3.1 Integration by Parts

1. Integration by Parts

Title

Calculus II · Section 3.1

Running the product rule backwards

2. What this lesson gives you

Objectives

Substitution undid the chain rule. This lesson undoes the product rule, and with it you can integrate a power times an exponential, a power times a sine, a logarithm, an inverse tangent, and products that seem to go round in circles.

• Derive the integration-by-parts formula from the product rule, sign and all
• Choose u and dv so the new integral is simpler than the old one, using LIATE as a guide
• Repeat the formula to bring a power of x down to a constant
• Solve for the integral when it reappears on the right-hand side
• Evaluate definite integrals by parts, and tell when substitution is the better tool

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 262 — learning objectives 3.1.1 to 3.1.3

Every differentiation rule you know has an integration technique hiding behind it. The chain rule, run backwards, became substitution. This lesson runs the product rule backwards, and the result is the single most used technique in the rest of this chapter.

The formula itself is one line, and you will derive it in three. The real skill is choosing what to feed it: which factor to call u and which to call dv. A good choice makes the problem collapse; a bad one makes it grow. Most of the lesson is about making that choice well, and about recognising the three ways the method can end: in one pass, in several passes down a ladder of powers, or in a loop that you escape by algebra.

At the end you will use the same idea with limits of integration, and you will learn when not to use it at all, because some products are really substitution problems in disguise.

3. Before anything new: the product rule

Warm-up

Write your answer before revealing. You need the product rule completely automatic for this lesson, including which factor is differentiated in each term.

The move of integrating both sides may feel strange, because you usually differentiate to get somewhere. But look at what it gives you. The left side is the derivative of a product, and integrating a derivative just returns the function, so it becomes f times g with no integral sign at all. The right side becomes two integrals.

Keep this line in view. Every step of the next few slides is just a rearrangement of it, and if you ever forget the integration-by-parts formula, including its sign, you can rebuild it from here in under a minute.

4. The formula and what it does

Section

Part 1

5. A product that substitution cannot touch

Concept

Substitution handles a product when one factor is the derivative of something inside the other.

\[ \int x\sin(x^2)\,dx = -\tfrac12\cos(x^2) + C \]

Change one symbol and substitution has nothing to grab: the x is no longer the derivative of anything inside the sine.

\[ \int x\sin x\,dx = \;? \]

There is no product rule for integrals. But the product rule for derivatives can be run backwards, and that is what integration by parts does.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 262 — introduction

Substitution works when the integrand contains a function of some inside expression together with the derivative of that inside expression. In x times the sine of x squared, the x outside is half the derivative of the x squared inside, so letting w be x squared makes everything fit.

Now take away the square. In x times sin x there is no inside expression whose derivative is sitting outside. The sine is just the sine of x, and the x is not its derivative. Every substitution you try leaves you with an integral no easier than before.

This is the gap the lesson fills. There is no rule saying the integral of a product is the product of the integrals; that is false. What there is, is a way to use the product rule for derivatives to exchange one integral for another. The rest of Part 1 builds that exchange.

6. Integrating the product rule

Concept

Let h be a product of two functions and differentiate it.

\[ h(x) = f(x)g(x) \quad\Longrightarrow\quad h'(x) = f'(x)g(x) + f(x)g'(x) \]

Integrate both sides. The integral of a derivative is the function itself.

\[ f(x)g(x) = \int g(x)f'(x)\,dx + \int f(x)g'(x)\,dx \]

Now solve for the second integral by moving the first one across the equals sign.

\[ \int f(x)g'(x)\,dx = f(x)g(x) - \int g(x)f'(x)\,dx \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 262 — derivation before Theorem 3.1

Follow the three lines slowly, because this derivation is short enough to redo on an exam whenever you doubt the formula.

The first line is the product rule, with h standing for the product. The second integrates both sides; on the left the integral of the derivative of the product gives the product back, and on the right you get two integrals, one containing f prime and one containing g prime.

The third line is pure algebra: move one of the integrals across the equals sign. That move is the only source of the minus sign in the final formula, which is why deriving it protects you from the most common error in this topic. Nothing has been evaluated yet. You have an identity that swaps the integral of f times g prime for f times g minus the integral of g times f prime.

7. The compact form: u and v

Concept

Rename the two functions and their differentials.

\[ u = f(x), \quad v = g(x), \quad du = f'(x)\,dx, \quad dv = g'(x)\,dx \]

\[ \int u\,dv = uv - \int v\,du \]

Integration by parts (Theorem 3.1) — If u and v are functions of x with continuous derivatives, then the integral of u dv equals u times v minus the integral of v du.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 262 — Theorem 3.1, equation 3.1

The function notation is correct but heavy, so the book renames things. Call the first function u and the second v. Then the differential of u is f prime of x dx and the differential of v is g prime of x dx, and the identity shrinks to the compact form in the middle of the slide.

Read the compact form as a sentence: the integral of u dv equals u v minus the integral of v du. Notice the symmetry. On the left, u is undifferentiated and v appears only as its differential. On the right, the roles swap: v is whole and u appears as its differential.

The hypothesis in Theorem 3.1, continuous derivatives, is there so that every integral in the formula makes sense. For the functions in this course it always holds on the intervals you will use.

8. Reading Theorem 3.1 piece by piece

Notation

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 262 — equation 3.1

Step through each annotation. The two roles are completely different jobs. You will differentiate u, so you want a u that gets simpler, or at least no worse, when you do. You will integrate dv, so dv has to be something you can actually integrate. And dv includes the dx; forgetting that is a common way to lose track.

The u v term is the reward. It has no integral sign, so it is simply written down in the answer. The minus integral of v du is the price, the new integral you still owe.

When you look at a formula like this, ask what each piece costs you. Here, the whole method succeeds or fails on that last piece: if the integral of v du is easier than the one you started with, you have made progress.

9. A trade, not an evaluation

Intuition

\[ \underbrace{\int u\,dv}_{\text{stuck}} = \underbrace{uv}_{\text{done}} - \underbrace{\int v\,du}_{\text{hopefully easier}} \]

The formula never finishes an integral by itself. It exchanges one integral for another, paying a finished product term as change.

Whether the exchange is worth making depends on your choice of u. A good choice makes the new integral easy; a bad one makes it worse, and then you swap the choice and trade again.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 262 — the advantage of the formula

This slide is the right mental model for everything that follows. Integration by parts does not evaluate anything. It takes an integral you are stuck on and offers you a different integral, plus a finished term.

Whether you should accept the offer depends entirely on the new integral. In a good trade, differentiating u removed the thing that made the original hard, a power of x or a logarithm, and the new integral is one you can do on sight. In a bad trade, the new integral is worse, and the right response is not to push on but to go back and choose u differently.

So after every application of the formula, pause and compare the new integral with the old one. Simpler means carry on. Worse means swap your choice. The same kind but with a lower power means apply the formula again. And the very same integral coming back means something special, which Part 4 deals with.

10. Integration by parts as area

Picture it

Plot u against v. The curve cuts the rectangle of width e and height 1 into two pieces: the area under the curve is the integral of u dv, the area beside it is the integral of v du. So one integral equals the rectangle minus the other.

\[ \underbrace{\int_1^e \ln v\,dv}_{1} = \underbrace{e\cdot 1 - 1\cdot 0}_{uv \text{ at the ends}} - \underbrace{\int_0^1 e^{u}\,du}_{e - 1} \]

This picture turns the formula into geometry. Instead of plotting against x, plot u against v, so the curve is traced out as x runs along. Here u is the natural log of v, running from the point where v is 1 and u is 0 to the point where v is e and u is 1.

