Teacher page: integration by inspection and substitution

The one thing to do with the animation

Make them check an answer before they trust one.

The checker differentiates the proposed answer numerically and sets it beside the integrand. Pick one, move the slider, and the two readouts stay locked together.

Then tell them the real use: every integral they produce can be checked this way in their head, in about five seconds, by differentiating it back. It is the single habit that turns this from guesswork into something they can be sure of.

n = −1 is now available

At Standard Level the power could not be −1 and 1/x was simply not integrated. Here it is, and it gives ln|x|.

That is why 1/(3x + 2) becomes a logarithm rather than a power, and it is the case most often fumbled by anyone still reaching for "add one to the power".

The answers

∫ sin(2x + 5) dx	−½cos(2x + 5) + c
∫ 4x sin(x²) dx	−2cos(x²) + c
∫ 1/(3x + 2) dx	⅓ln\|3x + 2\| + c
∫ (sin x / cos x) dx	−ln\|cos x\| + c
1. ∫ cos(3x) dx	B, ⅓sin(3x) + c.
2. By inspection	B, ∫2x e^x² dx, because the 2x is already there.
3. ∫₀¹ e^2x dx	3.195, from (e² − 1)/2.

Where the marks go

1 markIdentifying the inner function. Writing u = 3x + 2 and du = 3dx makes it explicit and is never wasted.

1 markThe dividing constant, which is where most answers go wrong.

1 markThe + c, or the limits applied correctly on a definite integral.

What each wrong answer tells you

They give	What it means
3sin(3x) (Q1)	Multiplied by the inside derivative instead of dividing. They are running the chain rule forwards.
sin(3x) (Q1)	Forgot the constant entirely. A five second differentiate-back catches it.
∫sin(x²) (Q2)	They think any inside function can be handled. Be clear: without the 2x there is no elementary answer at all, and none is expected.
6.389 (Q3)	Found e² − 1 and did not halve. The commonest slip once the method is right.
3.694 (Q3)	Did not subtract the lower limit. At x = 0 the antiderivative is a half, not zero.

Other things they will say

"Can I always substitute?" Only when the inner function's derivative is present as a factor, up to a constant. A constant you can fix by dividing; a missing variable you cannot.

"Why the modulus in ln?" Because the logarithm needs a positive argument and the expression may be negative. It costs nothing to write and is occasionally required.

"Does the calculator do this?" For a definite integral, yes, and that is a sensible check. For an indefinite one it will not give you the working, and the working is the marks.

A possible order

	What is happening
1	Differentiate three chain rule examples. Then write the answers on the left and ask them to get back.
2	The pattern, with the inner function named out loud every time.
3	The checker. Establish differentiating back as the default habit.
4	Linear insides, then the x² case where the factor must already be present.
5	The logarithm case and the modulus. Questions 1 to 3.
6	A mixed set where some are not doable, so they have to decide rather than grind.

Two things not to say

Do not set an exercise where every integral works. Students then substitute blindly, and the skill being assessed is partly deciding whether the pattern is there at all.

Do not skip the formal u and du for strong students. They can see the linear cases instantly and will then be stuck the first time the substitution is not obvious.

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.

Simple substitutionEvaluate the integral of 2x(x² + 1)³ from 0 to 1 using u = x² + 1.
du = 2x dx and u runs 1 to 2, so the integral of u³ is (16 − 1)/4 = 15/4 = 3.75.
A logarithmEvaluate the integral of x/(x² + 1) from 0 to 1.
Half the integral of 1/u gives 0.5 ln 2 = 0.347.

2In context

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

Change the limitsEvaluate the integral of cos 3x from 0 to π/6.
(1/3) sin 3x evaluated gives (1/3) sin(π/2) = 1/3 = 0.333.
Why the limits changeExplain why converting the limits is safer than substituting back at the end.
The new limits belong to u, so the answer comes straight out with no return step and no risk of evaluating a u expression at an x limit, which is the usual loss of marks.

3Challenge

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

Keep the modulusState the integral of 1/x and explain why the modulus sign matters.
ln|x| + c. Without the modulus the result is undefined for negative x, so an otherwise correct integral cannot be applied on an interval left of the origin. It costs nothing to write and is sometimes required explicitly.
Choose the substitutionFor the integral of x√(x² + 9), state the substitution and why it works.
u = x² + 9, because du = 2x dx and the x outside the root supplies exactly that factor. A substitution works when its derivative is already present, up to a constant.

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.