3.5 Other Strategies for Integration

1. Other Strategies for Integration

Title

Calculus II · Section 3.5

Choosing the technique, using a table, and trusting a machine

2. What this lesson gives you

Objectives

Sections 3.1 to 3.4 each handed you a technique together with integrals built for it. Real integrals arrive without a label. This lesson is about choosing: which technique an integrand is asking for, what to do when none fits, and how to check an answer that came from a table or a computer.

• Work through the techniques in order of cost, cheapest check first
• Rewrite an integrand algebraically before reaching for a method
• Use a table of integrals: match the form, then fix a, u and du
• Compare a computer algebra system's answer with yours, and explain why they can differ
• Recognise a non-elementary integral and stop searching for a formula

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, pp. 311-315 — learning objectives 3.5.1 and 3.5.2

Every section of this chapter so far came with a promise: the integrals at the end of it wanted the technique at the start of it. That promise made the exercises easier than real integrals are. When an integral shows up in a physics problem or on a cumulative exam, it does not say which section it came from.

The book's Section 3.5 is short. It shows you two tools, tables of integrals and computer algebra systems, and one important warning: two correct answers can look nothing alike. This deck keeps all of that and adds what the tools quietly assume you can already do, which is to recognise what kind of integral you are looking at.

By the end you should be able to glance at an integrand, name the cheapest technique that will work, put it into the shape a table expects, and decide confidently whether a machine's strange-looking answer agrees with yours.

3. Before anything new: five techniques, five signals

Warm-up

Write your list before you reveal the answer. If you can name the techniques but not the signals, that is exactly the gap this lesson fills: knowing how to do integration by parts is useless if you cannot tell when to do it.

Notice that every signal in the table is visible without calculus. A spare factor whose partner is its derivative. A product of two different kinds of function. Powers of sine and cosine. A square root of a quadratic with no helpful x outside. A fraction of polynomials. You can spot all of these in a couple of seconds.

Keep this table in mind for the whole deck. Almost every slide that follows is a variation on one question: which row does this integrand belong to, and is there a cheaper row that also works?

4. Selection is now the skill

Intuition

Inside Section 3.3 every integral wanted a trigonometric substitution, so you never had to decide. On an exam, or in a physics problem, nobody tells you which section an integral came from.

\[ \int \frac{x}{\sqrt{x^2+4}}\,dx \qquad \int \frac{dx}{\sqrt{x^2+4}} \]

These two differ by one factor of x, and that factor changes the right method completely: the first is a two-line substitution, the second needs a triangle or a table. Learning to see that difference quickly is worth more than any single new technique.

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5, pp. 503-507 — strategy for integration

Look hard at the two integrals on this slide. They differ by a single factor of x in the numerator, and that factor decides everything. With it, the numerator is half the derivative of the expression under the root, and a substitution finishes the job almost immediately. Without it, there is nothing for a substitution to grab, and you need a tangent substitution or a table.

This is why selecting a method is a skill in its own right. Students who know every technique perfectly still lose time, and sometimes lose the problem, because they commit to the first technique that comes to mind and push it through a page of algebra.

The rest of this part gives you an order in which to ask questions, so that you meet the cheap methods first and only pay for the expensive ones when you have to.

5. The order to try things

Section

Part 1

6. Five techniques, five signals

Concept

The rows are sorted by cost. When several techniques could work, the one higher in the table is almost always the one to use.

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5, pp. 503-507 — table of basic integration formulas

Read this table from top to bottom as a price list. A basic rule costs one line. A substitution costs two or three. Integration by parts means a small table of four quantities and sometimes a second round. Trig integrals need an identity before the substitution. Partial fractions means factoring, solving for constants and then integrating several pieces. Trig substitution needs a substitution, a triangle and a translation back.

The ordering matters because many integrals could be done by more than one row. The integral of x over the square root of x squared plus 4 can be done by a tangent substitution, and you would get the right answer, but it costs several times as much as a plain substitution.

So when you see an integrand, run down the table from the top and stop at the first row that fits. That habit is the whole strategy in one sentence.

7. Why try the cheap things first

Intuition

A check that takes five seconds and usually fails is still worth doing first, because when it succeeds it saves a page. Trying a trig substitution first and a plain substitution second wastes the page every time the plain one would have worked.

\[ \int \frac{x}{\sqrt{x^2+4}}\,dx: \quad \text{substitution: 3 lines} \quad \text{vs} \quad \text{tangent substitution: 8 lines} \]

Both routes reach the same answer. Order by cost is not about right and wrong; it is about not paying for power you do not need.

It can feel inefficient to check for easy methods on an integral that looks hard. The point is that the check itself is nearly free. Asking whether an inner derivative is present takes a few seconds, and when the answer is yes, you have saved a page of work.

The formula on the slide compares two routes to the same integral. The plain substitution takes three lines; the tangent substitution takes around eight, including drawing a triangle and converting back. Both are correct. The difference is only in how much you pay, and in how many chances you give yourself to make an arithmetic slip.

Think of the order as a filter. Cheap checks first catch most integrals early, and the expensive methods only ever see the integrals that genuinely need them.

8. Step one: simplify before you integrate

Concept

Many integrands become a basic rule after one line of algebra. Look for a product to expand, a fraction to split, or an identity to apply.

\[ \int \sqrt{x}\left(1 + \sqrt{x}\right)dx = \int\left(x^{1/2} + x\right)dx \]

\[ \int \frac{\tan\theta}{\sec^2\theta}\,d\theta = \int \sin\theta\cos\theta\,d\theta \]

\[ \int \frac{x^2 + 1}{x}\,dx = \int\left(x + \frac1x\right)dx \]

None of these needed a technique at all, only algebra you already knew.

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5, pp. 503-507 — step 1: simplify the integrand if possible

Before you choose any technique, look at the integrand as an algebra problem. The three examples here all look like they need something, and none of them does.

In the first, multiplying out the bracket turns a product of roots into a sum of two powers, each handled by the power rule. In the second, tangent over secant squared becomes sine over cosine times cosine squared, which simplifies to sine times cosine, a one-line substitution. In the third, dividing each term of the numerator by x splits the fraction into x plus one over x.

The lesson is that the form an integral is written in is often not the form it wants to be integrated in. One line of algebra first can make every later step unnecessary. Get into the habit of asking: can I expand this, split this, or rewrite this with an identity, before you ask anything else.

9. Step two: look for a derivative sitting in the integrand

Concept

A substitution works when some inner function appears together with its own derivative, up to a constant factor.

\[ \int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du, \qquad u = g(x) \]

\[ \int \frac{\ln x}{x}\,dx: \quad u = \ln x,\; du = \frac{dx}{x} \qquad \int x^2 e^{x^3}\,dx: \quad u = x^3 \]

Ask it as a question every time: is there a piece whose derivative is also here? A constant factor such as a 2 or a one third can be fixed, a missing x cannot.

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5, pp. 503-507 — step 2: look for an obvious substitution

Substitution is the reverse of the chain rule, so it works exactly when the integrand looks like the output of a chain rule: some function of an inner function, multiplied by the inner function's derivative.

In the two small examples, ln x appears together with one over x, which is its derivative, and x cubed sits in an exponent while x squared, its derivative up to a factor of 3, sits outside. In both cases the substitution is decided by looking.

The phrase up to a constant is important. If the derivative is present except for a factor such as 2 or one third, you can fix that factor by multiplying and dividing. What you cannot fix is a missing variable: if the derivative needs an x and there is no x in the integrand, the substitution fails and you move on to step three.

10. Step three: classify the integrand by its form

Concept

Each row is a shape you can recognise before any calculation. If an integrand fits a row, its technique is decided.

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5, pp. 503-507 — step 3: classify the integrand according to its form

If simplifying and substituting have both failed, the integrand's shape usually names the technique. This table is the decision you learned piece by piece in Sections 3.1 to 3.4, now gathered in one place.

