u$. Each substitution below comes with a range of $\theta$ chosen so that the trig factor it produces is non-negative on that range; that is the licence — and the only licence — for writing $\sqrt{a^{2}\cos^{2}\theta}=a\cos\theta$ without bars.
A dropped absolute value is the single most common way a correct method produces a wrong sign.
Constants of integration
Every indefinite integral ends in $+C$. When a step produces a fixed constant, such as the $-\ln 2$ that falls out of a reference triangle, we absorb it into $C$ and say so in the line where it happens.
Silently dropping a constant and silently absorbing it look identical on paper; only one of them is legitimate.
Logarithms
$\int \frac{du}{u}=\ln\lvert u\rvert+C$ carries bars. Over an irreducible quadratic the argument $x^{2}+bx+c$ is positive for every real $x$, so $\ln(x^{2}+bx+c)$ needs none.
Bars written everywhere look careful but hide whether you know why they are there.
Angles
Every trigonometric function here takes radians, and the ranges of $\theta$ are quoted in radians.
The derivative and antiderivative formulas used throughout are false in degrees.
Reporting an improper integral
A verdict is either "converges to $L$" with the number $L$, or "diverges". We write "diverges to $\infty$" when the limit runs off in one direction, because it carries more information than "the limit does not exist" — but $\infty$ is never treated as a number.
Treating $\infty$ as a number is how a divergent integral acquires a value.
Reporting an approximation
A numerical answer is reported with three things: which rule, which $n$, and an . A decimal on its own is not an answer to "approximate".
Without a bound there is no way to know whether the digits you wrote down mean anything.
Trading a root for a triangle
Everything so far has removed a root by cancelling it against something the integrand already contained. Here there is nothing to cancel against.
Solvable with what we have
$\int x\sqrt{9-x^{2}}\,dx$ — the spare $x$ is $-\tfrac12\,du$ for $u=9-x^{2}$
$\int\frac{x\,dx}{x^{2}+4}$ — again the numerator is half a $du$
$\int\frac{dx}{x^{2}+4}$ — a standard arctangent form, no root involved
Not solvable yet
$\int\sqrt{9-x^{2}}\,dx$
$\int\frac{dx}{x^{2}\sqrt{x^{2}+9}}$
$\int\frac{dx}{\sqrt{x^{2}-4}}$
Push $u=9-x^{2}$ through the first one anyway. Then $du=-2x\,dx$, so $dx=-\frac{du}{2x}$ and the integral becomes $-\frac12\int\frac{\sqrt{u}}{x}\,du$ — with an $x$ still sitting in it, which is $\sqrt{9-u}$. Nothing cancelled.
Why it fails
A substitution built from the integrand can only remove a factor the integrand already has. This one has no spare $x$ to spend, so no such substitution exists. The move that works comes from outside: choose $x$ so that an identity, rather than a cancellation, kills the root.
RuleTrigonometric substitution
Conditions
$a>0$, and the integrand contains exactly one of the three roots below
$\theta$ is restricted to the range listed, which is what makes the trig factor non-negative and lets the absolute value bars go
the answer is converted back to $x$ before it counts as an answer
A constant minus a square asks for sine, a constant plus a square asks for tangent, and a square with the constant taken away asks for secant. In each case the matching Pythagorean identity turns the root into $a$ times a single trigonometric function, and $dx$ travels with the substitution.
The substitution drawn as a picture. With $x=a\sin\theta$ the side opposite $\theta$ is $\textcolor{#1f6feb}{x}$ and the hypotenuse is $\textcolor{#6f42c1}{a}$, which forces the third side to be $\textcolor{#d1690a}{\sqrt{a^{2}-x^{2}}}$ — the root itself. That is how an answer in $\theta$ turns back into an answer in $x$.
Looks like this, but is not
$\displaystyle\int\frac{x\,dx}{\sqrt{9-x^{2}}}$ carries the same root, so it looks like a job for $x=3\sin\theta$.
It is not. The numerator already carries an $x$, so $u=9-x^{2}$ finishes it in one line: $-\frac12\int u^{-1/2}\,du=-\sqrt{9-x^{2}}+C$. The trigonometric route reaches the same place three steps later. The root alone does not select the tool — the root together with the absence of a spare $x$ does.
An x² in the denominator: ∫ dx/(x²√(x²+9))
A root of a sum, and no $x$ anywhere to help. This is the tangent template.
Differentiate the answer instead of redoing the integral: $\frac{d}{dx}\left[\frac{\sqrt{x^{2}+9}}{x}\right]=\frac{x^{2}/\sqrt{x^{2}+9}-\sqrt{x^{2}+9}}{x^{2}}=\frac{-9}{x^{2}\sqrt{x^{2}+9}}$, so the derivative of $-\frac19$ times it is $\frac{1}{x^{2}\sqrt{x^{2}+9}}$ — the integrand.
Three conversions ($x$, $dx$, the root), one $u$-substitution inside, one triangle on the way out.
The triangle is not a memory aid; it is the substitution itself, drawn.
The integral from the opening: ∫ √(9 − x²) dx
The one that defeated us on the first line of this section.
Given
$\displaystyle\int\sqrt{9-x^{2}}\,dx$
Find
the antiderivative, and the area under the quarter circle as a check
SolutionConvert
$x=3\sin\theta,\quad dx=3\cos\theta\,d\theta$
a constant minus a square is the sine template, with $a=3$
Independent check against geometry: over $[-3,3]$ the graph is the upper half of a circle of radius $3$, so the area must be $\frac12\pi(3)^{2}=\frac{9\pi}{2}$. The antiderivative gives $\left(\frac92\cdot\frac{\pi}{2}+0\right)-\left(\frac92\cdot\left(-\frac{\pi}{2}\right)+0\right)=\frac{9\pi}{2}$. It matches.
The hook is closed: the missing $x$ in front of the root is exactly what $dx=3\cos\theta\,d\theta$ supplies.
Checkpoint
§14.1 — reading the template off the root●●○○○
Thirty seconds, no calculation. Only the choice of substitution is being asked for.
Do not start integrating. Ask only which of the three roots this is.
Hint 2/4
A square with a constant subtracted from it, $\sqrt{x^{2}-a^{2}}$, is the secant template.
Hint 3/4
Here $a^{2}=25$, so $a=5$ and the substitution is $x=5\sec\theta$.
Hint 4/4
The substitution is $x=5\sec\theta$.