The orange region, under the curve and above the v-axis, has area equal to the integral of u dv. The green region, to the left of the curve and right of the u-axis, has area equal to the integral of v du. Together they fill the rectangle whose corner is at the end of the curve, which has area u times v there, e times 1.

So one integral is the rectangle minus the other, which is the parts formula exactly. The numbers check: the orange area is 1 and the green area is e minus 1, and they add to e. The corner at the start contributes nothing here because u is zero there; in general you subtract a small rectangle at the start, which is the evaluation from a to b you will meet in Part 5.

11. Example 3.1: x sin x

Worked example

Evaluate the integral with u equal to x and dv equal to sin x dx.

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

\[ u = x, \qquad dv = \sin x\,dx \]

\[ du = 1\,dx \]

\[ v = \int \sin x\,dx = -\cos x \]

\[ \int x\sin x\,dx = (x)(-\cos x) - \int(-\cos x)(1\,dx) \]

\[ = -x\cos x + \int \cos x\,dx \]

\[ = -x\cos x + \sin x + C \]

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

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, pp. 262-263 — Example 3.1

This is the model solution, and the layout is worth copying: name u and dv, compute du and v, substitute, simplify, integrate, check. Writing du and v out separately before substituting prevents most mistakes.

Watch the signs in the substitution step. The v is minus cos x, so the u v term is minus x cos x, and the traded integral has a minus in front of a minus cosine, which makes it plus the integral of cos x. That new integral is a basic one, so the trade was a good one: differentiating x turned it into the number one and removed the obstacle.

The figure shows the answer against the integrand. Where x sin x is positive, the antiderivative climbs; where it is negative, the antiderivative falls; and at x equal to pi, where the integrand crosses zero, the antiderivative peaks. The derivative check on the last line confirms the same thing algebraically, in one line. Get into the habit of that check; it catches almost every sign error.

12. What if u is the sine?

Prediction

\[ \int x\sin x\,dx: \quad u = \sin x, \quad dv = x\,dx \]

Commit to one option before revealing. The point of this slide is to see what a bad choice looks like, so that you recognise one on sight.

If the sine is u, then dv is x dx, and integrating x gives one half x squared. The formula is still perfectly valid, but the new integral is one half x squared cos x, which has a higher power of x than the original. You traded a hard integral for a harder one.

This is the most important diagnostic in the lesson. After one application, compare the powers. If the power of x went up, you chose u backwards. Go back and swap.

13. The other choice climbs instead of descending

Concept

\[ u = \sin x, \quad dv = x\,dx \quad\Longrightarrow\quad du = \cos x\,dx, \quad v = \tfrac12 x^2 \]

\[ \int x\sin x\,dx = \tfrac12 x^2\sin x - \int \tfrac12 x^2\cos x\,dx \]

True, and useless: the power of x went from one up to two. Another pass the same way would take it to three.

The lesson is not that the formula failed. It is that the choice of u decides whether the power of x falls or climbs, and you want it to fall.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 263 — analysis of Example 3.1

The book works through the bad choice in full, and it is worth seeing written out once. Every line is correct; the formula did exactly what it promises. It just handed you the integral of one half x squared cos x, which is further from done than where you started.

Notice the mechanism. A power of x gets simpler when differentiated and more complicated when integrated. Sines and cosines do neither; they just swap into each other. So the power of x should be on the differentiating side, which is u. Put it on the integrating side, as dv, and every pass raises the power by one.

In practice you will sometimes try a choice and discover it is bad. That is normal, and not wasted effort. Recognising the climb after one line and switching is exactly what experienced integrators do.

14. Why v needs no constant

Concept

Any antiderivative of dv is a legal v. Carry a general constant K through Example 3.1 and watch it vanish.

\[ \int x\sin x\,dx = x(-\cos x + K) - \int(-\cos x + K)\,dx \]

\[ = -x\cos x + Kx + \sin x - Kx + C = -x\cos x + \sin x + C \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 263 — analysis of Example 3.1

When you integrate dv to get v, you are finding an antiderivative, and every antiderivative comes with an arbitrary constant. The picture shows several of them for dv equal to sin x dx: the curves minus cos x plus K, all vertical shifts of each other, all with derivative sin x.

The algebra on the slide carries a general K through Example 3.1. The K appears twice, once in the u v term as K times x and once in the traded integral as minus the integral of K, which is minus K times x. They cancel, and the answer is the same as before.

So you may always take the simplest v, with no constant. The only constant you need is the C added once, at the very end, to the final answer.

15. Trap: the plus sign

Trap

16. Checkpoint 3.1: x times e to the 2x

Worked example

Use u equal to x and dv equal to e to the 2x dx.

\[ \int xe^{2x}\,dx \]

\[ u = x, \qquad dv = e^{2x}\,dx \]

\[ du = dx \]

\[ v = \int e^{2x}\,dx = \tfrac12 e^{2x} \]

\[ \int xe^{2x}\,dx = \tfrac12 xe^{2x} - \int \tfrac12 e^{2x}\,dx \]

\[ = \tfrac12 xe^{2x} - \tfrac14 e^{2x} + C \]

\[ \frac{d}{dx}\left(\tfrac12 xe^{2x} - \tfrac14 e^{2x}\right) = \tfrac12 e^{2x} + xe^{2x} - \tfrac12 e^{2x} = xe^{2x} \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 263 — Checkpoint 3.1

This is the same structure as Example 3.1, with an exponential instead of a sine. The power x is u, since differentiating it gives 1, and the exponential is dv, since integrating it gives another exponential.

The only care needed is the factor from the exponent. The antiderivative of e to the 2x is one half e to the 2x, not 2 times it, because the chain rule would multiply by 2 when you differentiate back. That half appears again in the traded integral, so the final exponential term carries one quarter.

The check is short: the product rule on one half x e to the 2x gives one half e to the 2x plus x e to the 2x, and the derivative of minus one quarter e to the 2x removes the one half e to the 2x. What is left is exactly the integrand.

17. Fill in Exercise 8: x cos x

Fill the middle

\[ \int x\cos x\,dx = x\sin x - \int \square\,dx = x\sin x + \square + C \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 270 — Exercise 8

Fill both blanks before checking. This is the mirror image of Example 3.1, with cosine in place of sine, and the signs work out differently, which is the reason it is worth doing.

With u equal to x, v is sin x, so the traded integral is the integral of sin x. That integral is minus cos x. Subtracting a minus cosine gives plus cos x. If you wrote minus cos x in the second blank, you have subtracted the integrand instead of its antiderivative, or dropped one sign; trace back and find which.

Whenever you finish an integral like this, differentiate: x sin x plus cos x has derivative sin x plus x cos x minus sin x, which is x cos x.

18. Choosing u and dv

Section

Part 2

19. What makes a good trade

Concept

Two requirements, one on each side of the formula.

• dv must be integrable: you have to find v, or the formula cannot even be written down.
• du should be simpler than u: differentiating u is what makes the traded integral easier.

Powers and logarithms improve when differentiated; exponentials and sines never get worse when integrated. So the first make good u, the second good dv.