Products of a polynomial with an exponential or a trig function go to integration by parts, with the polynomial as u because differentiating it eventually makes it disappear. Logarithms and inverse trig functions on their own also go to parts, with dv equal to dx. Powers of sine and cosine, or tangent and secant, go to the trig integral methods. Square roots of a quadratic with no spare x go to trig substitution. Ratios of polynomials go to long division if needed and then partial fractions.

Notice the words no spare x in the radical row. That is step two checking in again: a radical with its inner derivative nearby is a substitution, not a trig substitution.

11. Step four: when the first attempt stalls

Concept

If nothing fits, do not stare. Change the integrand, then run the list again.

• Try a substitution anyway, even if the derivative is not obviously present; the new integrand may be recognisable
• Manipulate: multiply by a conjugate, add and subtract the same thing, use an identity
• Relate to a problem you have done: the same shape with different constants
• Combine methods: a substitution often turns an integral into one for parts or partial fractions

\[ \int e^{\sqrt{x}}\,dx \;\xrightarrow{u = \sqrt x}\; 2\int u e^{u}\,du \;\xrightarrow{\text{parts}}\; 2(u - 1)e^{u} + C \]

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5, pp. 503-507 — step 4: try again

Some integrals fit no row as written. When that happens, the move is not to stare harder but to change the integrand and try again.

The example on the slide shows the most common version. The integral of e to the square root of x fits nothing: it is not a product, there is no inner derivative visible, and it is not a trig power or a radical of a quadratic. Substitute u equal to the square root of x anyway. Then x is u squared, dx is 2u du, and the integral becomes 2 times the integral of u times e to the u, which is a textbook case for parts.

Many real integrals are chains like this: a substitution that turns the problem into a recognisable form, then a second technique that finishes it. When you are stuck, ask what substitution would make the ugliest part simpler, and see what the integral turns into.

12. The strategy as a decision tree

Picture it

The four steps as a map. Most integrals leave at the first or second box; the expensive techniques on the right only see what is left over.

Follow the arrows from the left. The first box asks whether algebra alone can simplify the integrand. The second asks whether an inner function's derivative is present; if so, you go straight up to substitution and you are done.

If not, the third box classifies the form, and the four boxes on the right are the four heavier techniques of this chapter. The pictures in those boxes are the signals from the warm-up: a product such as x times e to the x, powers of sine and cosine, a root of a squared plus or minus x squared, and a ratio of polynomials.

The box at the bottom is where this section's new tools live: tables, computer algebra systems and numerical methods. It is at the bottom not because those tools are forbidden, but because using them well depends on everything above it. A table cannot help you if you cannot put your integrand into one of its shapes.

13. A radical that does not need a triangle

Worked example

Choose a method, then evaluate.

\[ \int \frac{x}{\sqrt{x^2+4}}\,dx \]

\[ \frac{d}{dx}\left(x^2 + 4\right) = 2x \]

\[ u = x^2 + 4, \quad du = 2x\,dx, \quad x\,dx = \tfrac12\,du \]

\[ \int \frac{x\,dx}{\sqrt{x^2+4}} = \frac12\int u^{-1/2}\,du \]

\[ \frac12\int u^{-1/2}\,du = \frac12\cdot 2u^{1/2} + C = u^{1/2} + C \]

\[ \int \frac{x\,dx}{\sqrt{x^2+4}} = \sqrt{x^2+4} + C \]

\[ \frac{d}{dx}\sqrt{x^2+4} = \frac{2x}{2\sqrt{x^2+4}} = \frac{x}{\sqrt{x^2+4}} \]

The square root of x squared plus 4 is the classic trigger for a tangent substitution, and that is exactly why this example is here. Before you reach for a triangle, step two asks whether the inside of the root has its derivative in the integrand. It does: the derivative of x squared plus 4 is 2x, and there is an x in the numerator.

So let u be the inside of the root. The x and the dx together become one half of du, and the integral turns into a power of u. The power rule gives u to the one half, and you replace u to finish.

The check is the chain rule in reverse. Differentiating the square root of x squared plus 4 brings down one over twice the root, multiplied by 2x, and the twos cancel to leave the integrand. Three lines of calculus, where a tangent substitution would have cost you a page.

14. Before you choose: the same radical without the x

Step zero

\[ \int \frac{dx}{\sqrt{x^2+4}} \]

Commit to an answer before you reveal it. The integrand looks almost identical to the last one, and the temptation is to repeat the same substitution.

Try it and see what goes wrong. With u equal to x squared plus 4, du is 2x dx, but there is no x in the numerator to absorb. You would have to divide by x, and x is not a constant, so the substitution does not simplify anything. Step two has failed.

Step three then looks at the form: a root of a variable squared plus a constant squared, with nothing on top. That is the tangent substitution, or equivalently a table entry whose answer is a logarithm. You will see both routes, and a computer's third route, in Checkpoint 3.21 later in the lesson.

15. Exercise 247: a quadratic that factors

Worked example

\[ \int \frac{dx}{x^2 + 6x} \]

\[ x^2 + 6x = x(x+6) \]

\[ \frac{1}{x(x+6)} = \frac{A}{x} + \frac{B}{x+6} \]

\[ 1 = A(x+6) + Bx \]

\[ x = 0: \; 1 = 6A, \; A = \tfrac16 \qquad x = -6: \; 1 = -6B, \; B = -\tfrac16 \]

\[ \int\left(\frac{1/6}{x} - \frac{1/6}{x+6}\right)dx = \frac16\ln|x| - \frac16\ln|x+6| + C \]

\[ \int \frac{dx}{x^2+6x} = \frac16\ln\left|\frac{x}{x+6}\right| + C \]

\[ \frac16\left(\ln\frac28 - \ln\frac17\right) = \frac16\ln\frac74 \approx 0.09327 \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 314 — Exercise 247

The integrand is a ratio of polynomials, the degree of the top is less than the degree of the bottom, and the bottom factors as x times x plus 6. That is the full signal for partial fractions, and there is no cheaper route: no inner derivative, since the derivative of the bottom is 2x plus 6 and the top is just 1.

Split into A over x plus B over x plus 6, clear denominators, and solve by putting x equal to each root in turn. That gives one sixth and minus one sixth. Each piece integrates to a logarithm, and the difference of two logarithms combines into the log of a quotient.

Look at the figure. The awkward curve is exactly the sum of one sixth of one over x and a small negative curve below the axis, which is the picture of what partial fractions does. The check uses a definite integral from 1 to 2: the formula gives 0.09327, and a numerical estimate of the area agrees to every digit shown.

16. Exercise 272: substitute first, then integrate by parts

Worked example

\[ \int e^{x}\cos^{-1}\left(e^{x}\right)dx \]

\[ u = e^{x}, \quad du = e^{x}\,dx \]

\[ \int e^{x}\cos^{-1}\left(e^{x}\right)dx = \int \cos^{-1} u\,du \]

\[ w = \cos^{-1} u, \quad dw = -\frac{du}{\sqrt{1-u^2}}, \quad v = u \]

\[ \int \cos^{-1} u\,du = u\cos^{-1} u + \int \frac{u\,du}{\sqrt{1-u^2}} \]

\[ \int \frac{u\,du}{\sqrt{1-u^2}} = -\sqrt{1-u^2} + C \]

\[ \int e^{x}\cos^{-1}\left(e^{x}\right)dx = e^{x}\cos^{-1}\left(e^{x}\right) - \sqrt{1 - e^{2x}} + C \]

\[ \frac{d}{dx}\left[e^{x}\cos^{-1}e^{x}\right] = e^{x}\cos^{-1}e^{x} - \frac{e^{2x}}{\sqrt{1-e^{2x}}} \]

\[ \frac{d}{dx}\left[-\sqrt{1-e^{2x}}\right] = +\frac{e^{2x}}{\sqrt{1-e^{2x}}} \quad\text{so the sum is } e^{x}\cos^{-1}e^{x} \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 314 — Exercise 272

This integral looks intimidating because it mixes an exponential with an inverse cosine. Run the strategy. Step two asks whether an inner function's derivative is present, and it is: the inverse cosine is applied to e to the x, and e to the x is sitting right in front.