Show solutionClassify the root
$\sqrt{x^{2}-a^{2}}\ \text{with}\ a^{2}=25$
the square comes first and the constant is taken away
$a=5$
the template constant is the square root of $25$
$x=5\sec\theta$
the third line of the rule
Answer $$x=5\sec\theta$$
Check
Sanity check on the identity: $25\sec^{2}\theta-25=25\tan^{2}\theta$, a perfect square, so the root does collapse. Trying $5\sin\theta$ instead would give $25\sin^{2}\theta-25=-25\cos^{2}\theta$, a negative number under a root.
⚠ Leaving the answer in θ
the trigonometric integral is the hard part, so finishing it feels like finishing the problem
Halve the coefficient of $x$, square it, add and subtract it: the quadratic becomes a perfect square plus whatever is left over, and shifting the variable by that half turns it into one of the three templates.
Looks like this, but is not
$\displaystyle\int\frac{dx}{\sqrt{3-2x-x^{2}}}$ has a minus sign in front of $x^{2}$, so it looks like the secant case $\sqrt{x^{2}-a^{2}}$.
Complete the square first and the shape changes: $3-2x-x^{2}=4-(x+1)^{2}$, which is a constant minus a square, so this is the sine template with $a=2$ and $u=x+1$. Read the shape after the rewrite, never before.
A root that hides a template: ∫ dx/√(x² + 2x + 5)
There is no bare square here yet, so no template applies yet.
Given
$\displaystyle\int\frac{dx}{\sqrt{x^{2}+2x+5}}$
Find
the antiderivative in $x$
SolutionMake a square appear
$x^{2}+2x+5=(x+1)^{2}+4$
half of $2$ is $1$; $1^{2}=1$ is added inside the square and removed from the constant, $5-1=4$
$u=x+1,\quad du=dx$
the shift costs nothing: the differential is unchanged
Differentiate: with $R=\sqrt{x^{2}+2x+5}$ the derivative is $\frac{1+(x+1)/R}{x+1+R}=\frac{(R+x+1)/R}{x+1+R}=\frac{1}{R}$, which is the integrand.
Completing the square never solves the integral; it only makes the integral belong to a family you already know.
Checkpoint
§14.2 — which template after the rewrite●●●○○
Thirty seconds. Complete the square in your head first, then classify.
Given
$\displaystyle\int\frac{dx}{\sqrt{3-2x-x^{2}}}$
Find
After completing the square, which substitution applies?
Hint 1/4
Rewrite what is under the root before you classify it. The minus sign in front of $x^{2}$ is the whole trap.
Hint 2/4
Factor $-1$ out of the $x$ terms first: $-(x^{2}+2x)+3$, then complete the square inside the bracket.
Hint 3/4
$-(x^{2}+2x+1)+1+3=4-(x+1)^{2}$, so with $u=x+1$ the root is $\sqrt{4-u^{2}}$.
Hint 4/4
That is a constant minus a square with $a=2$, so $u=2\sin\theta$.
Show solutionRewrite
$3-2x-x^{2}=-(x^{2}+2x)+3$
pull the minus out of the $x$ terms so the square can be completed in the usual direction
$=-\left[(x+1)^{2}-1\right]+3=4-(x+1)^{2}$
half of $2$ is $1$, and the $-1$ inside the bracket comes back out as $+1$
Classify
$\sqrt{4-u^{2}},\quad u=x+1$
a constant minus a square
$u=2\sin\theta$
the sine template with $a=2$
Answer $$u=x+1,\quad u=2\sin\theta$$
Check
Domain check, which is independent of the algebra: $3-2x-x^{2}\ge0$ holds exactly for $-3\le x\le1$, an interval of length $4$ centred at $-1$ — precisely what $4-(x+1)^{2}\ge0$ describes.
⚠ Completing the square without balancing
the square is added inside and easy to forget to remove outside
wrong$x^{2}+2x+5=(x+1)^{2}+5$
right$x^{2}+2x+5=(x+1)^{2}+4$
⚠ Forgetting the minus sign in front of x²
the rewrite is done on autopilot, in the direction practised most often
wrong$3-2x-x^{2}=(x-1)^{2}+2$
right$3-2x-x^{2}=4-(x+1)^{2}$
⚠ Shifting the variable but not the limits
in an indefinite integral the shift really is free, and the habit carries over to definite ones
Each factor of the bottom contributes one fraction for each of its powers; a linear factor contributes a plain constant on top, an irreducible quadratic contributes a linear expression on top.
Where the unknowns come from. The factor $\textcolor{#1f6feb}{(x-1)}$ sends one unknown; $\textcolor{#d1690a}{(x+2)^{2}}$ sends two, one for each power. Counting slots before solving anything is what stops the most common error in this block.
Looks like this, but is not
$\dfrac{3x+1}{x^{2}+1}$ is a ratio of polynomials with the numerator of lower degree, so it looks like a candidate for a decomposition.
There is nothing to decompose: $x^{2}+1$ has $b^{2}-4c=-4<0$, so it does not factor over the reals and the template gives back the same fraction. The move here is different — split the numerator into the part that is a multiple of $2x$ and the constant left over: $\frac{3x}{x^{2}+1}+\frac{1}{x^{2}+1}$, a logarithm plus an arctangent.
One linear factor, one irreducible quadratic
The mixed case: a plain constant over one factor, a linear numerator over the other.
Independent numerical check at a single point. The integrand at $x=0$ is $\frac{1}{(-1)(1)}=-1$. The derivative of the answer at $x=0$ is $\frac{2}{-1}-\frac{0}{1}+\frac{1}{1}=-1$. They agree.
Three unknowns, one root plugged in, two coefficients matched, three standard antiderivatives.
Split the quadratic piece into its $2x$ part and its constant part before integrating: the first is a logarithm, the second an arctangent, and they never mix.
A repeated linear factor: three slots, not two
The factor $(x+1)^{2}$ contributes two fractions on its own.
Point check at $x=0$: the integrand is $\frac{3}{(-1)(1)}=-3$, and the derivative of the answer is $\frac{3/2}{-1}-\frac{1/2}{1}-\frac{1}{1}=-\tfrac32-\tfrac12-1=-3$.
Only the first power of a repeated factor produces a logarithm; every higher power produces a rational function.
Checkpoint
§14.3 — counting the slots before solving●●●○○
Thirty seconds, no unknowns solved. Only the shape of the template is asked for.
Given
$\displaystyle\frac{2x+5}{(x+3)^{2}(x^{2}+4)}$
Find
Which template is the correct one?
Hint 1/4
Count the powers, do not solve for anything. Each power of each factor gets a slot.
Hint 2/4
A linear factor to the power $k$ gives $k$ slots with constant numerators; an irreducible quadratic gives a slot with a $Bx+C$ numerator for each of its powers.
Hint 3/4
Here the factors are $(x+3)^{2}$, a linear factor squared, and $x^{2}+4$, irreducible and to the first power only.
