IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 ~7 min read

Integrating Special Functions

Once you know differentiation, the integrals of sin, cos, e^x and 1/x almost write themselves — they’re just differentiation in reverse. The trick is the linear (ax + b) version, where you have to remember to divide by a. All four are in your formula booklet, so this is one of the easiest topics to bank marks on.

📘 What you need to know

Four standard integrals: ∫sin x dx, ∫cos x dx, ∫e^x dx, ∫(1/x) dx.
For each one, the linear (ax + b) version exists — just divide by a.
Trig integrals only work in radians. Always check your GDC’s mode.
The 1/x integral gives ln|x|, not ln(x). The bars matter.
All of these are in the formula booklet — you don’t need to memorise them, just spot when to use them.

The four standard integrals

These four are the building blocks. Each one is just the reverse of a derivative you already know.

The four standards — all in your formula booklet

trig

∫sin x dx = −cos x + c

⚠ note the minus sign

trig

∫cos x dx = sin x + c

no minus sign here

exp

∫e^x dx = e^x + c

stays the same — easy

log

∫1x dx = ln|x| + c

use modulus bars

Compare these with the derivatives you already know: differentiating sin gives cos, differentiating cos gives negative sin. Reverse it, and the negative jumps onto sin. That’s why ∫sin x dx = −cos x + c.

The linear (ax + b) shortcut

If the function inside is linear — like sin(2x + 3) or e^{5x − 1} — you use the same standard, then divide by a (the coefficient of x).

The four linear versions

∫sin(ax + b) dx = −1a cos(ax + b) + c ∫cos(ax + b) dx = 1a sin(ax + b) + c ∫e^{ax + b} dx = 1a e^{ax + b} + c ∫1ax + b dx = 1a ln|ax + b| + c

✓ in formula booklet

The “divide by a” idea

∫sin(3x + 1) dx → −13 cos(3x + 1) + c

apply the standard, then divide by the coefficient of x (here, 3)

🧠

“Same shape, divide by a”

The integral looks identical to the standard one — the (ax + b) sticks together. The only extra step is to divide by a (the coefficient of x) at the front. That’s it.

Trig integrals: always radians

📍

Radians, not degrees

Calculus with sin and cos only works in radians. If your GDC is in degree mode, the answers will be wrong. Check the mode setting before you start any trig calculus problem.

Why is it ln|x| and not ln(x)?

The function 1/x is defined for negative x too, but ln(x) only exists for positive x. The modulus bars |x| let the integral work for all x ≠ 0.

In an exam, you’ll lose marks for writing ln(x) instead of ln|x|. It’s a tiny detail with a big cost — train yourself to put the bars in every time.

Three-step method

Spot · adjust · integrate

Spot the function shape. Is it sin, cos, e to a power, or 1 over something? That tells you which standard.
Identify the linear bit (ax + b). Read off a — the coefficient of x.
Apply the standard, then divide by a. Don’t forget + c (it’s an indefinite integral).

Worked examples

WE 1

The basic standard — straight from the booklet

Find ∫cos x dx.

apply the standard ∫cos x dx is one of the four standards ∫cos x dx = sin x + c no minus sign — only sin’s integral has the minus!

WE 2

Linear sin — divide by a

Find ∫ 3 sin(2x + π/3) dx.

step 1 — pull constant out I = 3 ∫sin(2x + π/3) dxstep 2 — apply standard, divide by a = 2 a = 2, so multiply by −½ I = 3 × [−½ cos(2x + π/3)] + cstep 3 — simplify I = −3/2 cos(2x + π/3) + c always check by differentiating — you should get back to 3 sin(2x + π/3) ✓

WE 3

Combined trig — integrate term by term

Find ∫ (2 sin(4x) − 3 cos(2x)) dx.

step 1 — split the integral I = 2 ∫sin(4x) dx − 3 ∫cos(2x) dxstep 2 — apply each standard first: a = 4, sin → −cos, divide by 4 2 × [−¼ cos(4x)] = −½ cos(4x) second: a = 2, cos → sin, divide by 2 −3 × [½ sin(2x)] = −3/2 sin(2x)step 3 — combine I = −½ cos(4x) − 3/2 sin(2x) + c treat each term separately — it’s much cleaner than doing it all at once!

WE 4

The 1/(ax + b) type — gives ln

Find ∫ 12x + 5 dx.

spot the standard it’s the 1/(ax + b) shape → ln|…|apply with a = 2 I = ½ ln|2x + 5| + c I = ½ ln|2x + 5| + c don’t forget the modulus bars! “ln(2x + 5)” loses marks.

WE 5

Find f(x) given f'(x) and a known point

A curve has gradient function f'(x) = 33x + 2 + e^{4 − x}. Given f(1) = ln 10 − e³, find f(x).

step 1 — integrate term by term first term: ∫3/(3x + 2) dx pull the 3 out: 3 ∫1/(3x + 2) dx = 3 × ⅓ ln|3x + 2| = ln|3x + 2| second term: ∫e^(4 − x) dx, here a = −1 = −e^(4 − x)step 2 — combine f(x) = ln|3x + 2| − e^(4 − x) + cstep 3 — use f(1) = ln 10 − e³ sub x = 1: ln|5| − e³ + c = ln 10 − e³ c = ln 10 − ln 5 = ln 2f(x) = ln|3x + 2| − e^(4 − x) + ln 2 tidy form: ln|2(3x + 2)| − e^(4 − x), using log laws

💡 Top tips

Check the formula booklet — all four standards (and their linear versions) are in there. No need to memorise.
Always divide by a when the function is linear (ax + b). Forgetting this is the #1 mistake.
Differentiate to check. If you get back to the original, your integral is correct.
Pull out constants first: ∫3 sin(2x) dx = 3 ∫sin(2x) dx — much cleaner.
Modulus bars on ln — every single time. “ln|…|” never “ln(…)”.
Radians for trig — check your GDC mode.

⚠ Common mistakes

Forgetting the negative on ∫sin x dx = −cos x + c. (∫cos x dx = +sin x — no minus.)
Not dividing by a in the linear version. ∫sin(2x) dx is not −cos(2x) + c — it’s −½ cos(2x) + c.
Writing ln(x) instead of ln|x|. The bars are not optional.
Negative coefficient: ∫e^{4 − x} dx has a = −1, so divide by −1 → answer has a minus sign.
Forgetting + c on indefinite integrals. Easy mark to lose.
GDC in degrees for trig — you’ll get a totally wrong answer.

Next up: Reverse Chain Rule — what happens when the inside isn’t just (ax + b) but a more complex function. Same idea, just with a sharper eye for spotting patterns.

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