After the substitution the integral is simply the integral of inverse cosine of u, which is a known case for integration by parts. Take the inverse cosine as the part you differentiate, because its derivative is algebraic, and take dv to be du. The leftover integral, u over the square root of 1 minus u squared, is itself a quick substitution.

This is the chain-of-techniques pattern from step four: one technique turns the problem into another technique's territory. A table would also list the integral of inverse cosine of u directly. The check differentiates the final answer: the product rule produces a messy radical term, and the derivative of the second piece cancels it exactly.

17. Which technique goes first?

Sorting

Decide each one from its shape alone, as you would at the start of an exam problem. You are sorting by which question in the strategy list the integral answers yes to first.

Two of these are substitutions in disguise. In x over x squared plus 9, the top is half the derivative of the bottom, so you never need the arctangent or partial fractions. In x times 2 to the x squared, the x pairs with the derivative of x squared. The product x cubed times sine x is integration by parts, repeated three times. The odd powers, cosine cubed and tangent to the fifth, go to trig identities.

The two radicals both go to trig substitution, but for different reasons: the square root of 9 minus x squared has no x outside it, and one over x times the square root of x squared minus 1 has an x, but in the wrong place to help. The fraction with x squared minus 1 underneath factors, so it goes to partial fractions.

18. Trap: an arctangent that is really a logarithm

Trap

19. Put the strategy in order

Ranking

Arrange the questions from cheapest to most expensive. The reason for the order is not tradition; it is that each question costs more to answer than the one before, and each has a good chance of ending the problem.

Algebraic simplification comes first because it can remove the need for any technique. The derivative check comes next because it takes seconds and succeeds often. Classifying the form requires more thought, and changing the integrand with a substitution or rewrite is a genuine attempt that may take several lines.

A table or a computer is last. That is not because they are cheating; the book encourages both. It is because a table only helps once your integrand matches one of its shapes, which uses the same recognition as the earlier steps, and because an answer from either one still has to be checked.

20. Rewriting first

Section

Part 2

21. Algebra before calculus: four rewrites

Concept

Each rewrite changes how the integrand looks without changing its value, and the new look is one you already know how to integrate. None of them is a technique of integration; they are all algebra.

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5, pp. 503-507 — simplify the integrand; manipulate the integrand

These four rewrites cover most of what step one means in practice, and each has a clear signal. A fraction whose numerator is a sum can often be split term by term. A fraction whose top and bottom have the same degree can be turned into a whole number plus a remainder by adding and subtracting the right constant. A one plus or one minus a trig function in a denominator invites the conjugate. And a product of trig functions often has an identity waiting.

None of these is an integration technique. They are all algebra you learned before calculus, and that is exactly why they are easy to forget: in a calculus class your mind is primed for calculus.

The next few slides work through the conjugate, the add-and-subtract trick, and the identity in full, and show that the conjugate turns an integral with no visible route into two basic rules.

22. Exercise 278: multiply by the conjugate

Worked example

\[ \int \frac{dx}{1 + \sin x} \]

\[ \frac{1}{1 + \sin x} \]

\[ \frac{1}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x} = \frac{1 - \sin x}{1 - \sin^2 x} \]

\[ = \frac{1 - \sin x}{\cos^2 x} \]

\[ = \sec^2 x - \sec x\tan x \]

\[ \int\left(\sec^2 x - \sec x\tan x\right)dx = \tan x - \sec x + C \]

\[ \frac{d}{dx}(\tan x - \sec x) = \sec^2 x - \sec x\tan x = \frac{1 - \sin x}{\cos^2 x} = \frac{1}{1 + \sin x} \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 315 — Exercise 278

One over one plus sine x resists every technique. There is no inner derivative, it is not a product, it is not a power of sine, and there is no radical. When nothing fits, change the integrand.

Multiplying by one minus sine over itself changes nothing in value but turns the denominator into one minus sine squared, which the Pythagorean identity turns into cosine squared. The fraction then splits into secant squared minus secant times tangent, and both of those are derivatives you know by heart.

Look at the figure. The integrand is always positive, so the antiderivative always climbs. At pi over 2 the formula tan x minus sec x looks dangerous, since both tangent and secant blow up there, but their difference passes smoothly through zero. The check differentiates the answer and undoes the conjugate step, landing back on the integrand.

23. Add zero on top

Fill the middle

\[ \int \frac{x^2}{x^2+1}\,dx \]

\[ \frac{x^2}{x^2+1} = \frac{(x^2+1) - 1}{x^2+1} = 1 - \frac{\square}{x^2+1} \]

\[ \int \frac{x^2}{x^2+1}\,dx = x - \tan^{-1} x + C \]

Fill in the blank before you reveal it. The goal is to make the numerator contain a copy of the denominator, because the denominator divided by itself is one.

Writing x squared as x squared plus 1, minus 1, does exactly that. The fraction then splits into 1, minus one over x squared plus 1. The first piece integrates to x and the second is the arctangent, so the whole integral is x minus the arctangent of x, plus C.

This trick is long division in disguise. Whenever the degree of the top is at least the degree of the bottom, you could divide properly, but when the bottom is as simple as x squared plus 1, adding and subtracting a constant does the division in your head. Look for it whenever the top and bottom have the same degree.

24. One integral, three answers: sine times cosine

Worked example

\[ \int \sin x\cos x\,dx \]

\[ u = \sin x: \quad \int u\,du = \frac{\sin^2 x}{2} + C_1 \]

\[ u = \cos x: \quad -\int u\,du = -\frac{\cos^2 x}{2} + C_2 \]

\[ \int \tfrac12\sin 2x\,dx = -\frac{\cos 2x}{4} + C_3 \]

\[ \frac{\sin^2 x}{2} - \left(-\frac{\cos^2 x}{2}\right) = \frac{\sin^2 x + \cos^2 x}{2} = \frac12 \]

\[ \frac{\sin^2 x}{2} - \left(-\frac{\cos 2x}{4}\right) = \frac{2\sin^2 x + 1 - 2\sin^2 x}{4} = \frac14 \]

This is the smallest example of the phenomenon the book warns about, and it is worth doing all three routes yourself. You can substitute u equal to sine, or u equal to cosine, or rewrite with the double-angle identity. All three are legitimate, and all three give different-looking answers.

The figure shows why none of them is wrong. The three graphs are the same wave, slid up or down by fixed amounts: a quarter between two of them, a half between the outer two. Sliding a graph vertically does not change its slope anywhere, so all three have the same derivative, which is sine times cosine.

The checks make that precise. Subtract two answers and simplify: the Pythagorean identity collapses the first difference to one half, and the double-angle identity collapses the second to one quarter. Differences that are numbers are absorbed by the constant of integration.

25. Break the claim: different-looking means wrong

Counterexample

Write your counterexample before revealing it. The previous slide handed you several, so the real point here is the second half of the question: what single test settles such a dispute?

The answer is to differentiate both. If both derivatives equal the integrand, both answers are correct, however different they look. The reason is a fact from the Mean Value Theorem: two functions with equal derivatives on an interval differ by a constant there.

This matters in practice more than it seems. When your answer disagrees with the back of the book, the instinct is to assume you made a mistake and start again. Differentiating your answer first takes less time and may show that you were right all along, just in a different form.

26. Using a table

Section

Part 3

27. What a table of integrals is

Concept

A table of integrals is a catalogue of antiderivatives, written in a dummy variable u and constants a and b, and grouped by form: radicals of a squared minus u squared, rational functions of a plus bu, powers of trig functions, and so on. Appendix A of the book has over a hundred entries.

\[ \int \frac{du}{a^2 + u^2} = \frac1a\tan^{-1}\frac{u}{a} + C \qquad \int \frac{u\,du}{a + bu} = \frac{1}{b^2}\left(a + bu - a\ln|a + bu|\right) + C \]

Every entry was derived with the techniques of this chapter. A table saves you from repeating the derivation, not from recognising the form.