Hint 4/4
So the template is $\frac{A}{x+3}+\frac{B}{(x+3)^{2}}+\frac{Cx+D}{x^{2}+4}$.
Count check, independent of the template rules: the denominator has degree $4$, so the decomposition must carry exactly $4$ unknowns. $A,B,C,D$ is four.
⚠ A single constant over an irreducible quadratic
the linear factors all take one constant, and the pattern is applied one factor too far
Two structural tools are now on the shelf next to substitution and parts, and the expensive mistake is no longer executing badly but choosing badly.
MethodStrategy for integration
Conditions
the questions are asked in this order, and the first yes wins
a yes does not forbid a second look: many integrands admit several routes, and the order is about cost, not legality
$$\boxed{\text{simplify}\ \to\ \text{is a }du\text{ present?}\ \to\ \text{classify the shape}\ \to\ \text{table entry}\ \to\ \text{samples}}$$
Clean the integrand up first, then look for a substitution that is already there, then classify by shape into root, ratio, product or trigonometric power, and only when all of that fails reach for a table or for numerical sampling.
The order matters more than the list. Each question is cheaper to answer than the one below it, so working top down means you never pay for a triangle when a substitution would have done.
Looks like this, but is not
$\displaystyle\int\frac{x\,dx}{\sqrt{x^{2}-16}}$ contains $\sqrt{x^{2}-a^{2}}$, which the classification step sends straight to $x=4\sec\theta$.
The classification step is not the first step. The question above it — is a $du$ already present? — is answered yes here, since $u=x^{2}-16$ has $du=2x\,dx$ and the numerator is $\tfrac12\,du$. One line: $\sqrt{x^{2}-16}+C$. Skipping the cheap question and landing on the expensive one is the most common way to lose time in an exam.
Triage: three integrals, three different answers to "which tool"
Nothing here is hard once the tool is chosen. The exercise is the choosing.
Check $I_{2}$ at $x=0$ without redoing it: the integrand is $\frac{1}{-4}=-\tfrac14$, and the derivative of the answer is $0+\frac{9/4}{-2}+\frac{7/4}{2}=-\tfrac98+\tfrac78=-\tfrac14$.
Two of the three took one line each once the right question was asked first; the third is not a failure of effort.
An integral that resists everything is not always a hard integral. Sometimes it is a definite integral in disguise, and the answer is a number rather than a formula.
Checkpoint
§14.4 — the cheapest route, not the first one recognised●●○○○
Thirty seconds. All four listed techniques would eventually work on this integrand; only one of them is cheap.
Given
$\displaystyle\int\frac{x\,dx}{x^{2}+9}$
Find
Which technique gets there fastest?
Hint 1/4
Before classifying the shape, ask the cheaper question: is the numerator related to the derivative of the denominator?
Hint 2/4
If $u$ is the denominator and the numerator is a constant multiple of $du$, the integral is a logarithm and nothing else is needed.
Hint 3/4
Here $u=x^{2}+9$ gives $du=2x\,dx$, and the numerator $x\,dx$ is $\tfrac12\,du$.
Hint 4/4
So it is a substitution: the answer is $\tfrac12\ln(x^{2}+9)+C$.
Contrast as the check: remove the $x$ and the same denominator gives $\frac13\arctan\frac{x}{3}+C$ instead. One factor changed the entire answer, which is what makes the cheap question worth asking first.
⚠ Classifying the shape before looking for a du
the new tools are the memorable ones, so the eye jumps to the root or the ratio and skips the cheap test
A representative entry: an integrand of this exact shape, in whatever variable, has this antiderivative, with the same letter put everywhere the entry writes $u$ and the same constant everywhere it writes $a$.
Looks like this, but is not
A table gives $\int\frac{dx}{x^{2}-1}=\frac12\ln\left\lvert\frac{x-1}{x+1}\right\rvert+C$, while your own partial fractions gave $\frac12\ln\lvert x-1\rvert-\frac12\ln\lvert x+1\rvert+C$. Two different answers, so one of them looks wrong.
Neither is wrong: $\ln A-\ln B=\ln\frac{A}{B}$, so the two expressions are the same function. Antiderivatives are only determined up to shape and up to $+C$, and a printed answer that differs from yours is an invitation to check, not evidence of an error. Differentiating both is the fastest way to settle it.
Bending an integrand to fit an entry: ∫ x² dx / √(5 − 4x²)
The entry is in $u$ with a bare $u^{2}$ under the root; the integrand has $4x^{2}$.
Differentiate the answer. The first term gives $-\frac18\sqrt{5-4x^{2}}+\frac{x^{2}}{2\sqrt{5-4x^{2}}}$ and the second gives $\frac{5}{8\sqrt{5-4x^{2}}}$; over the common denominator the constants cancel, $-\frac{5-4x^{2}}{8}+\frac{x^{2}}{2}+\frac58=x^{2}$, leaving $\frac{x^{2}}{\sqrt{5-4x^{2}}}$.
One substitution, one copied line, one back-substitution — and the check took longer than the lookup.
A table entry is a template, exactly like a partial fraction template: it is matched, not solved.
Checkpoint
§14.5 — what else has to change when the variable does●●●○○
Thirty seconds. A student rewrites an integrand to match a table entry written in $u$, and changes only the visible occurrences of the variable.
Join the sample points with straight lines and add the trapezoids; or use the height at the middle of each strip and add the rectangles; or fit a parabola through every consecutive three points, which is where the weights one, four, two, four, one come from.
Why the overshoots here. On $[1,3]$ the graph of $\textcolor{#1f6feb}{y=1/x}$ bends upwards, so every $\textcolor{#d1690a}{\text{chord}}$ lies above it and the strips carry more area than the region does — with two strips, $0.0681$ too much.
Looks like this, but is not
More sample points always means a better answer, so $T_{100}$ must beat $S_{4}$.
Not necessarily. For $\int_{1}^{3}\frac{dx}{x}$ the trapezoidal error falls like $n^{-2}$ and Simpson's like $n^{-4}$: $T_{100}$ is off by about $3\times10^{-5}$ while $S_{4}$, on five function values instead of a hundred and one, is off by about $1.4\times10^{-3}$ — and $S_{20}$ already beats $T_{100}$ with a fifth of the work. Which rule you choose matters as much as how many points you spend.
rule, $n=4$
value
error
what the sign says
$T_{4}$
$1.1166667$
$+0.0180544$
chords lie above a curve that bends up
$M_{4}$
$1.0897547$
$-0.0088576$
the midpoint rectangle undershoots, and by about half as much
$S_{4}$
$1.1000000$
$+0.0013877$
parabolas follow the bend, so the error is an order smaller
$T_{4}$ and $S_{4}$ use exactly the same five heights — only the weights differ — and Simpson's is about thirteen times more accurate for it. That is the whole argument for learning the weights $1,4,2,4,1$.