There are exactly two requirements, and they come from the two jobs. The dv job is to be integrated, so dv must be something you can integrate; otherwise you cannot write v and the formula is useless. The u job is to be differentiated, and you want that to make things simpler.

The table shows how the common kinds of function behave. A power of x becomes a lower power when differentiated and a higher power when integrated. A logarithm becomes one over x when differentiated, which is algebraic and much friendlier, but its antiderivative is more complicated. Exponentials and sines stay the same kind of function either way.

So the natural pairing is: whatever improves when differentiated goes in u, and whatever does not get worse when integrated goes in dv. The next slide packages that into a five-letter list.

20. LIATE: an order for choosing u

Concept

Whichever kind in your integrand comes first in the list becomes u. The rest, together with dx, is dv. In Example 3.1 the algebraic x beats the trigonometric sine, so u was x.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, pp. 263-264 — the LIATE mnemonic

LIATE is a priority list, not a law. Scan your integrand for the kinds of function it contains, and whichever kind appears earliest in the list becomes u. Everything else, together with dx, becomes dv.

For x sin x the kinds are algebraic and trigonometric; A comes before T, so u is x. For x cubed times the log of x the kinds are logarithmic and algebraic; L comes before A, so u is the logarithm, even though the power is the more complicated-looking factor.

The book stresses that it is a guide that takes some of the guesswork out, not a guarantee. You will meet a case in a few slides where following it strictly leads you to an impossible dv. When that happens, the two requirements from the previous slide are what you fall back on.

21. Why LIATE works

Intuition

Read the list from the end. Exponentials and sines are the easiest functions to integrate, and integrating them never makes them worse, so they sit at the end: ideal dv.

Read it from the front. There is no simple integral of a lone logarithm or inverse tangent, so they cannot be dv at all; but their derivatives are algebraic, which is a real simplification. So they come first: ideal u.

\[ \frac{d}{dx}\ln x = \frac1x, \qquad \frac{d}{dx}\arctan x = \frac{1}{1+x^2} \]

Algebraic functions sit in the middle because they are easy both ways.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 264 — why the mnemonic works

The order is not arbitrary; it follows from the two requirements. Start at the end of the list. Exponentials and sines and cosines are about the easiest functions there are to integrate, and integrating them never makes them worse. That makes them ideal dv, so they go last.

Now the front. You have no simple formula for the integral of a lone logarithm or a lone inverse tangent, so they cannot serve as dv at all. But their derivatives are algebraic, as the formulas on the slide show, which is a genuine simplification. That makes them ideal u, so they go first.

Algebraic functions sit in the middle because they are easy in both directions. They should be u when paired with an exponential or a trig function, and dv when paired with a logarithm or an inverse trig function, which is exactly what their position says.

22. Example 3.2: ln x over x cubed

Worked example

\[ \int \frac{\ln x}{x^3}\,dx \]

\[ \int \frac{\ln x}{x^3}\,dx = \int x^{-3}\ln x\,dx \]

\[ u = \ln x, \qquad dv = x^{-3}\,dx \]

\[ du = \frac1x\,dx, \qquad v = -\tfrac12 x^{-2} \]

\[ = (\ln x)\left(-\tfrac12 x^{-2}\right) - \int\left(-\tfrac12 x^{-2}\right)\frac1x\,dx \]

\[ = -\tfrac12 x^{-2}\ln x + \tfrac12\int x^{-3}\,dx \]

\[ = -\tfrac12 x^{-2}\ln x - \tfrac14 x^{-2} + C \]

\[ = -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C \]

\[ \frac{d}{dx}\left(-\tfrac12 x^{-2}\ln x - \tfrac14 x^{-2}\right) = x^{-3}\ln x - \tfrac12 x^{-3} + \tfrac12 x^{-3} \]

\[ = x^{-3}\ln x = \frac{\ln x}{x^3} \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 264 — Example 3.2

The first move is to see a quotient as a product: one over x cubed is x to the minus three. Now the integrand has a logarithm and a power, and LIATE says the logarithm is u.

That choice is almost forced. You could not take the log as dv, because you would have to integrate it, and that is a harder problem than the one you started with. With u equal to ln x, du is one over x dx, which will merge with the power in v.

Watch the traded integral. v is minus one half x to the minus two, and multiplying by one over x gives x to the minus three. The logarithm is gone, and what remains is a power you integrate on sight. The picture checks the answer by its shape: the integrand is negative below x equal to 1 and positive above it, so the antiderivative must fall and then rise, with its lowest point at x equal to 1. It does, and the derivative check on the last two lines makes that exact.

23. Checkpoint 3.2: x ln x

Worked example

\[ \int x\ln x\,dx \]

\[ u = \ln x, \qquad dv = x\,dx \]

\[ du = \frac1x\,dx, \qquad v = \tfrac12 x^2 \]

\[ \int x\ln x\,dx = \tfrac12 x^2\ln x - \int \tfrac12 x^2\cdot\frac1x\,dx \]

\[ = \tfrac12 x^2\ln x - \tfrac12\int x\,dx \]

\[ = \tfrac12 x^2\ln x - \tfrac14 x^2 + C \]

\[ \frac{d}{dx}\left(\tfrac12 x^2\ln x - \tfrac14 x^2\right) = x\ln x + \tfrac12 x - \tfrac12 x = x\ln x \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 265 — Checkpoint 3.2

Here the power x is the dv, which may feel wrong after Example 3.1, where x was u. LIATE settles it: a logarithm outranks an algebraic function, because you cannot integrate the log on its own but you can differentiate it.

Integrating x does raise the power, to one half x squared, but that does not matter this time. The du is one over x dx, and it cancels one power of x from v, leaving one half x in the traded integral. The log has disappeared, which is what made the trade worthwhile.

The check uses the product rule on one half x squared ln x, which gives x ln x plus one half x. The derivative of minus one quarter x squared removes the one half x. Keep practising this check until you do it without being asked.

24. Exercise 7: ln x on its own

Worked example

A single function is still a product: multiply it by 1.

\[ \int \ln x\,dx = \int 1\cdot\ln x\,dx \]

\[ u = \ln x, \qquad dv = dx \]

\[ du = \frac1x\,dx, \qquad v = x \]

\[ \int \ln x\,dx = x\ln x - \int x\cdot\frac1x\,dx \]

\[ = x\ln x - x + C \]

\[ \frac{d}{dx}\left(x\ln x - x\right) = \ln x + x\cdot\frac1x - 1 = \ln x \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 270 — Exercise 7 and its hint

At first sight this integral has no product in it at all. The trick, given as a hint in the book, is to see ln x as one times ln x. The one is an algebraic function, and dv is just dx.

The logarithm must be u, since integrating it is exactly what you cannot yet do. Its derivative is one over x, and v, the integral of dx, is x. In the traded integral the x and the one over x cancel, leaving the integral of 1. The whole integral collapses in two lines.

Keep this move in your toolkit: whenever you meet a lone logarithm or a lone inverse trig function, set dv equal to dx. The same trick evaluates the integral of arctan x, which is the heart of Example 3.6, and the integral of arcsin x.

25. Match each integral to its u

Matching

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 270 — Exercises 1 to 5

Match each one before checking. These are the first four exercises in the book's set, which ask only for the choice, not the evaluation. Choosing well is the skill; the rest is bookkeeping.