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 311 — Tables of Integrals

A table of integrals is a reference list of antiderivatives, written with a dummy variable u and constants such as a and b, and grouped by the form of the integrand. The book's Appendix A has over a hundred entries, organised into families: integrals involving a squared plus u squared, the square root of a squared minus u squared, a plus bu, trig powers, exponentials and logarithms.

Two entries are on the slide. The first is the arctangent entry you already know. The second, for u over a plus bu, you could derive with long division, but having it written down saves the work.

The important thing to understand is what a table does not do for you. Every entry was derived with the techniques of this chapter, so a table replaces the derivation, not the recognition. You still have to see that your integrand belongs to a family, identify the constants, and adjust the differential.

28. Three radicals, three shapes, three entries

Picture it

A plus or minus sign decides which family of entries you need. On paper the three radicals differ by one symbol; on axes they are a circle, a valley and two separate branches.

Look at the three panels. They are drawn from three radicals that differ by a single sign, and they could hardly look more different: a semicircle, a smooth valley that never drops below 2, and two separate branches that do not exist at all between minus 2 and 2.

Tables file these as three separate families, for the same reason Section 3.3 used three different substitutions. The first matches sine, and its entries involve inverse sine. The second matches tangent, and its entries involve a logarithm of u plus the root. The third matches secant, and its entries involve the logarithm of the absolute value.

The practical lesson is to read the sign before you look anything up. Two of the traps later in this lesson come from matching the right numbers to the wrong family.

29. Matching an entry: find a, find u, fix du

Concept

To use an entry, write your integrand in the entry's exact shape. That takes three identifications, in this order.

1. a: the constant. In 16 minus e to the 2x, a squared is 16, so a is 4
2. u: the quantity that is squared. In 16 minus e to the 2x, u is e to the x
3. du: the differential of that u must be in the integral. If it is not, multiply and divide to make it appear, or the match fails

\[ 16 - e^{2x} = 4^2 - \left(e^{x}\right)^2 \qquad u = e^{x} \;\Longrightarrow\; du = e^{x}\,dx \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 311 — Example 3.36

Using a table correctly is a three-step identification, and the order matters. First find the constant a. Then find the quantity u that is being squared. Then check that du, the differential of that u, is actually present in your integral.

The example on the slide is the one from Example 3.36. Sixteen minus e to the 2x is 4 squared minus e to the x squared, so a is 4 and u is e to the x. Then du is e to the x dx, and you must be able to find exactly that in your integral.

The third step is where most mistakes happen. If u is 2x, then du is 2 dx, not dx, and you must divide by 2. If u is e to the x, you need a factor of e to the x next to the dx, and if it is not there you have to create it by multiplying and dividing. A table entry is a substitution in disguise, and every substitution needs its du.

30. Reading table entry 88

Notation

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 311 — the formula used in Example 3.36 (Appendix A, entry 88)

Tap each part of the entry. The family is set by the square root of a squared minus u squared: a constant first, the variable second, with a minus sign between. The u squared in the denominator is the second requirement: your integrand must have exactly that power of u underneath after substitution.

The du at the end is the part people forget. The entry is written with respect to u, so whatever u stands for, its differential must be sitting in your integral. On the right, the answer is written in the same letters, u and a, which you replace at the end.

Notice that the answer has two pieces, an algebraic part and an inverse sine. You could derive this with a sine substitution and a triangle, which would take most of a page. That page of work is what the table saves.

31. Example 3.36: a table entry in disguise

Worked example

\[ \int \frac{\sqrt{16 - e^{2x}}}{e^{x}}\,dx \]

\[ 16 - e^{2x} = 4^2 - \left(e^{x}\right)^2: \quad a = 4, \; u = e^{x} \]

\[ \int \frac{\sqrt{16 - e^{2x}}}{e^{x}}\,dx = \int \frac{\sqrt{16 - e^{2x}}}{e^{2x}}\,e^{x}\,dx \]

\[ = \int \frac{\sqrt{4^2 - u^2}}{u^2}\,du \]

\[ = -\frac{\sqrt{16 - u^2}}{u} - \sin^{-1}\frac{u}{4} + C \]

\[ = -\frac{\sqrt{16 - e^{2x}}}{e^{x}} - \sin^{-1}\frac{e^{x}}{4} + C \]

\[ F(1) - F(0) = 2.29897 \quad\text{and}\quad \int_0^1 \frac{\sqrt{16 - e^{2x}}}{e^{x}}\,dx \approx 2.29897 \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, pp. 311-312 — Example 3.36

At first glance nothing in the table looks like this integrand. The trick is to name the family from the radical: 16 minus e to the 2x is a constant squared minus a square, so a is 4 and u is e to the x.

Now du has to appear. It should be e to the x times dx, but the e to the x is in the denominator. Multiplying the top and bottom by e to the x fixes that: the denominator becomes e to the 2x, which is u squared, and e to the x dx stands alone as du. The integral is now exactly entry 88 with a equal to 4.

After applying the entry, put e to the x back everywhere. The book's printed answer has a u left in one denominator; it should be e to the x, as here. The figure shows that the integrand only exists to the left of ln 4, a restriction a computer would not mention. The check compares the antiderivative's change from 0 to 1 with a numerical estimate of the shaded area: both are 2.29897.

32. Find the error: the du that was never found

Error analysis

Tap each note in turn. The first part of this solution is right: 4x squared plus 25 really is 2x squared plus 5 squared, so the arctangent entry applies with a equal to 5 and u equal to 2x.

The mistake is in the differential. The entry integrates with respect to u, and du is 2 dx, but the integral only contains dx. So only half of the du is present, and a factor of one half has to come out in front. The correct answer has one tenth, not one fifth.

This is the single most common table error, and the check catches it instantly. Differentiate the wrong answer: the chain rule produces an extra two fifths, which combined with the one fifth gives 2 over 4x squared plus 25, twice the integrand. Whenever a table answer is off by a clean factor, look at du first.

33. Exercise 250: a coefficient inside the square

Worked example

\[ \int \frac{dx}{4x^2 + 25} \]

\[ \int \frac{du}{a^2 + u^2} = \frac1a \tan^{-1}\frac{u}{a} + C \]

\[ 4x^2 + 25 = (2x)^2 + 5^2: \quad u = 2x, \; a = 5 \]

\[ du = 2\,dx \;\Longrightarrow\; dx = \tfrac12\,du \]

\[ \int \frac{dx}{4x^2 + 25} = \frac12\int \frac{du}{u^2 + 5^2} \]

\[ = \frac12\cdot\frac15\tan^{-1}\frac{u}{5} + C \]

\[ = \frac{1}{10}\tan^{-1}\frac{2x}{5} + C \]

\[ \frac{1}{10}\cdot\frac{1}{1 + 4x^2/25}\cdot\frac25 = \frac{1}{25}\cdot\frac{25}{25 + 4x^2} = \frac{1}{4x^2 + 25} \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 314 — Exercise 250

This is the corrected version of the error you just analysed, done properly. The integrand has nothing on top and a sum of two squares underneath, so the arctangent entry is the right family.

The coefficient 4 is what makes it more than a direct lookup. Write 4x squared as the square of 2x, so u is 2x and a is 5. Then du is 2 dx, so dx is one half of du, and the one half comes out in front. The entry contributes one over a, one fifth, and the two constants multiply to one tenth.

The figure shows the integrand and its antiderivative side by side. Where the bell is tallest the antiderivative is steepest, and where the bell flattens the antiderivative levels off, at plus and minus pi over 20. The check differentiates the answer: the chain rule's two fifths meets the one tenth to give one twenty-fifth, and the compound fraction simplifies back to the integrand.

34. Reading the arctangent entry

Notation

This is the most-used entry in any table, so it is worth reading slowly. The a squared plus u squared underneath, with nothing on top, sets the family. An extra u on top would make the integral a logarithm instead, as the trap earlier in the lesson showed.