Simpson with four strips on ∫₀¹ e^(−x²) dx, with an error bound
The standard example of an integrand with no elementary antiderivative.
Given
$f(x)=e^{-x^{2}}$ on $[0,1]$, $n=4$
you may use $\lvert f^{(4)}(x)\rvert\le 12$ on $[0,1]$
Find
the approximation $S_{4}$ and a bound on its error
Independent check of the size, without recomputing the sum: $e^{-x^{2}}$ falls from $1$ to $0.3679$ across $[0,1]$ and is concave near the right end, so the integral must sit between the two crude bounds $0.3679$ and $1$, and nearer the top than the bottom. $0.747$ does. The true value is $0.7468241$, so the actual error is $3.1\times10^{-5}$ — comfortably inside the bound, as a bound should be.
A bound is a promise about the worst case, not a prediction: the real error here is eight times smaller than the guarantee.
Checkpoint
§14.6 — the one restriction Simpson's rule carries●●○○○
Thirty seconds. A calculation is set up with five subintervals and the weights are being written out.
Given
$\displaystyle\int_{0}^{1}e^{-x^{2}}\,dx$ with $n=5$
Find
What is wrong, and what should be done?
Hint 1/4
Look at the number of subintervals against the way the parabolas are fitted.
Hint 2/4
Simpson's rule fits one parabola to every consecutive three sample points, which uses up two subintervals at a time — so $n$ has to be even.
Hint 3/4
With $n=5$ the last parabola would have only one subinterval left to sit on, and the weight pattern $1,4,2,4,\ldots,4,1$ does not close.
Hint 4/4
Use an even $n$ — $n=4$ or $n=6$ — or switch to the trapezoidal or midpoint rule, which accept any $n$.
a parabola is determined by three points, and three consecutive sample points span two subintervals
$n=5\ \text{is odd}$
so the subintervals cannot be paired off and one is left over
$\text{use}\ n=4\ \text{or}\ n=6$
or a rule that works one subinterval at a time
Answer $$n\ \text{must be even}$$
Check
Check the weight pattern instead of the geometry: $1,4,2,4,\ldots,4,1$ must start and end with $1$ and alternate $4,2$ in between. With five subintervals the pattern would read $1,4,2,4,2,1$, ending on the wrong weight.
⚠ Running Simpson's rule with an odd n
the formula can be written down for any $n$, and nothing in the arithmetic complains
The definition of an improper integral was settled last time; what was missing was a way to compute the inner integral, and that is what the last three blocks have supplied.
MethodThe two step protocol
Conditions
name why the integral is improper, and at which endpoint, before anything else
if both endpoints are bad, split at any convenient interior point and require both halves to converge
a limit that runs to $\pm\infty$, or fails to settle, means divergence
Replace the bad endpoint with a letter, do an ordinary definite integral with the techniques of this section, and only then let the letter move to where it was not allowed to be.
Two tails that look alike and behave differently. $\textcolor{#d1690a}{1/x}$ and $\textcolor{#1f6feb}{1/x^{2}}$ both flatten towards the axis, but only the second encloses a finite total area — which is why a convergence verdict can never be read off the shape of a graph.
Looks like this, but is not
$\displaystyle\int_{0}^{1}\frac{dx}{x^{1/2}}$ has an integrand that blows up at $0$, and $\int_{1}^{\infty}\frac{dx}{x^{1/2}}$ diverges, so this one should diverge too.
It converges, to $2$. The two $p$ conditions point in opposite directions: $\int_{1}^{\infty}x^{-p}\,dx$ converges when $p>1$, while $\int_{0}^{1}x^{-p}\,dx$ converges when $p<1$. Near infinity a large $p$ means fast decay; near zero a small $p$ means a mild blow-up. Same family, opposite danger, opposite condition.
A partial fraction that makes an infinite tail computable
The decomposition is what turns this into a limit anyone can take.
the two logarithms are combined before the limit is taken, which is the whole trick: separately they would each run to infinity
$=\ln\frac{t}{t+1}-\ln\frac12$
evaluating at both ends
Let t move
$\lim_{t\to\infty}\ln\frac{t}{t+1}=\ln 1=0$
the ratio tends to $1$ and the logarithm is continuous there
$\text{value}=0+\ln 2$
so the tail converges
Answer $$\ln 2$$
Check
An independent upper bound: for $x\ge1$ we have $x+1>x$, so $\frac{1}{x(x+1)}<\frac{1}{x^{2}}$, and $\int_{1}^{\infty}x^{-2}\,dx=1$. The answer must therefore be a positive number below $1$, and $\ln 2\approx0.693$ is.
One decomposition, one combination of logarithms, one limit.
Combine the logarithms before taking the limit. Two divergent pieces can hide a convergent difference, and splitting them destroys the information.
A verdict without an antiderivative
Nothing here integrates in closed form, and nothing needs to.
Answer $$\text{converges, and its value is at most }2$$
Check
Sanity check on the size: the integrand at $x=2$ is $1/\sqrt7\approx0.378$ and it decays faster than $x^{-3/2}$, so a total of at most $2$ is the right order. A verdict that had come out larger than the dominating integral would have been self-contradictory.
A comparison proves convergence but does not produce a value; when the question says decide, that is enough, and when it says evaluate, it is not.
Checkpoint
§14.7 — which p condition applies at which end●●○○○
Thirty seconds. The singularity is at the left endpoint, not at infinity.
Given
$\displaystyle\int_{0}^{1}\frac{dx}{x^{2/3}}$
Find
Does it converge, and if so to what?
Hint 1/4
Ask first where the trouble is. That decides which of the two $p$ conditions you are allowed to use.
Hint 2/4
$\int_{0}^{1}x^{-p}\,dx$ converges exactly when $p<1$; the condition at infinity is the opposite one.
Hint 3/4
Here $p=\tfrac23<1$, so it converges, and $\int_{t}^{1}x^{-2/3}\,dx=\bigl[3x^{1/3}\bigr]_{t}^{1}=3-3t^{1/3}$.
Hint 4/4
Letting $t\to0^{+}$ gives $3$.
Show solutionLocate the trouble and cut it off
$x^{-2/3}\to\infty\ \text{as}\ x\to0^{+}$
the integrand, not the interval, is what is unbounded here
the power rule with exponent $-\tfrac23+1=\tfrac13$
$\lim_{t\to0^{+}}\left(3-3t^{1/3}\right)=3$
the cube root is continuous at $0$
Answer $$3$$
Check
Cross-check with the rule rather than the calculation: $p=\tfrac23<1$, and the condition for convergence at $0$ is exactly $p<1$. The two agree, and $\int_{1}^{\infty}x^{-2/3}\,dx$ with the same $p$ diverges — same function, other end, opposite verdict.