For the exponential and the cosine, the power of x or y is u, because algebraic comes before trigonometric and exponential. For the logarithm and the inverse tangent, the power is dv, because the log and the arctan cannot be integrated directly but differentiate into something algebraic.

The fifth exercise, e to the 3x times sin 2x, is the case where the book says it usually does not matter: both factors are the kind that stays the same. That integral goes round in a loop, as you will see in Part 4.

26. Find the error: v from the wrong operation

Error analysis

Read the worked line first and see whether you can spot the problem before revealing the notes. The choice of u is right, and the structure of the formula is right.

The error is in v. The writer differentiated e to the 3x, getting 3 e to the 3x, instead of integrating it, which gives one third e to the 3x. The two operations point in opposite directions for an exponential with a coefficient: differentiating multiplies by 3, integrating divides by 3.

The derivative check catches it at once. Differentiating the wrong answer gives nine times the integrand, because the factor of 3 was multiplied in twice instead of divided out twice. Whenever your check is off by a constant factor like this, look for a v or a traded integral computed in the wrong direction.

27. Before integrating t cubed times e to the t squared

Step zero

\[ \int t^3 e^{t^2}\,dt \]

Write your answer before revealing. This slide exists to stop you applying LIATE without thinking.

LIATE would make t cubed the u and e to the t squared dt the dv. But then you would have to integrate e to the t squared, and that function has no antiderivative made of elementary functions. The choice fails the first requirement: dv must be integrable.

The fix is to share the t's differently. Keep one t with the exponential. Then dv is t times e to the t squared, which substitution integrates at once, and u is t squared, which still simplifies when differentiated. LIATE is a guide; the two requirements are the rule.

28. Example 3.4: when LIATE does not quite work

Worked example

\[ \int t^3 e^{t^2}\,dt \]

\[ u = t^2, \qquad dv = t e^{t^2}\,dt \]

\[ du = 2t\,dt \]

\[ v = \int t e^{t^2}\,dt = \tfrac12\int e^{w}\,dw = \tfrac12 e^{t^2} \]

\[ \int t^3 e^{t^2}\,dt = \tfrac12 t^2 e^{t^2} - \int \tfrac12 e^{t^2}\cdot 2t\,dt \]

\[ = \tfrac12 t^2 e^{t^2} - \int t e^{t^2}\,dt \]

\[ = \tfrac12 t^2 e^{t^2} - \tfrac12 e^{t^2} + C \]

\[ \frac{d}{dt}\left(\tfrac12 t^2 e^{t^2} - \tfrac12 e^{t^2}\right) = te^{t^2} + t^3 e^{t^2} - te^{t^2} = t^3 e^{t^2} \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, pp. 265-266 — Example 3.4

With the split from the previous slide, the example runs smoothly. u is t squared, with du equal to 2t dt, and v comes from a quick substitution: with w equal to t squared, the integral of t e to the t squared is one half e to the w.

In the traded integral something nice happens. The one half from v and the 2t from du combine into t times e to the t squared, which is exactly the integral you just did for v. So the second integral costs nothing.

The picture confirms the shape. The integrand is an odd function, negative for negative t and positive for positive t, so its antiderivative must fall and then rise, with the turn at t equal to zero. The antiderivative is even, and its lowest point is minus one half at t equal to zero. The derivative check on the last line makes it exact.

29. Not every product needs parts

Counterexample

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 262 — introduction

Write your example before revealing. The claim sounds reasonable, and it is exactly the overgeneralisation that costs time on exams.

The example comes straight from the start of the section: x times the sine of x squared. The x in front is half the derivative of the x squared inside, so substitution with w equal to x squared finishes it in two lines. Integration by parts on it, with u equal to x, would need you to integrate the sine of x squared as dv, which is impossible in elementary terms.

So before you reach for parts, spend five seconds looking for an inside function whose derivative is sitting outside. If you find one, substitute. Parts is for products of genuinely different kinds of function.

30. Repeated application

Section

Part 3

31. Each pass lowers the power by one

Concept

With a power of x as u and an exponential or sine as dv, one pass trades the power for a power one lower, and leaves the other factor the same kind of function.

\[ \int x^n e^{ax}\,dx = \frac{1}{a}x^n e^{ax} - \frac{n}{a}\int x^{n-1}e^{ax}\,dx \]

So the power walks down a ladder, one rung per pass, until it reaches zero and the last integral is a plain exponential.

\[ x^2 \;\xrightarrow{d/dx}\; 2x \;\xrightarrow{d/dx}\; 2 \;\xrightarrow{d/dx}\; 0 \]

A power n needs n passes. It cannot go on forever, because the power drops every time and cannot drop below zero.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 265 — applying the formula more than once

The first line is the general version of what you did in Checkpoint 3.1. With the power as u and the exponential as dv, one pass produces a finished term plus an integral with the power one lower. The exponential stays an exponential, picking up a factor of one over a.

So the process is a ladder. A power of two becomes a power of one, then a constant, then the derivative is zero and there is nothing left to trade. Each pass walks down one rung.

That is also why the process has to stop. The power is a whole number that goes down by exactly one each time, and it cannot go below zero. A power of n needs n passes, no more. When you see a power of x multiplied by an exponential or a sine or cosine, you can predict the amount of work before you start.

32. Example 3.3: x squared times e to the 3x

Worked example

\[ \int x^2 e^{3x}\,dx \]

\[ u = x^2, \quad dv = e^{3x}\,dx, \quad du = 2x\,dx, \quad v = \tfrac13 e^{3x} \]

\[ \int x^2 e^{3x}\,dx = \tfrac13 x^2 e^{3x} - \int \tfrac23 xe^{3x}\,dx \]

\[ u = x, \quad dv = \tfrac23 e^{3x}\,dx, \quad du = dx, \quad v = \tfrac29 e^{3x} \]

\[ = \tfrac13 x^2 e^{3x} - \left(\tfrac29 xe^{3x} - \int \tfrac29 e^{3x}\,dx\right) \]

\[ \int \tfrac29 e^{3x}\,dx = \tfrac{2}{27}e^{3x} \]

\[ = \tfrac13 x^2 e^{3x} - \tfrac29 xe^{3x} + \tfrac{2}{27}e^{3x} + C \]

\[ \left[\left(\tfrac23 x - \tfrac29\right) + 3\left(\tfrac13 x^2 - \tfrac29 x + \tfrac{2}{27}\right)\right]e^{3x} \]

\[ = \left(\tfrac23 x - \tfrac29 + x^2 - \tfrac23 x + \tfrac29\right)e^{3x} = x^2 e^{3x} \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 265 — Example 3.3

The first pass follows LIATE: x squared is u, e to the 3x is dv. The new integral has only x to the first power in it, so this was progress, but it is not finished; you need a second pass.

The second pass is the key moment. Make the same kind of choice as the first time: the power is u again. And put the whole result of the second pass inside brackets before you do anything else, because the formula subtracts all of it. The book takes dv as two thirds e to the 3x on this pass, which just carries the constant along.

Removing the brackets flips both signs inside them, which is why the final term is plus two twenty-sevenths. The check factors out e to the 3x: the derivative of the bracket plus three times the bracket, which collapses to x squared, exactly the integrand.