The one over a in front and the u over a inside do different jobs. The u over a sets the width: the arctangent does most of its rising while u is within a few multiples of a. The one over a in front sets the height, and it is what makes a wide bell with a large a give a small antiderivative.

Because the arctangent itself levels off at plus and minus pi over 2, the antiderivative levels off at plus and minus pi over 2a. So the total area under the whole bell, from minus infinity to infinity, is pi over a. That is a finite area under a curve that never touches zero, which you will meet again with improper integrals in Section 3.7.

35. Slide a and watch the arctangent

Tweak it

Before moving the slider, predict what happens to the curve when a doubles. Then drag it and compare.

Two things change at once. The slope at the origin is the integrand's value at zero, which is one over a squared, so doubling a makes the curve four times less steep in the middle. And the heights where the curve levels off are plus and minus pi over 2a, so doubling a halves them.

This is the one over a and the u over a from the previous slide made visible. If you ever forget which way round the constants go in the arctangent entry, this picture is a way to reconstruct it: a small a means a tall, narrow bell, whose antiderivative rises steeply to high limits.

36. Exercise 279: complete the square, then look it up

Worked example

\[ \int \frac{dx}{x^2 + 2x + 10} \]

\[ x^2 + 2x + 10 \;\text{ is not yet }\; u^2 + a^2 \]

\[ x^2 + 2x + 10 = (x^2 + 2x + 1) + 9 = (x+1)^2 + 3^2 \]

\[ u = x + 1, \quad a = 3, \quad du = dx \]

\[ \int \frac{du}{u^2 + 3^2} = \frac13\tan^{-1}\frac{u}{3} + C \]

\[ \int \frac{dx}{x^2 + 2x + 10} = \frac13\tan^{-1}\frac{x+1}{3} + C \]

\[ \frac13\cdot\frac{1}{1 + (x+1)^2/9}\cdot\frac13 = \frac{1}{9 + (x+1)^2} = \frac{1}{x^2 + 2x + 10} \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 315 — Exercise 279

The denominator is a quadratic, but it does not factor over the real numbers, so partial fractions is out, and its stray 2x term keeps it from matching the arctangent entry directly. Completing the square is the rewrite that makes it match.

Half of 2 is 1, so add and subtract 1: the quadratic becomes x plus 1, squared, plus 9. Now u is x plus 1, a is 3, and du is just dx, since shifting by a constant does not change the differential. The entry gives one third of the arctangent of u over 3.

The figure shows why this costs nothing: the new bell is exactly the bell for one over x squared plus 9, slid one unit to the left. Same shape, same height, different centre. The check differentiates the answer; the one third in front and the one third from the chain rule multiply to one ninth, and the compound fraction returns the original quadratic.

37. Exercise 244: an entry with two constants

Worked example

\[ \int_0^4 \frac{x}{1 + 2x}\,dx \]

\[ \int \frac{u\,du}{a + bu} = \frac{1}{b^2}\left(a + bu - a\ln|a + bu|\right) + C \]

\[ u = x, \quad a = 1, \quad b = 2 \]

\[ F(x) = \tfrac14\left(1 + 2x - \ln(1 + 2x)\right) \]

\[ F(4) - F(0) = \tfrac14\left(9 - \ln 9\right) - \tfrac14(1 - 0) \]

\[ = 2 - \tfrac14\ln 9 = 2 - \tfrac12\ln 3 \approx 1.4507 \]

\[ \frac{x}{1+2x} = \frac12 - \frac{1}{2(1+2x)} \]

\[ \int_0^4\left(\frac12 - \frac{1}{2(1+2x)}\right)dx = 2 - \tfrac14\ln 9 \quad\checkmark \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 314 — Exercise 244

The integrand is x over a linear expression, so it belongs to the a plus bu family, whose entries have two constants instead of one. Here u is simply x, a is 1 and b is 2.

The entry gives the antiderivative as one over b squared times a bracket, and with b equal to 2 the factor in front is one quarter. At the upper limit 4 the linear expression is 9; at 0 it is 1 and the logarithm vanishes. The result simplifies to 2 minus one half ln 3, about 1.4507.

The check avoids the table entirely. Long division, or the add-and-subtract trick, splits x over 1 plus 2x into one half minus a simple fraction. Integrating that from 0 to 4 gives the same 2 minus one quarter ln 9. When a table and a hand calculation agree, you can trust both, and you have also seen that the table entry is nothing more than this division done once and written down.

38. Match each integrand to its table family

Matching

For each integral, identify the family, then the constants, before you look at the options. You are practising the three-step identification: find a, find u, and check du.

The first needs completing the square: x squared plus 6x plus 13 is x plus 3, squared, plus 2 squared. The second has a constant squared minus a variable squared under the root, with u equal to 3x, so dx is one third of du. The third is 3 times u over a plus bu, with a equal to 7 and b equal to 2.

The last one is the easiest to misfile. The square root of x squared minus 16 has the variable first, so it belongs to the secant family, whose entry is a logarithm, not an arcsine. That sign is the whole difference, and the next trap is about exactly this mistake.

39. Trap: matching the letters instead of the shape

Trap

40. Computer algebra systems

Section

Part 4

41. Two correct antiderivatives can look different

Concept

If two functions have the same derivative on an interval, their difference has derivative zero there, and a function with zero derivative on an interval is constant (the Mean Value Theorem's corollary).

\[ F'(x) = G'(x) \;\Longrightarrow\; \frac{d}{dx}\big(F(x) - G(x)\big) = 0 \]

\[ \;\Longrightarrow\; F(x) = G(x) + K \quad \text{on the interval} \]

So a table, a computer and you can all be right with three different-looking formulas, as long as they differ by a constant. The constant can even be zero, when the formulas are secretly identical.

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 311 — remark before Example 3.36

This slide is the mathematical reason behind everything in this part of the lesson. If two functions have the same derivative on an interval, then their difference has derivative zero on that interval. And a function whose derivative is zero everywhere on an interval is constant there, which is a consequence of the Mean Value Theorem from Calculus I.

So two correct antiderivatives of the same integrand always differ by a constant, on each interval where they are defined. That constant is what the plus C absorbs. The book puts it this way: as long as the difference in the two antiderivatives is a constant, they are equivalent.

Keep the special case in mind. The constant can be zero, which means the two formulas are secretly the same function written differently. Example 3.38 is exactly that case, and the inverse hyperbolic sine on the next slides is another.

42. Reading the equivalence test

Notation

Tap through the pieces. The left side is the test you can always run: differentiate both answers. The right side is the conclusion: on that interval, the answers differ by a constant.

The words on I matter more than they look. Some integrands live on two separate pieces of the number line, such as one over the square root of x squared minus 4, which exists for x at most minus 2 and for x at least 2. On a domain like that, the constant may be different on each piece, and both answers are still correct.

The last note is a caution about graphs. If you plot two answers and they look parallel, that is good evidence, and computer algebra systems will draw it for you. But a picture is only accurate to its pixels. The derivative is what proves it.

43. The book's opening pair: inverse sinh as a logarithm

Worked example

The book found two antiderivatives of one over the square root of 1 plus x squared: the inverse hyperbolic sine, and a logarithm. Show they are the same function.

\[ y = \sinh^{-1} x \iff x = \sinh y = \frac{e^{y} - e^{-y}}{2} \]

\[ 2x = e^{y} - e^{-y} \]

\[ 2x e^{y} = e^{2y} - 1 \]

\[ w^2 - 2x\,w - 1 = 0 \]

\[ w = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = x \pm \sqrt{x^2 + 1} \]

\[ e^{y} = x + \sqrt{x^2 + 1} \]

\[ \sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right) \]

\[ \sinh^{-1} 1 = 0.881374 \quad\text{and}\quad \ln\left(1 + \sqrt2\right) = 0.881374 \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 311 — the two antiderivatives of one over root 1 plus x squared

The book opens Section 3.5 by recalling that a tangent substitution and a hyperbolic sine substitution gave two different answers for the integral of one over the square root of 1 plus x squared: a logarithm, and the inverse hyperbolic sine. This slide proves they are the same function, by solving for the inverse directly.