⚠ Substituting infinity into a bracket
the bracket notation makes both ends look like numbers to be plugged in
right$\int_{-1}^{1}\frac{dx}{x^{2}}=\int_{-1}^{0}+\int_{0}^{1}\ \text{and both halves diverge}$
Running a trigonometric substitution
a root of a quadratic is present and the numerator has no spare factor to be absorbed into $du$
Name the pattern
Write the expression under the root as $a^{2}-x^{2}$, $a^{2}+x^{2}$ or $x^{2}-a^{2}$, completing the square first if there is a linear term. $a$ is the square root of the constant.
Substitute three things
Replace $x$, replace $dx$, and simplify the root by the matching identity. Leaving $dx$ behind is the most common single error.
Do the trigonometric integral
What is left is a trigonometric integral of the kind handled in the previous section: peel a factor, use an identity, or recognise a standard form such as $\int\sec\theta\,d\theta$.
Draw the triangle
The substitution itself gives the triangle: $x=a\sin\theta$ means opposite $x$, hypotenuse $a$. Read every trigonometric function of $\theta$ off it.
Convert and check
Rewrite the answer in $x$, add $+C$, and differentiate it once to confirm you get the integrand back.
Where it goes wrong
stopping while the answer still contains $\theta$
converting $x$ but not $dx$
dropping absolute value bars without naming the range of $\theta$
reaching for a triangle when a plain substitution was available
Integrating the four kinds of partial fraction piece
after a decomposition, when each fraction has to be turned into an antiderivative
Plain linear
$\displaystyle\int\frac{A}{x-a}\,dx=A\ln\lvert x-a\rvert+C$. Bars, always: the argument changes sign across $a$.
Repeated linear
$\displaystyle\int\frac{A}{(x-a)^{k}}\,dx=\frac{-A}{(k-1)(x-a)^{k-1}}+C$ for $k\ge2$. A power, not a logarithm.
Irreducible quadratic — split the numerator
$\displaystyle\frac{Bx+C}{x^{2}+a^{2}}$ splits into the part that is a multiple of the derivative $2x$, giving $\frac{B}{2}\ln(x^{2}+a^{2})$, and the constant left over, giving $\frac{C}{a}\arctan\frac{x}{a}$.
Irreducible quadratic with a linear term
Complete the square first, shift with $u=x+\frac{b}{2}$, then use step 3 on the shifted expression.
Where it goes wrong
a logarithm written for a repeated factor's higher power
the $\frac{1}{a}$ dropped from the arctangent
bars written around $x^{2}+a^{2}$, which is always positive
Deciding an improper integral
an endpoint is infinite, or the integrand blows up somewhere on the closed interval
Locate the trouble
Say out loud which endpoint is bad and why. If a blow-up sits strictly inside the interval, split there first; both halves have to converge.
Try for an antiderivative
Replace the bad endpoint by $t$ and integrate with the techniques of this section. Combine logarithms before taking the limit.
If no antiderivative comes, compare
Find $g$ with $0\le f\le g$ and a known verdict for $\int g$; a convergent $g$ drags $f$ with it. In the other direction, a divergent minorant forces divergence. The comparator is usually a $p$ integral.
State the verdict properly
Either "converges to $L$" with the number, or "diverges". A comparison gives the first word only, never the number.
Where it goes wrong
evaluating a bracket at $\infty$
using the $p>1$ condition at a singularity at $0$
comparing with an approximation instead of a genuine inequality
answering "converges to $2$" when only a bound of $2$ was proved
A sign that lets the denominator factor: ∫ dx/(x² − 4)
Minus four, so the bottom factors and the template applies.
The same two symbols with one sign changed: one integral is two logarithms, the other is a single arctangent, and no amount of algebra turns either into the other.
How to tell them apart
Compute $b^{2}-4c$ before writing anything. Negative means irreducible, which means arctangent and logarithm of the whole quadratic; non-negative means it factors, which means a partial fraction template.
With an x on top: ∫ x dx/√(x² − 16)
The numerator is half a $du$, so nothing else is needed.
Given
$\displaystyle\int\frac{x\,dx}{\sqrt{x^{2}-16}}$
Find
the antiderivative
SolutionSpot the du
$u=x^{2}-16,\quad du=2x\,dx$
the numerator is exactly $\tfrac12\,du$
$\tfrac12\int u^{-1/2}\,du=u^{1/2}+C$
power rule
Answer $$\sqrt{x^{2}-16}+C$$
Check
Differentiate: $\frac{x}{\sqrt{x^{2}-16}}$, the integrand.
Without it: ∫ dx/√(x² − 16)
The same root, no $x$ to spend, so the triangle is unavoidable.
Differentiate: with $R=\sqrt{x^{2}-16}$, $\frac{1+x/R}{x+R}=\frac{(R+x)/R}{x+R}=\frac1R$.
One factor of $x$ separates a one line substitution from a full trigonometric substitution with a triangle at the end.
How to tell them apart
Before classifying the root, ask whether the numerator is a constant multiple of the derivative of what is under it. If it is, the root never needs removing at all.
Scaffolding comes off
The common skeleton
Check properness: is the numerator degree lower than the denominator degree? If not, divide.
Factor the denominator completely over the real numbers.
Write one slot for every power of every factor, with $Bx+C$ on top of any irreducible quadratic.
Clear denominators, then plug in the roots of the linear factors before matching any coefficients.
Integrate slot by slot: logarithm, power, or logarithm plus arctangent.
Check by putting one convenient number into both the integrand and the derivative of your answer.
1 · fully worked
Two distinct linear factors, fully worked
Every step of the skeleton written out, with its reason.
Given
$\displaystyle\int\frac{5x-4}{x^{2}-x-2}\,dx$
Find
the antiderivative
SolutionProperness and factoring
$\deg(5x-4)=1<2=\deg(x^{2}-x-2)$
proper already, so no division
$x^{2}-x-2=(x-2)(x+1)$
two numbers multiplying to $-2$ and adding to $-1$
At $x=0$: the integrand is $\frac{-4}{-2}=2$, and the derivative of the answer is $\frac{2}{-2}+\frac{3}{1}=-1+3=2$.
Roots first, coefficient matching only for what the roots cannot reach.