33. The bookkeeping table

Picture it

Differentiate down the left column until you hit zero, integrate down the right column the same number of times, then multiply along the diagonals with alternating signs. Each diagonal is one pass of the formula, done in a single line.

When you need several passes with the same kind of choice each time, this table does all of them at once. It is sometimes called the tabular method, and it is just a compact way of writing what you did on the previous slide.

Differentiate u down the left column until it reaches zero. Integrate dv down the right column the same number of times. Then multiply along each diagonal arrow, starting with a plus and alternating signs. The three diagonals here give one third x squared e to the 3x, minus two ninths x e to the 3x, plus two twenty-sevenths e to the 3x, exactly the answer of Example 3.3.

The alternating signs are the minus sign of the parts formula, applied once per pass. The table stops because the left column hits zero, which only happens when u is a power of x. So use it for the ladder cases, and not for the circular cases of Part 4.

34. Checkpoint 3.3: x squared sin x

Worked example

\[ \int x^2\sin x\,dx \]

\[ u = x^2, \quad dv = \sin x\,dx, \quad du = 2x\,dx, \quad v = -\cos x \]

\[ \int x^2\sin x\,dx = -x^2\cos x + \int 2x\cos x\,dx \]

\[ u = 2x, \quad dv = \cos x\,dx, \quad du = 2\,dx, \quad v = \sin x \]

\[ \int 2x\cos x\,dx = 2x\sin x - \int 2\sin x\,dx = 2x\sin x + 2\cos x \]

\[ \int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x + C \]

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

\[ = x^2\sin x \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 267 — Checkpoint 3.3

Another ladder with two rungs: the power of x is u on both passes, and the sine then the cosine are dv.

The signs are the part to watch. On the first pass v is minus cos x, so the finished term is minus x squared cos x and the traded integral is plus the integral of 2x cos x. On the second pass v is sin x, and the traded integral is minus the integral of 2 sin x, which is plus 2 cos x.

Differentiating the answer produces six terms, and the check works only if every one of them cancels except x squared sin x. The two terms with 2x cos x cancel, and so do the two with 2 sin x. If you get a leftover term in a check like this, it points straight at the pass where a sign went wrong.

35. Trap: dropping the brackets on the second pass

Trap

36. Exercises 48 and 34: a reduction formula

Worked example

Do the pass once for a general power n, then reuse it.

\[ \int x^n e^x\,dx \]

\[ u = x^n, \quad dv = e^x\,dx, \quad du = nx^{n-1}\,dx, \quad v = e^x \]

\[ \int x^n e^x\,dx = x^n e^x - n\int x^{n-1}e^x\,dx \]

\[ \int x^3 e^x\,dx = x^3 e^x - 3\int x^2 e^x\,dx \]

\[ = x^3 e^x - 3\left(x^2 e^x - 2\int xe^x\,dx\right) \]

\[ = x^3 e^x - 3x^2 e^x + 6\left(xe^x - e^x\right) \]

\[ = \left(x^3 - 3x^2 + 6x - 6\right)e^x + C \]

\[ \left(3x^2 - 6x + 6\right)e^x + \left(x^3 - 3x^2 + 6x - 6\right)e^x = x^3 e^x \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, pp. 270-271 — Exercises 34 and 48

Instead of redoing the same pass for every power, do it once for a general power n. The result, Exercise 48 in the book, is called a reduction formula because the new integral has the power reduced by one.

Then apply it as many times as you need. For x cubed e to the x, use it with n equal to 3, then 2, then 1, each time substituting the new line into the old. Notice the coefficients building up: 3, then 3 times 2, then 3 times 2 times 1. That is why the constant at the end is 6.

The final check is elegant. For any polynomial times e to the x, the derivative is the derivative of the polynomial plus the polynomial itself, all times e to the x. Here that adds three x squared minus six x plus six to the bracket, and everything except x cubed cancels.

37. Count the passes

Comparison

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 270 — Exercises 10, 15, 7 and 19

Fill in each blank before checking. The rule you are practising is simple: a power of n needs n passes when it is paired with an exponential, a sine or a cosine.

Two rows are different. The integral of ln x needs only one pass, because with dv equal to dx the traded integrand is just the constant 1. And e to the x sin x never reaches a constant at all, because neither factor gets simpler when differentiated; two passes bring the original integral back and you solve for it.

Being able to predict the length of a calculation before you start is valuable. It tells you when a long calculation is expected and when it means you have chosen badly.

38. Which ending does it need?

Sorting

Sort all six before checking. You are not evaluating anything, only predicting how integration by parts will finish, which is exactly the judgement you need at the start of every problem.

A power of x times an exponential or a trig function walks down the ladder, one pass per power. A lone logarithm or inverse trig function, or one multiplied by a power, differentiates into something algebraic after one pass and then finishes with a basic integral or a substitution.

The loops are the products where nothing ever simplifies: an exponential times a sine or cosine, which just keep turning into each other, and a trig function of ln x, where the ln x differentiates into one over x and gets cancelled by v equal to x every time. Those need Part 4.

39. The circular case

Section

Part 4

40. When the original integral comes back

Concept

Sometimes two passes of parts return the very integral you started with. Give it a name, I, and the result is an equation.

\[ I = A(x) - I \]

That is not a dead end. Add I to both sides and divide by two.

\[ 2I = A(x) \quad\Longrightarrow\quad I = \tfrac12 A(x) + C \]

The only failure is a coefficient that cancels the I completely, which is exactly what a careless second pass produces (see the trap below).

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, pp. 266-267 — Example 3.5

Some integrals refuse to simplify. Differentiating a sine gives a cosine, and differentiating a cosine gives back a minus sine, so after two passes you are looking at the integral you started with. It seems as if you have gone in a circle.

The escape is to treat the integral as an unknown. Call it I. The two passes produce an equation in which I appears on both sides, with some finished terms, here called A of x, alongside. That is a linear equation in I, and you solve it the way you solve any linear equation.

The only way it can fail is if the I terms cancel completely, leaving a true but useless equation. That happens when you undo the first pass on the second one, which the trap later in this part shows you how to avoid.

41. Example 3.5: sin of ln x

Worked example

\[ I = \int \sin(\ln x)\,dx \]

\[ u = \sin(\ln x), \quad dv = dx, \quad du = \frac{\cos(\ln x)}{x}\,dx, \quad v = x \]

\[ I = x\sin(\ln x) - \int \cos(\ln x)\,dx \]

\[ u = \cos(\ln x), \quad dv = dx, \quad du = -\frac{\sin(\ln x)}{x}\,dx, \quad v = x \]

\[ \int \cos(\ln x)\,dx = x\cos(\ln x) + \int \sin(\ln x)\,dx \]

\[ I = x\sin(\ln x) - x\cos(\ln x) - I \]

\[ 2I = x\sin(\ln x) - x\cos(\ln x) \]

\[ I = \tfrac12 x\sin(\ln x) - \tfrac12 x\cos(\ln x) + C \]

\[ \tfrac12\sin(\ln x) + \tfrac12\cos(\ln x) - \tfrac12\cos(\ln x) + \tfrac12\sin(\ln x) \]

\[ = \sin(\ln x) \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, pp. 266-267 — Example 3.5 and its analysis

This integral appears to have only one function in it, so the first move is the one from Exercise 7: the other factor is 1, and dv is dx. The sine of ln x must be u, because if you could integrate it you would not need parts.