The key move is to see a quadratic. After clearing the fraction and multiplying by e to the y, the equation is a quadratic in e to the y, and the quadratic formula gives two roots. Only one is possible: e to the y is always positive, and x minus the square root of x squared plus 1 is always negative, because the root is bigger than x in size.

The figure shows the inverse as the reflection of sinh in the line y equals x, with the point at x equal to 1 marked. The check evaluates both sides at 1: both give 0.881374. In this case the constant difference between the two antiderivatives is zero; they are identical.

44. Example 3.37: will the machine agree with you?

Prediction

\[ \int \frac{dx}{\sqrt{x^2 - 4}} \]

By a secant substitution you got the first answer below. A computer algebra system returns the second.

\[ \ln\left|\frac{\sqrt{x^2-4}}{2} + \frac{x}{2}\right| + C \qquad \ln\left|\sqrt{x^2-4} + x\right| + C \]

Decide before you reveal. The two answers look similar, and the obvious instinct is either that they are identical or that one of them has a mistake.

Look at the first answer's structure. Both terms inside the absolute value have a 2 underneath, so the whole thing is the second answer divided by 2. And the logarithm of something divided by 2 is the logarithm of that thing minus ln 2. A logarithm turns a constant factor inside into a constant added outside.

So the answers differ by ln 2, a constant, and both are correct. This is the pattern you will see most often when comparing a machine's answer with yours: a factor inside a logarithm that became a constant outside it.

45. Example 3.37: a machine's logarithm and yours

Worked example

\[ \int \frac{dx}{\sqrt{x^2 - 4}} \]

\[ \text{CAS: } \ln\left|\sqrt{x^2-4} + x\right| + C \]

\[ \text{by hand: } \ln\left|\frac{\sqrt{x^2-4}}{2} + \frac{x}{2}\right| + C \]

\[ \ln\left|\frac{\sqrt{x^2-4}}{2} + \frac{x}{2}\right| = \ln\left|\frac{\sqrt{x^2-4} + x}{2}\right| \]

\[ = \ln\left|\sqrt{x^2-4} + x\right| - \ln 2 \]

\[ \text{by hand} - \text{CAS} = -\ln 2 \approx -0.693 \]

\[ \int_3^5 \frac{dx}{\sqrt{x^2-4}} = \ln\frac{\sqrt{21} + 5}{\sqrt5 + 3} \approx 0.60438 \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 312 — Example 3.37

This is the book's Example 3.37 worked in full. The computer gives one logarithm; the secant substitution from Section 3.3, with its triangle, gives another in which both terms have a 2 underneath.

Combining over one denominator and splitting the logarithm of the quotient shows that your answer is the machine's minus ln 2, about 0.693. Since that is a constant, it disappears into the plus C.

The figure makes it visible. On both branches of the domain, your answer's curve is the machine's curve slid down by the same amount, which is what a constant difference looks like. The check uses the fact that a constant cancels in a definite integral: whichever antiderivative you use, the integral from 3 to 5 comes out to 0.60438, and a numerical estimate agrees.

46. Example 3.38 by hand: an odd power of sine

Worked example

First the answer you would get with the Section 3.2 method, so there is something to compare the machine with.

\[ \int \sin^3 x\,dx \]

\[ \sin^3 x = \left(1 - \cos^2 x\right)\sin x \]

\[ u = \cos x, \quad du = -\sin x\,dx \]

\[ \int (1 - u^2)(-du) = \int\left(u^2 - 1\right)du = \frac{u^3}{3} - u + C \]

\[ \int \sin^3 x\,dx = \frac13\cos^3 x - \cos x + C \]

\[ \frac{d}{dx}\left(\tfrac13\cos^3 x - \cos x\right) = -\cos^2 x\sin x + \sin x = \sin^3 x \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 313 — Example 3.38

Before comparing with the machine, you need your own answer, and this is the Section 3.2 method for an odd power of sine. Keep one sine for the du, and turn the remaining sine squared into one minus cosine squared.

Then the substitution u equal to cosine turns everything into a polynomial in u. The minus sign from du flips the bracket, and the power rule does the rest. Replace u by cosine x to get one third cosine cubed minus cosine.

The check differentiates that answer. The chain rule gives minus cosine squared times sine from the first term, and the second term gives plus sine. Factoring out sine leaves one minus cosine squared, which is sine squared, so the result is sine cubed, exactly the integrand.

47. Example 3.38: turning the machine's answer into yours

Worked example

\[ \text{CAS: } \int \sin^3 x\,dx = \frac{1}{12}\left(\cos 3x - 9\cos x\right) + C \]

\[ \cos 3x = \cos x\cos 2x - \sin x\sin 2x \]

\[ = \cos x\left(2\cos^2 x - 1\right) - \sin x\left(2\sin x\cos x\right) \]

\[ = 2\cos^3 x - \cos x - 2\cos x\left(1 - \cos^2 x\right) \]

\[ \cos 3x = 4\cos^3 x - 3\cos x \]

\[ \frac{1}{12}\left(4\cos^3 x - 3\cos x - 9\cos x\right) = \frac13\cos^3 x - \cos x \]

\[ \tfrac13\cos^3 2 - \cos 2 = 0.392124 = \tfrac{1}{12}\left(\cos 6 - 9\cos 2\right) \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 313 — Example 3.38 and Figure 3.12

Wolfram Alpha answers with cos 3x, which looks unrelated to your cosine cubed. The way to compare them is to rewrite cos 3x in terms of cos x, using identities you already know.

Write 3x as x plus 2x and use the addition formula, then the double-angle formulas for cos 2x and sin 2x. Replace the sine squared that appears with one minus cosine squared, and everything collapses to the triple-angle formula: cos 3x equals 4 cosine cubed minus 3 cosine. Substituting that into the machine's answer, the nine cosines and three cosines combine to twelve, and the twelve cancels.

So the two answers are identical: the constant difference is zero. The figure, the book's Figure 3.12, draws both, and the dashed machine curve never leaves the thick hand curve. The check evaluates both formulas at x equal to 2; both give 0.392124.

48. Checkpoint 3.21: the radical with a plus sign

Worked example

\[ \int \frac{dx}{\sqrt{x^2 + 4}} \]

\[ \int \frac{du}{\sqrt{u^2 + a^2}} = \ln\left|u + \sqrt{u^2 + a^2}\right| + C \]

\[ \int \frac{dx}{\sqrt{x^2 + 4}} = \ln\left(x + \sqrt{x^2 + 4}\right) + C \]

\[ \text{CAS: } \sinh^{-1}\frac{x}{2} + C \]

\[ \sinh^{-1}\frac{x}{2} = \ln\left(\frac{x}{2} + \sqrt{\frac{x^2}{4} + 1}\right) = \ln\frac{x + \sqrt{x^2+4}}{2} \]

\[ \sinh^{-1}\frac{x}{2} = \ln\left(x + \sqrt{x^2+4}\right) - \ln 2 \]

\[ \ln\left(2 + \sqrt8\right) - \ln 2 = \sinh^{-1} 1 \approx 0.88137 \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 313 — Checkpoint 3.21

This is the integral from the step-zero slide, now finished. The form is a root of a variable squared plus a constant squared, with nothing on top, which is the tangent family. The table entry gives the logarithm of x plus the root directly, with a equal to 2 and no substitution needed.

A computer answers with the inverse hyperbolic sine of x over 2. To compare, use the logarithm formula for inverse sinh that you derived earlier, with x over 2 in place of x. Simplifying the root and combining over a denominator of 2 gives your logarithm minus ln 2.

