Teacher page: integration by substitution and by parts (AA 5.16)

The one thing to do with the figure

Show the wrong choice first, deliberately.

Toggle to u = sin x and let them watch x² appear in the new integral. The method has not failed; the choice has, and the result is worse than the start.

Then switch back. The rule that falls out is simple: choose u to be the part that gets simpler when differentiated.

ln x is the exception

It has no simple integral, so it must be u, with dv = dx and v = x. That is how ∫ln x dx = x ln x − x comes out.

It is the one case where the power of x is not u, and students who have memorised "x is always u" get stuck.

The answers

∫x sin x dx	−x cos x + sin x + C.
∫ln x dx	x ln x − x + C.
∫x²e^x dx	e^x(x² − 2x + 2) + C, after two applications.
∫₀¹xe^x dx	exactly 1.
1, 2, 3	B, u = x; B, x ln x − x + C; 1.

Where the marks go

1 markStating u, dv/dx, du/dx and v explicitly.

1 markApplying the formula correctly, including the minus.

1 markRepeating it the right number of times, or solving for the repeated integral.

1 markOn a definite integral, applying the limits to the uv term as well.

What each wrong answer tells you

They give	What it means
x ln x with no minus x	The commonest wrong answer to Q2. Differentiating x ln x gives ln x + 1, and the extra 1 has to be removed.
Choosing u = sin x	Not wrong in principle, but it makes the integral worse. Show the toggle rather than saying so.
Stopping halfway on x²e^x	One application leaves ∫2xe^x, which still needs parts.
Limits on the integral only	The uv term is evaluated at the limits too. Expensive and common.

Other things they will say

"How do I choose u?" Whichever gets simpler when differentiated. A power of x eventually disappears; a sine or exponential never does.

"What if the integral comes back?" For ∫e^xsin x it does, after two applications. Call it I, rearrange and solve. Recognising that is what the question is testing.

"Will I be given a substitution?" Yes, whenever the integral is not already in reverse chain rule form. Use the one you are given.

A possible order

	What is happening
1	Try ∫x sin x with the tools they have. Let it fail.
2	The formula, then both choices side by side with the toggle.
3	The choosing rule, and ln x as the exception.
4	Repeated parts on x²e^x.
5	The integral that comes back to itself.
6	Questions 1 to 3, and a definite one for the limits point.

Two things not to say

Do not give the choosing rule before showing a bad choice. The rule is memorable only once they have seen what it prevents.

Do not skip the definite example. Forgetting the limits on the uv term is a distinct error that only appears there.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

Standard caseEvaluate the integral of xe^x from 0 to 1.
[xe^x − e^x] from 0 to 1 = 0 − (−1) = 1.
With a logarithmEvaluate the integral of ln x from 1 to e.
[x ln x − x] from 1 to e = 0 − (−1) = 1.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

With trigonometryEvaluate the integral of x sin x from 0 to π.
[−x cos x + sin x] from 0 to π = π.
Choose uFor the integral of x²ln x, state which factor should be u and why.
u = ln x. Differentiating ln x simplifies it to 1/x, while integrating it does not help. The rule of thumb is to differentiate the awkward factor and integrate the easy one.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

When it returns to itselfApply parts twice to the integral of e^xsin x and explain what to do when the original reappears.
The original integral comes back on the right with a minus sign. Treat it as an unknown, collect it on the left and divide: the answer is e^x(sin x − cos x)/2. From 0 to π that gives 12.07. The reappearance is the method working, not failing.
The wrong choiceShow what happens to the integral of xe^x if u = e^x is chosen instead.
You get x²e^x/2 minus the integral of x²e^x/2, which is worse than the original. The choice is not arbitrary: one way reduces the power of x and the other raises it.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.