In the first pass, du has a one over x in it, and v is x, so they cancel, leaving the integral of cos of ln x. That is no simpler, but it is the same kind of integral, so apply parts again with the same kind of choice. The second pass returns the integral of sin of ln x with a plus sign, and substituting it back gives I on both sides.

From there it is algebra: add I to both sides, divide by 2, and only then add the constant. The two-panel picture shows why no simpler route exists: the integrand oscillates ever more slowly as x grows, and the antiderivative climbs and falls in step, peaking exactly where the integrand crosses zero at e to the pi. The derivative check is how you should always settle any doubt about a circular answer.

42. Going round the loop

Picture it

Follow the arrows: two passes carry you back to where you started, but with extra terms picked up on the way. Those terms are the answer; the returning I is what lets you isolate them.

The diagram draws the argument of Example 3.5 as a journey. You start at I, the first pass takes you to a finished term minus the integral of cos of ln x, and the second pass takes you to two finished terms minus I.

The arrow back to the start is the moment of recognition: the integral on the right is the one you began with. Instead of treating that as failure, you treat it as an equation. The finished terms you picked up on the way round are what remains when you solve it, and that is the answer.

Keep the picture in mind as a test. If your loop comes back with I and a coefficient of plus one on the right, the I terms cancel and you have gone round without collecting anything. That is the signal that you switched roles on the second pass.

43. Reading the circular equation

Notation

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 267 — Example 3.5

Step through the annotations. The left side is the unknown. The two finished terms on the right came from the two passes, one each. The last term is the returning integral, and its coefficient decides everything.

Here the coefficient is minus one, so moving it to the left gives two I, and dividing by 2 finishes the job. In other problems the coefficient can be different; for example, an exponential with a coefficient in the exponent times a sine gives a returning integral with some other multiple in front, and you divide by one plus that multiple instead.

The one coefficient that cannot be solved is plus one, because then the I terms cancel. That is the failure the next trap is about.

44. Why solving for I is legitimate

Intuition

I stands for a family of antiderivatives, and every line of the argument is an equation between such families. Adding I to both sides and halving is ordinary algebra on them.

Constants are the one subtlety. Two antiderivatives of the same function can differ by a constant, so the equation really holds up to a constant; that is why you add C only at the very end, once, to the solved-for answer.

And the answer is checkable like any other: differentiate it and you get the integrand back. That check is the real guarantee.

It can feel like cheating to solve for an integral as if it were a number. It is not, but it is worth understanding why.

The symbol I stands for the collection of all antiderivatives of the integrand. Each pass of parts is an honest identity between such collections, so the final equation is true, and adding I to both sides and halving is legitimate algebra. The only subtlety is that antiderivatives are defined up to a constant, so the equation really holds up to a constant. That is exactly why you leave out C while solving and put it in once at the end.

And if any doubt remains, the derivative check removes it. An answer whose derivative is the integrand is an antiderivative, however it was found.

45. Exercise 19: e to the x times sin x

Worked example

\[ I = \int e^x\sin x\,dx \]

\[ u = \sin x, \quad dv = e^x\,dx, \quad du = \cos x\,dx, \quad v = e^x \]

\[ I = e^x\sin x - \int e^x\cos x\,dx \]

\[ u = \cos x, \quad dv = e^x\,dx, \quad du = -\sin x\,dx, \quad v = e^x \]

\[ \int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx = e^x\cos x + I \]

\[ I = e^x\sin x - e^x\cos x - I \]

\[ I = \tfrac12 e^x(\sin x - \cos x) + C \]

\[ \tfrac12 e^x(\sin x - \cos x) + \tfrac12 e^x(\cos x + \sin x) = e^x\sin x \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 270 — Exercise 19

This is the classic circular integral. Both factors are the kind that never simplify, so LIATE lets you choose either as u. Here the sine is u; choosing the exponential would work equally well, as long as you make the same choice on both passes.

The first pass turns the sine into a cosine. The second pass, with the cosine as u, turns it back into a sine, with a plus sign that comes from the minus sine derivative meeting the minus in the formula. Substituting gives I equal to e to the x sin x minus e to the x cos x minus I, and halving finishes it.

The picture shows the answer behaving as an antiderivative must. Every zero of the integrand, at minus pi and at zero, is a turning point of the antiderivative, and the antiderivative climbs steeply exactly where the integrand is largest. The derivative check adds the bracket to its own derivative and gets two sin x, which the one half turns into sin x.

46. Trap: switching roles on the second pass

Trap

47. Put the circular method in order

Ranking

Drag the steps into order before checking. Most of the order is forced: you cannot recognise the returning integral before you have done the two passes, and you cannot solve for it before you have recognised it.

The two places people go wrong are the second pass and the constant. The second pass must use the same kind of u as the first, or the loop collapses. And the constant goes in after solving, not during, because it belongs to the final family of antiderivatives.

The derivative check comes last, and for circular integrals it is especially worth doing, because the solving step can hide a sign error that the check exposes immediately.

48. Definite integrals, and when not to use parts

Section

Part 5

49. Parts with limits

Concept

Integrate the product rule from a to b instead. The Fundamental Theorem turns the integral of a derivative into an evaluation.

\[ \int_a^b (uv)'\,dx = uv\Big|_a^b = \int_a^b v\,du + \int_a^b u\,dv \]

Solve for the integral of u dv exactly as before.

\[ \int_a^b u\,dv = uv\Big|_a^b - \int_a^b v\,du \]

Theorem 3.2 — If u and v have continuous derivatives on the interval from a to b, the definite integral of u dv equals uv evaluated from a to b, minus the definite integral of v du.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 267 — Theorem 3.2, equation 3.2

Nothing new is needed for definite integrals: run the same derivation with limits on every integral. The Fundamental Theorem of Calculus says the integral from a to b of the derivative of u v is u v evaluated at b minus its value at a. That replaces the product term.

Solving for the integral of u dv then gives Theorem 3.2. It has the same shape as before, but the finished term is now a number, u v evaluated between the limits, and the traded integral is a definite integral over the same interval.

The practical rule is short. Keep the limits attached to both pieces throughout. You can either evaluate as you go, which is what the book does, or find the antiderivative first and evaluate once at the end. Both give the same number.

50. Reading Theorem 3.2 piece by piece

Notation

Step through the annotations. The limits a and b are values of x, the original variable, on all three pieces. That matters if you also substitute somewhere, because then limits must be changed; in a pure parts calculation they never change.

The evaluated term is where most of the marks are lost. It means u times v at the upper limit minus u times v at the lower limit, and it is a number. Leaving it as an expression is a common error.

Finally, notice what is missing: there is no constant. A definite integral is a number, and any constant you might have put into v would cancel between the two ends.

51. Estimate the area under arctan x

Estimation

\[ \text{Area} = \int_0^1 \tan^{-1} x\,dx \]

\[ \frac{\pi}{8} \approx 0.393 < \text{Area} < 0.5 \]

Commit to an estimate before revealing. Estimating first is not a formality; it is how you will know whether the exact answer on the next slide is believable.

The curve arctan x bends downward on this interval, so it lies above the straight chord from the origin to its value at 1, which is about 0.785. The triangle under that chord has area about 0.393, so the region has more area than that. The curve also lies below the line y equal to x, whose triangle has area one half. So the answer is trapped between about 0.39 and 0.5.