Notice that this is the same pattern as Example 3.37: a factor of 2 inside a logarithm became a constant outside. The check evaluates the definite integral from 0 to 2. The constant cancels, and both forms give 0.88137, which is also the inverse sinh of 1 from the earlier slide.

49. What a CAS will not tell you

Concept

A computer is fast and rarely wrong about the derivative, but it answers the question it assumed you asked. Read its answer the way you would read a classmate's: check it, and put back what it left out.

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 312 — Computer Algebra Systems

A computer algebra system is a remarkable tool, and the book encourages you to use one. But it answers the question it assumed you asked, and its assumptions are often not yours.

Read down the table. Many systems leave out the plus C. Some drop absolute values inside logarithms because they assume the variable is positive, or work with complex numbers where the logarithm of a negative is allowed. A system chooses one form of the answer, which may use functions your course writes differently, such as inverse sinh instead of a logarithm. It may answer with a special function such as erf, which tells you no elementary answer exists. And it rarely mentions the domain, such as the restriction to x at most ln 4 in Example 3.36.

So treat a machine's answer the way you would treat a classmate's: check it by differentiating, and put back what it left out.

50. Two sound ways to compare answers

Intuition

When your answer and a book's or machine's disagree in form, there are exactly two arguments that settle it, and a third that only suggests.

1. Differentiate both. If each gives back the integrand, both are correct. This always works
2. Subtract and simplify. If the difference is a number, the answers agree; Examples 3.37 and 3.38 did this
3. Graph both (Figure 3.12). Parallel curves are strong evidence, but a picture proves nothing

\[ \frac{d}{dx}\left[\ln\left(x + \sqrt{x^2+4}\right)\right] = \frac{1 + \frac{x}{\sqrt{x^2+4}}}{x + \sqrt{x^2+4}} = \frac{1}{\sqrt{x^2+4}} \]

When two answers disagree in form, you have two arguments that actually settle the matter, and a third that only suggests.

Differentiating both always works, because it tests exactly the property that defines an antiderivative. The formula on the slide does this for the logarithm answer to Checkpoint 3.21: the chain rule gives a compound fraction, and multiplying top and bottom by the root collapses it to one over the root, the integrand. Subtracting and simplifying is the second test, and it is the one Examples 3.37 and 3.38 used: if the difference simplifies to a number, the answers agree.

Graphing is the third, and it is useful for a quick look, especially with a machine that draws for you. But parallel curves on a screen are only evidence. Two curves can look parallel and differ by a tiny slope, so a proof needs one of the first two methods.

51. Which pairs of answers agree?

Comparison

Fill in each blank by subtracting the two answers and simplifying. Where the difference is a constant, the two are the same family of antiderivatives.

The first row is the sine and cosine pair from earlier, and the Pythagorean identity gives one half. The second is Example 3.37, where the difference is minus ln 2. The tangent and secant squared row uses the identity that one plus tangent squared is secant squared, so the difference is minus one half. The row with ln of 2x and ln x uses a log rule: ln 2x is ln 2 plus ln x.

The last row is the one that does not agree. The arctangent of x and the arctangent of 2x are different functions whose difference changes as x changes, so they are antiderivatives of different integrands. A different constant inside a function is not the same as a constant added outside.

52. When nothing will work

Section

Part 5

53. Integrals with no elementary antiderivative

Concept

Elementary functions are those built from powers, roots, exponentials, logarithms and trig functions (and their inverses) by arithmetic and composition. Every continuous function has an antiderivative, but for some of them no elementary formula exists. This was proved by Liouville in the 1830s, so no amount of cleverness will find one.

\[ \int e^{-x^2}\,dx, \qquad \int \frac{\sin x}{x}\,dx, \qquad \int \frac{dx}{\ln x}, \qquad \int \sqrt{1 + x^4}\,dx \]

Each of these defines a new function, named and tabulated because it turns up so often: the error function, the sine integral, the logarithmic integral. A CAS will answer with those names.

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5, pp. 503-507 — can we integrate all continuous functions?

Every continuous function has an antiderivative: the Fundamental Theorem of Calculus builds one as an integral with a variable upper limit. What is not guaranteed is that the antiderivative can be written using the functions you know.

The four integrals on the slide have been proved to have no elementary antiderivative. That is a theorem, going back to Joseph Liouville in the 1830s, not a statement that nobody has been clever enough yet. No substitution, no identity and no table will ever turn e to the minus x squared into a formula made of powers, roots, exponentials, logs and trig functions.

The practical lesson is to recognise these and stop. If a computer returns the error function, the sine integral or the logarithmic integral, it is telling you exactly this. And as the next slides show, not having a formula does not stop a definite integral from having a perfectly good value.

54. Exercise 274: one factor of x changes everything

Worked example

\[ \int_0^1 x e^{-x^2}\,dx \]

\[ u = -x^2, \quad du = -2x\,dx, \quad x\,dx = -\tfrac12\,du \]

\[ x = 0 \to u = 0, \qquad x = 1 \to u = -1 \]

\[ -\frac12\int_0^{-1} e^{u}\,du = -\frac12\left(e^{-1} - 1\right) \]

\[ = \frac12\left(1 - \frac1e\right) \approx 0.31606 \]

\[ \frac{d}{dx}\left(-\tfrac12 e^{-x^2}\right) = -\tfrac12 e^{-x^2}(-2x) = x e^{-x^2} \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 314 — Exercise 274

Compare this with the integral of e to the minus x squared, which has no elementary antiderivative. Here there is one extra factor of x, and that factor is the derivative of the exponent up to a constant. So step two succeeds: substitute u equal to minus x squared.

Change the limits as you go. When x is 0, u is 0, and when x is 1, u is minus 1. The integral becomes minus one half times the integral of e to the u from 0 to minus 1, and the answer is one half of one minus one over e, about 0.31606.

The figure puts the two integrals side by side. The area under x times e to the minus x squared has an exact formula; the area under the bell, 0.7468, does not. The check differentiates minus one half e to the minus x squared: the chain rule brings down minus 2x, and the minus one half turns it into x times e to the minus x squared.

55. Trap: inventing an antiderivative for the bell

Trap

56. Estimate the area under the bell from 0 to 1

Estimation

\[ \int_0^1 e^{-x^2}\,dx \]

Choose an answer before you reveal. You have no formula, so the only tools are the curve's heights and some sense of its shape.

The curve starts at height 1 when x is 0 and falls to one over e, about 0.37, when x is 1. So the area over that unit interval must be somewhere between 0.37 and 1. That already rules out two of the options.

To choose between the others, notice the bell's shape near 0: it is flat at the top, and it stays close to 1 for a while before dropping. So the area is closer to the top than to the bottom. A numerical rule, which Section 3.6 teaches, gives 0.7468. Getting within a few hundredths by reasoning about the shape is a genuinely useful skill, because it catches errors in any numerical or machine answer.

57. A function with no formula, drawn anyway

Picture it

Each point on the rising curve is a definite integral computed numerically. The curve is perfectly well defined, it approaches the square root of pi over 2, and Section 3.6 will show you how the numbers are computed.

\[ \int_0^1 e^{-x^2}\,dx \approx 0.7468, \qquad \int_0^{\infty} e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2} \approx 0.8862 \]

Look at the rising curve. Every point on it is a definite integral, the area under the bell from 0 up to that x, computed numerically. There is no formula for this curve in terms of elementary functions, and yet here it is, drawn accurately.

The curve rises quickly at first, because the bell is tall near 0, and then levels off, because the bell's tail is thin. It approaches a ceiling at the square root of pi over 2, about 0.8862. That exact value can be proved, even though the curve itself has no elementary formula.

This is the right way to think about non-elementary integrals. They are not mysterious or undefined; they are just new functions. Statisticians tabulate this one as the error function, and Section 3.6 will show you the numerical rules that compute its values.