2 · you write the reasoning
Same skeleton, easier numbers, and this time the reasons are yours to supply. Work out why each line is allowed before opening the model answers: $\displaystyle\int\frac{3\,dx}{x(x+3)}$.
reasoning
The denominator is already factored into two distinct linear pieces, and the numerator has degree $0<2$, so the template is legal with one constant per factor.
reasoning
Multiplying both sides by $x(x+3)$ clears every denominator at once and leaves an identity between polynomials, true for every $x$.
reasoning
Putting $x=0$ annihilates the $Bx$ term, so $A$ falls out on its own — cheaper than expanding and matching coefficients.
reasoning
Putting $x=-3$ annihilates the $A(x+3)$ term for the same reason, and gives $B$ in one line.
reasoning
Each slot is a plain linear factor, so each integrates to a logarithm with bars; the answer can also be written as $\ln\left\lvert\frac{x}{x+3}\right\rvert+C$.
3 · find the buried error
Harder than the last one: an irreducible quadratic joins in. Below is a student's full solution to $\displaystyle\int\frac{2x^{2}+x+4}{x(x^{2}+4)}\,dx$. The template and the unknowns are right. Two of the numbered lines are not.
the two buried errors (2)
⚠ step 4
the factor $\tfrac12$ is missing: the numerator is $\tfrac12\,du$, not $du$, since $du=2x\,dx$
the numerator is recognised as "the derivative of the denominator" and the constant that makes that statement exact is skipped
Same skeleton, no steps written out, one irreducible quadratic in the denominator. Hints are free and cost nothing.
Given
$\displaystyle\int\frac{4x}{(x-1)(x^{2}+1)}\,dx$
Find
Evaluate the integral, and check your answer at one value of $x$.
Hint 1/4
The denominator is already factored, and the numerator degree is lower than the denominator degree, so you can go straight to a template. Decide first how many unknowns it must carry.
Hint 2/4
One slot per power: $\frac{A}{x-1}+\frac{Bx+C}{x^{2}+1}$. Clear denominators, put $x=1$ first, then match the coefficient of $x^{2}$ and the constant.
Hint 3/4
With $4x=A(x^{2}+1)+(Bx+C)(x-1)$: at $x=1$, $4=2A$ so $A=2$; the $x^{2}$ coefficients give $A+B=0$; the constants give $A-C=0$.
Hint 4/4
$A=2$, $B=-2$, $C=2$, so the integral is $2\ln\lvert x-1\rvert-\ln(x^{2}+1)+2\arctan x+C$.
At $x=0$ the integrand is $\frac{0}{(-1)(1)}=0$, and the derivative of the answer is $\frac{2}{-1}-0+\frac{2}{1}=0$. A check at a point where the answer is $0$ is weak on its own, so check $x=2$ too: the integrand is $\frac{8}{(1)(5)}=1.6$, and the derivative is $\frac{2}{1}-\frac{4}{5}+\frac{2}{5}=2-0.8+0.4=1.6$.
Full exam-style question
Final-style question: four parts, four different decisionsexam format
Each part stands on its own and each one is worth the same. The marks are in naming the technique and finishing the conversion, not in heroic algebra.
Four independent checks, one per part. (a) differentiating $-\frac{\sqrt{4-x^{2}}}{4x}$ returns $\frac{1}{x^{2}\sqrt{4-x^{2}}}$. (b) at $x=0$ the integrand is $1$ and the derivative of the answer is $\tfrac12+0+\tfrac12=1$. (c) for $x\ge3$, $x^{2}-4\ge\tfrac59x^{2}$, so the value is at most $\tfrac95\int_{3}^{\infty}x^{-2}\,dx=0.6$, and $\tfrac14\ln5=0.402$ sits below it. (d) the true value is $0.7468241$, so the actual error is $3.6\times10^{-4}$, inside the bound.
Roughly twenty lines in total, and every one of the four decisions was made in the first line of its part.
Notice that (c) is a partial fraction problem wearing an improper integral's clothes. The improper part costs one extra line at the start and one at the end.
Practice
A · concept 3 questions
1§14.1 — a root does not by itself pick the tool●●○○○
One sentence of the kind that opens a quiz. Decide, and be ready to justify with a single integral rather than with a rule.
Given
Claim: "An integrand containing $\sqrt{x^{2}+a^{2}}$ can only be handled by a trigonometric substitution."
Find
True or false, with a counterexample if false.
Hint 1/4
Try to break the claim before believing it: can you build one integral with that root that falls to something cheaper?
Hint 2/4
A plain substitution works whenever the numerator is a constant multiple of the derivative of what is under the root.
Hint 3/4
Take $\int\frac{x\,dx}{\sqrt{x^{2}+9}}$: with $u=x^{2}+9$, $du=2x\,dx$, so the integral is $\sqrt{x^{2}+9}+C$ in one line.
Hint 4/4
One counterexample settles it: the claim is false.
Test it on one case rather than trusting the algebra: for $f(x)=x^{3}$ on $[0,1]$, $S_{2}=\frac{1/2}{3}\left[0+4(0.125)+1\right]=\frac{1.5}{6}=0.25$, and $\int_{0}^{1}x^{3}\,dx=\tfrac14$. Exact, as promised.
The same argument shows the trapezoidal rule is exact for straight lines, since there the second derivative vanishes.
B · computation 5 questions
1§14.1 — secant template, all the way back to x●●●○○
A trigonometric substitution with the third template. The marks are split between the conversion in and the conversion out.
Given
$\displaystyle\int\frac{dx}{x^{2}\sqrt{x^{2}-9}}$, for $x>3$
Find
(a) Name the substitution and write $dx$ and $\sqrt{x^{2}-9}$ in terms of $\theta$.
(b) Reduce the integrand and integrate it.
(c) Convert back to $x$ and state the antiderivative.
Hint 1/4
Classify the root first. Which of the three templates has the square first and the constant subtracted?
Hint 2/4
$\sqrt{x^{2}-a^{2}}$ goes with $x=a\sec\theta$, and then $\sec^{2}\theta-1=\tan^{2}\theta$ collapses the root.
Hint 3/4
With $a=3$: $x=3\sec\theta$, $dx=3\sec\theta\tan\theta\,d\theta$, $\sqrt{x^{2}-9}=3\tan\theta$, and $x^{2}=9\sec^{2}\theta$.
Hint 4/4
Everything collapses to $\frac19\int\cos\theta\,d\theta=\frac19\sin\theta+C=\frac{\sqrt{x^{2}-9}}{9x}+C$.
Show solutionConvert
$x=3\sec\theta,\quad 0\le\theta<\tfrac{\pi}{2}$
the root is $\sqrt{x^{2}-a^{2}}$ with $a=3$, and $x>3$ keeps $\theta$ in the first quadrant where $\tan\theta\ge0$
A second, independent check on the sign: $f(x)=\frac{1}{1+x^{2}}$ has $f''(0)=-2<0$, so the graph bends downwards over most of $[0,1]$ and chords lie below it — the trapezoidal value should therefore be an underestimate, and $0.7827941<0.7853982$ confirms it.