Only one option fits. This kind of bracketing costs a few seconds and catches wrong signs and dropped terms, which on a definite integral would usually push the answer outside the bracket.

52. Example 3.6: the area under arctan x

Worked example

Find the area of the region under y equal to arctan x, above the x-axis, over the interval from 0 to 1.

\[ \text{Area} = \int_0^1 \tan^{-1} x\,dx \]

\[ u = \tan^{-1} x, \quad dv = dx, \quad du = \frac{dx}{x^2+1}, \quad v = x \]

\[ \text{Area} = x\tan^{-1} x\Big|_0^1 - \int_0^1 \frac{x}{x^2+1}\,dx \]

\[ x\tan^{-1} x\Big|_0^1 = 1\cdot\frac{\pi}{4} - 0 = \frac{\pi}{4} \]

\[ \int_0^1 \frac{x}{x^2+1}\,dx = \tfrac12\ln(x^2+1)\Big|_0^1 = \tfrac12\ln 2 \]

\[ \text{Area} = \frac{\pi}{4} - \frac12\ln 2 \approx 0.4388 \]

\[ 0.393 < 0.4388 < 0.5 \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, pp. 267-268 — Example 3.6 and Figure 3.2

The inverse tangent cannot be integrated directly, so this is the ln x trick again: u is the inverse tangent, dv is dx, and v is x. The du is one over x squared plus one, dx.

Theorem 3.2 gives a boundary term and a traded integral, both over the interval from 0 to 1. The boundary term is 1 times arctan of 1, which is a quarter of pi, minus zero. The traded integral, x over x squared plus 1, is a substitution: the numerator is half the derivative of the denominator, so it gives one half the log of x squared plus one, which is one half ln 2 over this interval.

The exact area is a quarter of pi minus one half ln 2, about 0.4388. The picture shows it sitting between the two triangles, just as the estimate required, and Simpson's rule computed numerically agrees to four decimal places. Two independent checks, both passed.

53. Example 3.7: a volume by shells

Worked example

Revolve the region under f(x) equal to e to the minus x, from x equal to 0 to 1, about the y-axis.

\[ V = 2\pi\int_0^1 xe^{-x}\,dx \]

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

\[ \int_0^1 xe^{-x}\,dx = -xe^{-x}\Big|_0^1 + \int_0^1 e^{-x}\,dx \]

\[ = -\frac1e + \left(1 - \frac1e\right) = 1 - \frac2e \]

\[ V = 2\pi\left(1 - \frac2e\right) = 2\pi - \frac{4\pi}{e} \approx 1.6603 \]

\[ \frac{\pi}{e} + \frac{\pi}{3}\left(1 - \frac1e\right) \approx 1.8177 > 1.6603 \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 269 — Example 3.7 and its analysis

Look at the figure first. The region under e to the minus x from 0 to 1 is being turned about the y-axis, and the natural slice is a thin vertical strip, which sweeps out a cylindrical shell. The shell at position x has radius x and height e to the minus x, so the volume integral is two pi times the integral of x e to the minus x.

That integrand is a power times an exponential, one rung of the ladder, so one pass of parts finishes it. The boundary term is minus x e to the minus x from 0 to 1, which is minus one over e. The traded integral of e to the minus x is one minus one over e. Together they give one minus two over e.

The book's reasonableness check compares the solid with a cylinder plus a cone that contain it, whose volume is about 1.8177, and the answer, about 1.6603, is a bit less, as it should be. Simpson's rule on the integral gives the same 1.6603, which is the sharper check.

54. Checkpoint 3.4: x cos x from 0 to pi over 2

Worked example

\[ \int_0^{\pi/2} x\cos x\,dx \]

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

\[ = x\sin x\Big|_0^{\pi/2} - \int_0^{\pi/2}\sin x\,dx \]

\[ x\sin x\Big|_0^{\pi/2} = \frac{\pi}{2}\cdot 1 - 0 = \frac{\pi}{2} \]

\[ \int_0^{\pi/2}\sin x\,dx = -\cos x\Big|_0^{\pi/2} = 0 + 1 = 1 \]

\[ \int_0^{\pi/2} x\cos x\,dx = \frac{\pi}{2} - 1 \approx 0.5708 \]

\[ \frac{d}{dx}(x\sin x + \cos x) = x\cos x, \quad \left(\tfrac{\pi}{2} + 0\right) - (0 + 1) = \tfrac{\pi}{2} - 1 \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 269 — Checkpoint 3.4

This is Exercise 8 with limits, so the antiderivative is already familiar: u is x, v is sin x, and the traded integral is the integral of sin x.

Evaluate each piece separately. The boundary term, x sin x, is a half pi times 1 at the top and zero at the bottom. The traded integral of sin x from 0 to a half pi is exactly 1. The answer is a half pi minus 1, about 0.571.

The picture shows a hump rising to about 0.56 and falling back to zero at a half pi, about 1.57 wide. A rough area of two thirds of width times height is about 0.59, close to the exact 0.571. The final line checks the antiderivative by differentiating it and confirms the number by evaluating it at the two ends.

55. Exercise 40: substitute first, then parts

Worked example

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

\[ x = w^2, \quad dx = 2w\,dw \]

\[ \int_0^1 e^{\sqrt x}\,dx = \int_0^1 2w e^{w}\,dw \]

\[ u = 2w, \quad dv = e^w\,dw, \quad du = 2\,dw, \quad v = e^w \]

\[ = 2we^{w}\Big|_0^1 - \int_0^1 2e^{w}\,dw \]

\[ = 2e - (2e - 2) = 2 \]

\[ 1 < 2 < e, \qquad \text{Simpson: } 2.0000 \]

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 271 — Exercise 40

The book's hint is to substitute first. The square root inside the exponential is the obstacle, so let w be the square root of x. Then x is w squared, dx is 2w dw, and because the square root of 0 is 0 and of 1 is 1, the limits do not even change.

After the substitution a product has appeared: 2w times e to the w. That is a power times an exponential, so one pass of parts finishes it. The boundary term is 2e, the traded integral is 2e minus 2, and the answer is exactly 2.

Techniques combine. Substitution can create a product that needs parts, and parts can leave an integral that needs substitution, as in Example 3.6. The check here is numeric: the integrand runs from 1 to e on an interval of length 1, so the answer must lie between 1 and about 2.72, and Simpson's rule agrees with 2 to four decimals.

56. Substitution or parts: read the integrand first

Concept

Two integrals that differ by one symbol need different tools.

\[ \int xe^{x^2}\,dx: \quad w = x^2 \;\Longrightarrow\; \tfrac12 e^{x^2} + C \]

\[ \int xe^{x}\,dx: \quad u = x, \; dv = e^x\,dx \;\Longrightarrow\; (x - 1)e^x + C \]

Look first for a factor that is, up to a constant, the derivative of something inside another factor. If you find one, substitute. If the factors are unrelated kinds, a power with an exponential, a logarithm with a power, use parts.

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 271 — Exercises 52 to 57

These two integrals differ by a single exponent, and they need completely different tools. In x times e to the x squared, the x is half the derivative of the x squared in the exponent, so substitution finishes it in one line. In x times e to the x, the x is not the derivative of anything in the exponent, so it needs parts.

