IB Maths AA HL
Topic 5 — Calculus
Paper 1 & 2
~9 min read
Integrating Trigonometric, Exponential & Reciprocal Functions
The next batch of standard antiderivatives: sin and cos give back each other (with a sign flip), sec² gives tan, ex is its own antiderivative, and 1/x gives ln|x|. For linear arguments (ax + b), divide by a in front. All these are in the formula booklet — but recognising the pattern matters more than memorising.
📘 What you need to know
- Trig basics: ∫sin x dx = −cos x + c; ∫cos x dx = sin x + c; ∫sec²x dx = tan x + c.
- Exponential: ∫ex dx = ex + c (ex is its own derivative AND its own antiderivative).
- Reciprocal: ∫(1/x) dx = ln|x| + c. This is the n = −1 case where the power rule fails.
- Linear argument trick — divide by a: for any of these standard integrals, if x is replaced by (ax + b), the answer gets a factor of 1/a out front.
- Angles in radians: trig integration ONLY works with radians. Check your GDC mode.
- The sign on sin: ∫sin x dx has a MINUS sign (−cos x). ∫cos x dx has NO sign change. Don’t mix them up.
- Modulus on ln: always write ln|·| (with absolute value bars) — ln is only defined for positive arguments, but the antiderivative formula uses |·| to cover both signs.
- Formula booklet: ∫sin x, ∫cos x, ∫ex, ∫(1/x) are all listed; ∫sec²x is not, but the derivative d/dx(tan x) = sec²x is — use that “the other way round”.
The three standard integrals
Trig (radians only)
∫ sin x dx = −cos x + c
∫ cos x dx = sin x + c
∫ sec²x dx = tan x + c
Exponential & reciprocal
∫ ex dx = ex + c
∫ 1x dx = ln|x| + c
Why the minus on ∫sin? The derivative of cos x is −sin x. To go backwards from sin x to its antiderivative, you need −cos x so that differentiating gives −(−sin x) = sin x. The minus signs cancel.
Linear arguments (ax + b) — divide by a
If the argument x is replaced by a linear expression (ax + b), each formula picks up a factor of 1/a. This follows from the reverse chain rule (full discussion in the next note). For now, just apply the pattern:
| Integral | Result |
|---|
| ∫ sin(ax + b) dx | −(1/a) cos(ax + b) + c |
| ∫ cos(ax + b) dx | (1/a) sin(ax + b) + c |
| ∫ sec²(ax + b) dx | (1/a) tan(ax + b) + c |
| ∫ eax + b dx | (1/a) eax + b + c |
| ∫ 1/(ax + b) dx | (1/a) ln|ax + b| + c |
| ∫ a/(ax + b) dx | ln|ax + b| + c (special case: numerator IS the derivative of denominator) |
Sanity check: differentiate your answer. The chain rule will multiply by the derivative of (ax + b), which is a — that exactly cancels the 1/a you put out front. The result should be the integrand you started with.
🧭 Recipe — integrate a trig / exp / reciprocal function
- Identify the family: is it sin / cos / sec² / e(…) / 1/(…) ?
- Look at the argument: is it just x, or is it a linear expression (ax + b)?
- Apply the standard antiderivative: sin → −cos, cos → sin, sec² → tan, e(…) → e(…), 1/(…) → ln|…|.
- If the argument is (ax + b), multiply by 1/a (and keep any coefficient that was already on the outside).
- Add “+ c“, then verify by differentiating.
Worked examples
WE 1The three foundational integrals — direct lookup
Find each indefinite integral.
(a) ∫sin x dx (b) ∫ex dx (c) ∫(1/x) dx
(a) ∫sin x dx — use the standard formula
∫sin x dx = −cos x + c
(b) ∫e^x dx — e^x is its own antiderivative
∫e^x dx = e^x + c
(c) ∫(1/x) dx — the n = −1 special case
∫(1/x) dx = ln|x| + c
(a) −cos x + c | (b) e^x + c | (c) ln|x| + c
verify by differentiating each: d/dx(−cos x) = sin x ✓ d/dx(e^x) = e^x ✓ d/dx(ln|x|) = 1/x ✓
WE 2Trig with linear arguments (ax + b)
Find ∫(3 cos(4x + 2) − 5 sin(2x − 1)) dx.
Integrate term by term
For 3 cos(4x + 2): cos goes to sin, divide by a = 4
∫3 cos(4x + 2) dx = 3 · (1/4) sin(4x + 2) = (3/4) sin(4x + 2)
For −5 sin(2x − 1): sin goes to −cos, divide by a = 2
∫(−5) sin(2x − 1) dx = (−5) · (−(1/2)) cos(2x − 1) = (5/2) cos(2x − 1)
(double negative: minus on −5, minus on −cos → plus)
Combine and add + c
= (3/4) sin(4x + 2) + (5/2) cos(2x − 1) + c
∫(3 cos(4x + 2) − 5 sin(2x − 1)) dx = (3/4) sin(4x + 2) + (5/2) cos(2x − 1) + c
verify: d/dx of result = 3 cos(4x + 2) − 5 sin(2x − 1) ✓
WE 3sec² with linear argument
Find ∫6 sec²(2x + π/4) dx.