Multiplying $T_{4}$ by $4$ gives $3.1312$, an approximation of $\pi$ from five divisions and no calculator.
5§14.7 — an infinite tail after completing the square●●●○○
Two ideas from this section in one integral: a quadratic that has to be rewritten, and an endpoint that has to be replaced by a letter.
the lower endpoint contributes $\arctan1$, not $\arctan0$, because of the shift
Take the limit
$\lim_{t\to\infty}\arctan(t+1)=\frac{\pi}{2}$
the horizontal asymptote of the arctangent
$\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$
so the tail converges
Answer $$\frac{\pi}{4}$$
Check
Independent bound: for $x\ge1$, $x^{2}+2x+2>x^{2}$, so the tail beyond $1$ is less than $\int_{1}^{\infty}x^{-2}\,dx=1$, and on $[0,1]$ the integrand is at most $\tfrac12$. The total is therefore below $1.5$, and $\frac{\pi}{4}\approx0.785$ sits under it.
Shifting inside an arctangent moves the endpoint values too; that is where the $\frac{\pi}{4}$ came from, not from the upper limit.
C · exam level 3 questions
1§14.4 — an odd power on top changes the route●●●●○
Exam level, and the trap is the strategy rather than the algebra: the root looks like a trigonometric substitution, and it is not.
(c) Evaluate at the endpoints and give an exact answer.
Hint 1/4
Look at the two factors of the denominator and decide how many unknowns the template must carry before touching the numerator.
Hint 2/4
One linear factor and one irreducible quadratic: $\frac{A}{x+1}+\frac{Bx+C}{x^{2}+1}$, and the quadratic slot splits into a logarithm part and an arctangent part.
Independent numerical check with the other half of this section: Simpson's rule with $n=2$ on the same integral uses the heights $f(0)=3$, $f(0.5)=2.1333333$ and $f(1)=1.25$, giving $S_{2}=\frac{0.5}{3}(12.7833333)=2.1305556$. The exact answer is $2.1367822$, so the two agree to within $0.3$ percent — as they should, since the integrand is smooth and $n$ is small.
An exact answer and a numerical one are not rivals: the second is the cheapest way to catch an algebra slip in the first.
3§14.6 — how many strips does the promise need●●●●○
Exam level. The bound is run backwards: instead of computing an error, you are asked how much work buys a guarantee.
Given
$\displaystyle\int_{1}^{3}\frac{dx}{x}$ by the trapezoidal rule
$f''(x)=\dfrac{2}{x^{3}}$, so $\lvert f''\rvert\le2$ on $[1,3]$
required: $\lvert E_{T_{n}}\rvert<10^{-3}$
Find
What is the smallest $n$ that the bound guarantees?
Hint 1/4
You are solving an inequality for $n$, not approximating anything. Write the bound first with the numbers in it.
Hint 2/4
$\lvert E_{T_{n}}\rvert\le\frac{K(b-a)^{3}}{12n^{2}}$ with $K=2$ and $b-a=2$.
Hint 3/4
$\frac{2\cdot 2^{3}}{12n^{2}}=\frac{16}{12n^{2}}=\frac{4}{3n^{2}}<10^{-3}$, so $n^{2}>\frac{4000}{3}=1333.3$.
the inequality flips direction because $n^{2}$ moves to the other side
$n>36.51$
taking the positive square root
$n=37$
the smallest whole number above it; any larger $n$ also works
Answer $$n=37$$
Check
Check the promise at $n=37$: $\frac{4}{3(37)^{2}}=\frac{4}{4107}=9.74\times10^{-4}<10^{-3}$, and at $n=36$ it is $\frac{4}{3888}=1.03\times10^{-3}$, which fails. So $37$ really is the smallest.
A guaranteed $n$ is almost always far larger than the $n$ you actually need: at $n=4$ the true error here is already $0.018$, and the bound claimed $0.083$.
D · interleaved 3 questions
1§14 — mixed practice, type not given●●●○○
From here on the type of the question is not announced. Read the integrand and decide for yourself which tool it wants.
Size check without redoing it: on $[0,\pi/2]$ the integrand is non-negative and never exceeds $0.186$, its largest value, reached near $x\approx0.89$; so the integral is below $0.186\cdot\frac{\pi}{2}\approx0.29$. And $\frac{2}{15}=0.133$ sits comfortably under that.
Peeling works whenever one of the two powers is odd; when both are even, the half-angle identities are the only way in.
2§14 — mixed practice, type not given●●●●○
Two sections meet in this one: something has to be integrated, and then something has to be taken to a limit.
Decide whether the integral converges, and evaluate it if it does.
Hint 1/4
Deal with the two difficulties separately: first the antiderivative, then the endpoint. Which technique produces an antiderivative of a logarithm times a power?
Hint 2/4
Integration by parts with $u=\ln x$ and $dv=x^{-2}\,dx$, so $v=-\frac1x$; then the infinite endpoint is handled by $\lim_{t\to\infty}$.
Independent bound: for $x\ge e$ we have $\ln x\le x^{1/2}$, so the tail beyond $e$ is at most $\int_{e}^{\infty}x^{-3/2}\,dx=\frac{2}{\sqrt e}\approx1.21$, and since $\ln x\le1$ on $[1,e]$ the piece there is at most $\int_{1}^{e}x^{-2}\,dx=1-\frac1e\approx0.63$. A finite total below $1.9$, consistent with the value $1$.
When an improper integral needs parts, take the limit only after the boundary term and the remaining integral have both been written down.
3§14 — mixed practice, type not given●●●●○
A geometry question that turns into one of this week's integrals halfway through.
Given
The region under $y=\dfrac{1}{\sqrt{x^{2}+9}}$ from $x=0$ to $x=4$ is rotated about the $x$ axis.
Find
(a) Set up the volume as an integral.
(b) Evaluate it exactly.
Hint 1/4
Write the volume before thinking about techniques at all: what is the radius of the disc at a given $x$?
Hint 2/4
Discs perpendicular to the axis give $V=\pi\int_{a}^{b}\left[R(x)\right]^{2}dx$, and squaring the radius here removes the root entirely.
Hint 3/4
$R(x)=\frac{1}{\sqrt{x^{2}+9}}$, so $\left[R(x)\right]^{2}=\frac{1}{x^{2}+9}$ and $V=\pi\int_{0}^{4}\frac{dx}{x^{2}+9}$.
Hint 4/4
That is the arctangent form with $a=3$: $V=\frac{\pi}{3}\arctan\frac43$.