The general test is on the slide. Look for an inside function whose derivative, up to a constant multiple, is sitting outside as a factor. If you find one, substitute. If the factors are simply different kinds of function, a power and an exponential, a logarithm and a power, use parts.

Try substitution first because it is faster when it works, and it fails quickly when it does not.

57. Substitution or parts?

Sorting

Sort all seven before checking, using the test from the previous slide on each one.

The substitution cases all have an inside function with its derivative outside. In the log squared over x, the one over x is the derivative of ln x. In the two exponentials of quadratics, the x outside is a constant times the derivative of the quadratic. In x squared times the sine of a cubic, the x squared is a constant times the derivative of the cubic.

The parts cases have no such relationship. A power times a logarithm, a power times an exponential of x, a power times a sine of x: in each, the factors are unrelated, and LIATE tells you which is u. Exercise 56 and Exercise 57 make the contrast sharp: x squared times sin x needs parts, while x squared times the sine of a cubic needs substitution.

58. Slide a and watch the area level off

Tweak it

\[ \int_0^{x} te^{-at}\,dt = \frac{1 - (1 + ax)e^{-ax}}{a^2} \]

Start with a equal to 1 and watch the curve. The area under t times e to the minus t climbs and then flattens toward 1. Now slide a down toward one half: the curve climbs higher and more slowly, flattening toward 4. Slide a up to 3: it flattens almost at once, near one ninth.

The formula under the slider, found by one pass of parts, explains the pattern. As x grows, the exponential term dies away, however large the factor in front of it, and the area approaches one over a squared.

So halving the decay rate multiplies the total area by four. The power of x in front delays the decay, and the slower the exponential, the longer the delay lasts. You will meet this same integral, with x running off to infinity, when improper integrals arrive in Section 3.7.

59. How far does the particle go?

Real world

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 271 — Exercise 62

Write your answer before revealing. Because the velocity is never negative, the distance travelled is simply the integral of the velocity from 0 to 2 seconds, the shaded area in the picture.

The integrand is t squared times an exponential, so it is a two-rung ladder: u is t squared on the first pass and 2t on the second. The antiderivative is minus the quantity t squared plus 2t plus 2, times e to the minus t. At 2 seconds that is minus 10 times e to the minus 2; at 0 it is minus 2. The difference is 2 minus 10 over e squared, about 0.647 feet.

The picture gives you a check. The velocity rises to about 0.54 at 2 seconds, so the shaded region is less than a rectangle 2 wide and 0.54 tall, about 1.08, and because the curve starts at zero and bends upward first, the area should be well under that rectangle. Just over half a foot is right.

60. Putting it together

Section

Part 6

61. Pattern: integrating a product

Pattern

1. Look for substitution first. If a factor is the derivative of an inside function, substitute.
2. Otherwise choose u by LIATE, and make sure dv, with its dx, can be integrated.
3. Write the formula with its minus sign, and bracket everything a second pass produces.
4. Read the new integral. Simpler: finish. Lower power: repeat. The original again: solve for I.
5. With limits, evaluate uv at both ends and keep the limits on the traded integral.
6. Differentiate your answer to check it.

This is the order in which to think about any integral that is a product. It starts with the cheapest check and ends with the one that guarantees your answer.

Substitution comes first because it is faster when it applies, and the signature is easy to spot: an inside function with its derivative outside. If that is absent, choose u by LIATE, and check that you can integrate dv before you commit.

After each pass, read the new integral. It tells you what to do next: finish it, repeat with the same kind of choice, or, if the original has returned, solve for it. With limits, evaluate the finished term at both ends and keep the limits on the traded integral. Whatever route you took, differentiate your answer at the end. It is the only step that tells you whether you are right.

62. Check: choosing u

Check

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 270 — Exercise 14

The choice of u is standard: x is algebraic, cos 3x is trigonometric, so u is x. The work is in the constants from the 3 in the argument.

Integrating cos 3x gives one third sin 3x, dividing by the 3. The traded integral is one third of the integral of sin 3x, which is minus one ninth cos 3x. Subtracting that gives plus one ninth cos 3x. If you picked the option with a minus sign, you lost one of the two minus signs. If you picked the one with 3x, you differentiated where you should have integrated.

63. Check: the circular case

Check

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 270 — Exercise 20

Keep the trig factor as u on both passes, as in Exercise 19. The first pass gives e to the x cos x plus the integral of e to the x sin x. The second pass turns that integral into e to the x sin x minus the original.

So I equals e to the x cos x plus e to the x sin x minus I, and halving gives the answer. The option without the one half forgot the division, and the I equals I option describes what happens only if you switch roles on the second pass. Differentiating the correct answer returns e to the x cos x, which settles it.

64. Check: a definite integral

Check

Each wrong option is a specific, common slip, so if you picked one, find which step you skipped.

With u equal to ln x and dv equal to dx, the boundary term is x ln x from 1 to e, which is e times 1 minus 1 times 0, so e. The traded integral is the integral of 1 from 1 to e, which is e minus 1. The answer is their difference, 1. Reporting e forgets the traded integral; reporting e minus 1 reports the traded integral instead of subtracting it; reporting 0 evaluates the antiderivative at the top and forgets the bottom, where it is minus 1.

65. Explain the circular trick

Explain it to yourself

Write your explanation before revealing. If you can explain both halves, you understand the circular case well enough to never be caught by it.

The first half is about what I means. It is a family of antiderivatives, and each pass of parts produces a true equation between such families, so solving that equation is ordinary algebra; the constant goes in at the end because the equation only holds up to a constant.

The second half is about why a swapped second pass is fatal. Integration by parts with the roles reversed is the same identity read the other way, so it exactly undoes the first pass, and the result is I equals I. Nothing is wrong with it, but nothing can be learned from it.

66. Exit ticket

Exit ticket

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

OpenStax Calculus Volume 2, §3.1 Integration by Parts §3.1, p. 270 — Exercise 10, with limits

This combines the lesson's main skills in one problem: choose u, predict the number of passes, carry out the ladder, and evaluate between limits.

The power x squared is u, since algebraic comes before exponential, and a power of two needs two passes. The antiderivative is x squared minus 2x plus 2, all times e to the x, which you can check by adding the bracket to its derivative. At 1 the bracket is 1, giving e; at 0 it is 2, giving 2. The answer is e minus 2, about 0.718.

The integrand rises from 0 to e across the interval and bends upward, so an area a little under one is sensible.

67. Recap

Recap

\[ \int u\,dv = uv - \int v\,du \]

Next, Section 3.2 meets products of powers of sines and cosines, where trigonometric identities do the work that parts did here.

Stewart, Calculus: Early Transcendentals 8e, §7.1 Integration by Parts §7.1, pp. 472-478 — the same material in Stewart

One formula, and a handful of situations that tell you how to use it. For a product of unrelated kinds of function, choose u by LIATE and trade. For a lone logarithm or inverse trig function, let dv be dx. For a power times an exponential or a trig function, repeat, one pass per power, or use the table. For an exponential times a trig function, or a trig function of ln x, go round the loop twice and solve for the integral.

With limits, the finished term is evaluated at both ends and the traded integral keeps the same limits. And when a factor is the derivative of an inside function, it was never a parts problem at all.

The next section, trigonometric integrals, handles products of powers of sines and cosines. There, identities do most of the work that parts did here, but parts comes back for integrals such as sec cubed.