Recall: ∫sec²(ax + b) dx = (1/a) tan(ax + b) + c
Here a = 2, so divide by 2
∫6 sec²(2x + π/4) dx = 6 · (1/2) tan(2x + π/4) + c
= 3 tan(2x + π/4) + c
∫6 sec²(2x + π/4) dx = 3 tan(2x + π/4) + c
∫sec² isn’t in the formula booklet directly — use d/dx(tan x) = sec² x from the booklet, “the other way round”
WE 4Exponential with linear arguments
Find ∫(e3x − 1 + 4e−2x) dx.
Integrate term by term using ∫e^(ax+b) dx = (1/a) e^(ax+b) + c
For e^(3x − 1): a = 3, divide by 3
∫e^(3x − 1) dx = (1/3) e^(3x − 1)
For 4e^(−2x): a = −2, divide by −2 (which is multiplying by −1/2)
∫4 e^(−2x) dx = 4 · (−1/2) e^(−2x) = −2 e^(−2x)
Combine and add + c
= (1/3) e^(3x − 1) − 2 e^(−2x) + c
∫(e^(3x − 1) + 4 e^(−2x)) dx = (1/3) e^(3x − 1) − 2 e^(−2x) + c
careful: dividing by NEGATIVE a flips the sign. ∫e^(−2x) dx = −(1/2) e^(−2x), not (1/2) e^(−2x)
WE 5Reciprocal of a linear expression — multiple terms
Find ∫(52x + 3 − 45x − 1) dx.
Use ∫(1/(ax+b)) dx = (1/a) ln|ax+b| + c on each term
For 5/(2x + 3): a = 2, coefficient 5
∫ 5/(2x + 3) dx = 5 · (1/2) ln|2x + 3| = (5/2) ln|2x + 3|
For −4/(5x − 1): a = 5, coefficient −4
∫ (−4)/(5x − 1) dx = (−4) · (1/5) ln|5x − 1| = −(4/5) ln|5x − 1|
Combine and add + c
= (5/2) ln|2x + 3| − (4/5) ln|5x − 1| + c
= (5/2) ln|2x + 3| − (4/5) ln|5x − 1| + c
always write absolute value bars around ln arguments — ln is only defined for positive numbers, but |·| covers both sides
WE 6Curve from gradient function — find + c
A curve has gradient function dy/dx = cos(2x) + 13x + 1 and passes through the point (0, 5). Find an expression for y.
Step 1 — integrate dy/dx
y = ∫(cos(2x) + 1/(3x + 1)) dx
= (1/2) sin(2x) + (1/3) ln|3x + 1| + c
Step 2 — substitute (0, 5) to find c
at x = 0: y = (1/2) sin(0) + (1/3) ln|1| + c
= 0 + 0 + c = c
→ c = 5
Write out the answer
y = (1/2) sin(2x) + (1/3) ln|3x + 1| + 5
y = (1/2) sin(2x) + (1/3) ln|3x + 1| + 5
sin(0) = 0 and ln|1| = 0, so the constant becomes obvious. Using x = 0 always makes the algebra simplest when it’s available
💡 Top tips
- Verify by differentiating — for any of these patterns, you can check your answer by differentiating it back and confirming you get the integrand.
- Radians only for trig — IB calculus always uses radians. Set your GDC to RAD mode.
- For (ax + b) arguments, just divide by a. The intuition: the chain rule on the OUTSIDE produces an extra factor of a, so the 1/a out front cancels it.
- Coefficients ride along — ∫k f(x) dx = k ∫f(x) dx. A constant in front of the integrand stays put through integration.
- Booklet shortcut: ∫(a/(ax + b)) dx = ln|ax + b| + c directly (no 1/a needed) because the numerator IS the derivative of the denominator.
⚠ Common mistakes
- Forgetting the minus on ∫sin: ∫sin x dx = −cos x + c, NOT cos x. The sign flip catches many students.
- Missing the 1/a for linear arguments: writing ∫sin(3x) dx = −cos(3x) is WRONG. You need −(1/3) cos(3x).
- Sign error with negative a: ∫e−2x dx = (1/(−2)) e−2x = −(1/2) e−2x, NOT (1/2) e−2x.
- Forgetting the absolute value in ln|…| — examiners do penalize “ln(…)” without the modulus bars.
- Degrees instead of radians in trig integrands — instant 0 marks if you compute sin(2) thinking it’s 2°.
Up next: Reverse Chain Rule. So far the inside has always been (ax + b) — linear. What if the inside is something more general, like x² + 1, or 5x² − 2? You spot the pattern by recognising that the derivative of the inside appears (possibly with adjustment) as a factor outside. This is “integration by inspection” — and once you spot the trick, integrals like ∫2x cos(x²) dx become instant.
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