Show solutionSet up
$V=\pi\int_{0}^{4}\left[R(x)\right]^{2}dx$
the region touches the axis of rotation, so the cross sections are discs and not washers
Bracket the answer with two cylinders instead of recomputing: the radius falls from $\frac13$ at $x=0$ to $\frac15$ at $x=4$, so the volume lies between $\pi(\tfrac15)^{2}(4)=0.503$ and $\pi(\tfrac13)^{2}(4)=1.396$. The answer $0.9711$ sits between them.
Whenever a radius carries a root, square it before choosing a technique — half of these problems stop being root problems at that moment.
Mistake ledger (20 entries)
⚠ Leaving the answer in θ
the trigonometric integral is the hard part, so finishing it feels like finishing the problem
$a>0$, and the integrand contains exactly one of the three roots below; $\theta$ is restricted to the range listed, which is what makes the trig factor non-negative and lets the absolute value bars go; the answer is converted back to $x$ before it counts as an answer
$\deg P<\deg Q$; if not, divide first and template only the remainder; $Q$ is factored completely over the real numbers, into linear factors and quadratics with $b^{2}-4c<0$; every power of every factor gets its own slot
Strategy for integration
$\boxed{\text{simplify}\ \to\ \text{is a }du\text{ present?}\ \to\ \text{classify the shape}\ \to\ \text{table entry}\ \to\ \text{samples}}$
the questions are asked in this order, and the first yes wins; a yes does not forbid a second look: many integrands admit several routes, and the order is about cost, not legality
the entry is written in its own variable, usually $u$, and in its own constant, usually $a$; your integrand has to be brought into that exact shape by a substitution before the entry may be copied; the entry's own restriction — here $\lvert u\rvert
$\Delta x=\frac{b-a}{n}$ and $x_{i}=a+i\,\Delta x$ for $i=0,1,\ldots,n$; $\bar x_{i}$ is the midpoint of the $i$-th subinterval; $n$ must be even for $S_{n}$, because the parabolas are fitted two subintervals at a time
name why the integral is improper, and at which endpoint, before anything else; if both endpoints are bad, split at any convenient interior point and require both halves to converge; a limit that runs to $\pm\infty$, or fails to settle, means divergence
each carries $+C$; the middle one carries $\frac1a$ and the first carries bars
Check yourself
Close the page and write down, from memory: the three roots and the substitution each one asks for, the template slot an irreducible quadratic gets, the one restriction Simpson's rule carries, and the two $p$ conditions with the end of the interval each belongs to.
Say which of the three templates $\sqrt{16-9x^{2}}$ asks for, and what $a$ is.
c-trig-substitution
Rewrite $5-4x-x^{2}$ as a constant minus a square, and name the shift.
c-complete-square
Write the template for $\frac{1}{(x-2)^{3}(x^{2}+1)}$ without solving for anything, and say how many unknowns it must carry.
c-partial-fractions
Explain in one sentence why $\int\frac{x\,dx}{\sqrt{9-x^{2}}}$ and $\int\frac{dx}{\sqrt{9-x^{2}}}$ take different routes.
c-strategy
State the two separate constants that appear when a table entry written in $u$ is applied to an integrand written in $3x$.
c-tables
Write the Simpson weights for $n=6$ and say why $n=5$ is not allowed.
c-numerical
Decide $\int_{0}^{1}x^{-3/2}\,dx$ and $\int_{1}^{\infty}x^{-3/2}\,dx$ in your head, and say why the answers differ.
Replacing $x$ by $a\sin\theta$, $a\tan\theta$ or $a\sec\theta$ so that a Pythagorean identity collapses a root of a quadratic to a single trigonometric term.
reference trianglereferans üçgen
The right triangle read straight off the substitution, used to turn an answer written in $\theta$ back into one written in $x$.
completing the squarekareye tamamlama
Rewriting $x^{2}+bx+c$ as $\left(x+\frac{b}{2}\right)^{2}+\left(c-\frac{b^{2}}{4}\right)$, which exposes one of the three substitution templates.
proper rational functiondüzgün rasyonel fonksiyon
A ratio of polynomials whose numerator has strictly lower degree than its denominator; only these may be given a partial fraction template.
Writing a proper rational function as a sum of fractions whose denominators are powers of the linear and irreducible quadratic factors of the original denominator.
irreducible quadraticindirgenemez kuadratik
A quadratic $x^{2}+bx+c$ with $b^{2}-4c<0$, so it has no real roots and cannot be factored further over the real numbers.
integral tablosu
A printed list of antiderivatives written in a generic variable $u$ and a generic constant $a$, applied by matching an integrand to an entry through a substitution.
indirgeme formülü
A table entry that expresses an integral in terms of the same integral with a smaller exponent, applied repeatedly until a basic form is left.
sayısal integrasyon
Approximating $\int_{a}^{b}f$ by evaluating $f$ at finitely many points; the midpoint, trapezoidal and Simpson rules.
trapezoidal ruleyamuk kuralı
Joining consecutive sample points with straight lines: weights $1,2,2,\ldots,2,1$ over $\frac{\Delta x}{2}$, with error falling like $n^{-2}$.
Simpson's ruleSimpson kuralı
Fitting a parabola through each consecutive three sample points: weights $1,4,2,4,\ldots,4,1$ over $\frac{\Delta x}{3}$, valid only for even $n$, with error falling like $n^{-4}$.
error boundhata sınırı
A guaranteed ceiling on how far an approximation can be from the true value; the actual error is usually far smaller.
improper integralgenelleştirilmiş integral
A definite integral with an infinite endpoint or an unbounded integrand, defined as the limit of ordinary definite integrals.
p integralp integrali
The two anchor families $\int_{1}^{\infty}x^{-p}\,dx$, convergent exactly when $p>1$, and $\int_{0}^{1}x^{-p}\,dx$, convergent exactly when $p<1$.
comparison testkarşılaştırma testi
Inheriting a convergence verdict from a larger or smaller function whose integral is already known; it gives a verdict, never a value.
What comes next
That is the last block of the course. What is left is the exam, and the useful thing to do with these four tools now is to practise the choosing rather than the executing: shuffle problems from every section of the term together, cover the labels, and give yourself ten seconds per integral to name the technique before you write a single line.
Sources
James Stewart, Calculus, Metric Version, Ninth Edition — sections 7.3, 7.4, 7.5 and 7.6 The templates and rules are the ones this book states; every worked function here is a different one.
Course syllabus, week 14: Techniques of Integration 7.3, 7.4, 7.5, 7.6 The assessment weights quoted on the card come from the same syllabus.
Every numerical value on this page Recomputed independently before publication, including the approximation tables, the error bounds and the values of the definite integrals.