58. Exercise 294: an arc length that needs a table

Worked example

Find the length of the curve y equal to x squared over 4 from x equal to 0 to 8.

\[ y' = \frac{x}{2}, \quad L = \int_0^8 \sqrt{1 + \frac{x^2}{4}}\,dx \]

\[ u = \frac{x}{2}: \quad L = 2\int_0^4 \sqrt{1 + u^2}\,du \]

\[ \int \sqrt{a^2 + u^2}\,du = \frac{u}{2}\sqrt{a^2 + u^2} + \frac{a^2}{2}\ln\left(u + \sqrt{a^2 + u^2}\right) + C \]

\[ L = 2\left[\frac{u}{2}\sqrt{1 + u^2} + \frac12\ln\left(u + \sqrt{1 + u^2}\right)\right]_0^4 \]

\[ L = 4\sqrt{17} + \ln\left(4 + \sqrt{17}\right) \]

\[ L \approx 16.492 + 2.095 = 18.587 \]

\[ \text{chord} = \sqrt{8^2 + 16^2} = \sqrt{320} \approx 17.889 < 18.587 \]

OpenStax Calculus Volume 2, §3.5 Other Strategies for Integration §3.5, p. 315 — Exercise 294

Arc length is a common source of integrals that need a table, because the formula wraps a square root around one plus the derivative squared. Here the derivative is x over 2, so the integrand is the square root of 1 plus x squared over 4.

Substituting u equal to x over 2 tidies the radical into the square root of 1 plus u squared, at the cost of a factor of 2 from dx equal to 2 du. That radical is in the a squared plus u squared family, and the table entry has two pieces: an algebraic part and a logarithm. Evaluating from 0 to 4, both pieces vanish at 0, and the length is 4 root 17 plus the log of 4 plus root 17, about 18.587.

The figure draws the parabola to true scale with the straight chord between its ends. The check uses the fact that no path between two points is shorter than the straight line: the chord is about 17.889, a little less than 18.587, as it must be. A numerical estimate of the original integral also gives 18.587.

59. Where the bell lives: the normal distribution

Real world

Answer the question before you reveal it. This is where the non-elementary integral from Part 5 turns up in daily life: the normal distribution, which describes heights, test scores and measurement errors, is a rescaled version of e to the minus x squared.

The shaded area in the figure is the probability of landing within one standard deviation of the mean, and it is a definite integral with no elementary antiderivative. Its value, 0.6827, is the famous sixty-eight percent in the sixty-eight, ninety-five, ninety-nine point seven rule.

So how can a statistics textbook print a table of these values? Because a definite integral is a number, and numbers can be computed to any accuracy by numerical methods. Someone computed these areas carefully once and printed them. A normal table is really a table of integrals for one non-elementary function.

60. Putting it together

Section

Part 6

61. Pattern: facing an unfamiliar integral

Pattern

1. Simplify first. Expand, split a fraction, use an identity, or multiply by a conjugate.
2. Look for an inner derivative. If it is there up to a constant, substitute and stop.
3. Classify the form. Product: parts. Trig powers: identities. Radical of a quadratic: trig substitution. Rational: long division, then partial fractions.
4. Stuck? Change the integrand with a substitution or a rewrite and run the list again.
5. Table or CAS: match a, u and du exactly; then differentiate the answer, and restore C, absolute values and the domain.
6. No elementary answer? A definite integral still has a value: compute it numerically.

This is the whole lesson as a checklist you can carry into any exam. Read it top to bottom and notice that each step is cheaper than the one after it.

Simplify first, because algebra can remove the need for any technique. Look for an inner derivative next, because substitution is fast and succeeds often. Then classify by form, and let the form choose among parts, trig identities, trig substitution and partial fractions. If nothing fits, change the integrand with a substitution or a rewrite and go around again.

Tables and computers are the fifth step, and they come with their own discipline: match a, u and du exactly, check the answer by differentiating, and restore the constant, the absolute values and the domain. And if the integral has no elementary answer, stop searching and compute the definite value numerically.

62. Check: choosing the first move

Check

Run the strategy on this integral in order. Simplifying does nothing useful. Then ask whether an inner function's derivative is present. The inner function is x squared, its derivative is 2x, and there is an x in front, so the answer is yes.

Substitution with u equal to x squared turns the integral into one half the integral of e to the u, and the answer is one half e to the x squared plus C. Two lines.

The distractors are all instructive. Integration by parts would require integrating e to the x squared on its own, which has no elementary antiderivative, so it stalls at once. A table entry for u times e to the u does not match, because the exponent is x squared. And the non-elementary warning is about e to the minus x squared alone; the extra factor of x is exactly what rescues it.

63. Check: reading a table entry

Check

Identify a, u and du in that order. Nine x squared is the square of 3x, so u is 3x, and 16 is 4 squared, so a is 4.

The differential du is 3 dx, so dx is one third of du, and that one third comes out in front. The entry then contributes one over a, which is one quarter, and the arctangent of u over a, which is 3x over 4. The two constants multiply to one twelfth.

Each wrong option makes a specific slip. One forgets the one third from the differential, which the derivative check would catch as an answer three times too big. One turns u over a upside down. One keeps the one third but forgets the one over a. Differentiating your answer catches all three.

64. Check: comparing antiderivatives

Check

Do not decide by how the answers look. Either differentiate both, or subtract them and simplify.

Using the double-angle identity, cos 2x is one minus two sine squared, so minus cos 2x over 4 equals sine squared over 2 minus one quarter. The two answers differ by the constant one quarter, which the plus C absorbs. Differentiating either one gives sine x times cosine x, which confirms it.

The wrong options reflect common beliefs. One assumes that a different-looking form means a mistake. One assumes machines are always right and people are not. The last forgets that antiderivatives are only ever determined up to a constant, which is the whole reason for writing plus C.

65. Explain the constant

Explain it to yourself

Write your explanation before you reveal the model answer. A good answer has two parts: why the forms can differ, and why differentiating is a proof while a graph is not.

For the first part, think about where a constant can hide. Inside an identity, as sine squared plus cosine squared equal one. Inside a logarithm, as ln 2x equals ln x plus ln 2. Or in the choice of substitution, since substituting sine or cosine in the same integral gives answers a constant apart.

For the second part, the key idea is the Mean Value Theorem: equal derivatives on an interval force a constant difference. That is a logical argument, so it proves the answers agree. A graph shows curves that look parallel, which is convincing but only as accurate as the picture.

66. Exit ticket

Exit ticket

\[ \int \frac{x}{x^4 + 9}\,dx \]

This one combines two steps of the strategy. The x on top is half the derivative of x squared, so a substitution comes first. After it, the integral is one half the integral of du over u squared plus 9, which is the arctangent entry with a equal to 3.

The answer is one sixth of the arctangent of x squared over 3. The check differentiates it: the arctangent gives 9 over 9 plus x to the fourth, the chain rule gives 2x over 3, and with the one sixth in front everything simplifies to the integrand.

If you got one third instead of one sixth, you forgot the one half from the substitution; if you got one half, you forgot the one over a from the entry. Those are the two constants every table problem asks you to track.

67. Recap

Recap

\[ F' = G' \text{ on an interval} \;\Longrightarrow\; F = G + K \]

Next, Section 3.6 builds the numerical rules, the midpoint, trapezoidal and Simpson rules, that give values when formulas run out.

Stewart, Calculus: Early Transcendentals 8e, §7.5 Strategy for Integration §7.5-7.6, pp. 503-513 — strategy for integration; tables and computer algebra systems

This lesson added no new technique of integration. What it added is a way of choosing among the ones you have: simplify, look for an inner derivative, classify by form, and when stuck, change the integrand and try again.

It also added two tools and a discipline for using them. A table of integrals needs your integrand in its exact shape, with a, u and du identified. A computer algebra system is fast but may drop the constant, the absolute values and the domain. Answers from either can look different from yours and still be right, as long as the difference is a constant, and differentiating is the test that settles it.

Finally, some integrals have no elementary antiderivative, and recognising them saves you from searching forever. Their definite values still exist, and Section 3.6 builds the numerical rules, the midpoint, trapezoidal and Simpson rules, that compute them.