Integration is the highest-density topic on IB Mathematics Analysis & Approaches HL. It carries the largest mark allocation in the calculus strand, it is examined on every Paper 1, Paper 2 and Paper 3, and it gates almost all of the modelling and applied questions later in the syllabus (kinematics, differential equations, statistics, Maclaurin series). The HL extension over SL is huge: integration by substitution, integration by parts (including cyclic IBP), partial fractions, and volumes of revolution are all AHL content that students see for the first time at HL.
This cheatsheet condenses every formula, technique, and exam trap from SL 5.5, 5.10–5.11 and AHL 5.15–5.17 into one revisable page. The hardest part of HL Integration is not memorising rules — it is choosing the right technique fast under exam pressure. The Exam Attack Plan in §10 is built specifically for that decision. Scroll to the bottom for the printable PDF and the full notes booklet.
§1 — Standard Integrals SL 5.5, 5.10, AHL 5.15
Core table
| $f(x)$ | $\int f(x)\,dx$ |
|---|---|
| $x^n$ $(n \neq -1)$ | $\dfrac{x^{n+1}}{n+1} + C$ |
| $\dfrac{1}{x}$ | $\ln|x| + C$ |
| $e^x$ | $e^x + C$ |
| $a^x$ | $\dfrac{a^x}{\ln a} + C$ |
| $\sin x$ | $-\cos x + C$ |
| $\cos x$ | $\sin x + C$ |
| $\tan x$ | $\ln|\sec x| + C$ |
| $\sec^2 x$ | $\tan x + C$ |
| $\csc^2 x$ | $-\cot x + C$ |
| $\sec x \tan x$ | $\sec x + C$ |
| $\sec x$ | $\ln|\sec x + \tan x| + C$ |
Inverse trig & AHL
| $f(x)$ | $\int f(x)\,dx$ |
|---|---|
| $\dfrac{1}{1+x^2}$ | $\arctan x + C$ |
| $\dfrac{1}{a^2 + x^2}$ | $\dfrac{1}{a}\arctan\dfrac{x}{a} + C$ |
| $\dfrac{1}{\sqrt{1 - x^2}}$ | $\arcsin x + C$ |
| $\dfrac{1}{\sqrt{a^2 - x^2}}$ | $\arcsin\dfrac{x}{a} + C$ |
Linear composite rule
$$\int f(ax+b)\,dx = \frac{F(ax+b)}{a} + C$$
e.g. $\int e^{3x}\,dx = \dfrac{e^{3x}}{3} + C$, $\int\sin(2x)\,dx = -\dfrac{\cos 2x}{2} + C$.
§2 — Integration by Inspection (Reverse Chain Rule) SL 5.10
Three-step method
- Spot the inner function $g(x)$.
- Compute $g'(x)$.
- Check whether $g'(x)$ appears in the integrand (possibly times a constant $k$). If yes, integrate directly.
Core rule: $\int k \cdot g'(x) \cdot f(g(x))\,dx = k \cdot F(g(x)) + C$.
Pattern table
| Pattern | Antiderivative ($+C$) |
|---|---|
| $\int [g]^n \cdot g'\,dx$ | $\dfrac{[g]^{n+1}}{n+1}$ |
| $\int \dfrac{g'}{g}\,dx$ | $\ln|g|$ |
| $\int e^g \cdot g'\,dx$ | $e^g$ |
| $\int \sin(g)\cdot g'\,dx$ | $-\cos(g)$ |
| $\int \cos(g)\cdot g'\,dx$ | $\sin(g)$ |
| $\int \sec^2(g)\cdot g'\,dx$ | $\tan(g)$ |
| $\int \dfrac{g'}{\sqrt{1 - g^2}}\,dx$ | $\arcsin(g)$ |
| $\int \dfrac{g'}{1 + g^2}\,dx$ | $\arctan(g)$ |
Key examples
- $\int 2x(x^2 + 1)^4\,dx = \dfrac{(x^2+1)^5}{5} + C$
- $\int \dfrac{3x^2}{x^3 + 7}\,dx = \ln|x^3 + 7| + C$
- $\int 4x\sin(x^2)\,dx = -2\cos(x^2) + C$
- $\int \tan x\,dx = \ln|\sec x| + C$
- $\int \dfrac{e^x}{1 + e^{2x}}\,dx = \arctan(e^x) + C$
- $\int \dfrac{2x}{1 + x^4}\,dx = \arctan(x^2) + C$
Verify: differentiate your answer — you should get the integrand back.
§3 — Integration by Substitution AHL 5.16
Method — Indefinite
- Choose $u = g(x)$ (the inner / messy part).
- $du = g'(x)\,dx \Rightarrow dx = \dfrac{du}{g'(x)}$.
- Eliminate every $x$ from the integrand.
- Integrate in $u$.
- Back-substitute $u \to g(x)$, add $+C$.
Method — Definite
Same steps 1–4, plus: change the limits $x = a \Rightarrow u = g(a)$, $x = b \Rightarrow u = g(b)$. With converted limits, no back-substitution is needed.
Half-angle substitution (given in exam): $t = \tan(x/2) \Rightarrow \sin x = \dfrac{2t}{1 + t^2}$, $dx = \dfrac{2}{1 + t^2}\,dt$.
§4 — Integration by Parts AHL 5.16
Formula & LIATE
$$\int u\,dv = uv - \int v\,du$$
LIATE priority for choosing $u$: Log > Inverse trig > Algebraic > Trig > Exponential.
Key results
- $\int \ln x\,dx = x\ln x - x + C$
- $\int x e^x\,dx = (x - 1)e^x + C$
- $\int x \sin x\,dx = \sin x - x\cos x + C$
- $\int \arcsin x\,dx = x\arcsin x + \sqrt{1 - x^2} + C$
- $\int \arctan x\,dx = x\arctan x - \tfrac{1}{2}\ln(1 + x^2) + C$
Cyclic IBP & reduction
Cyclic IBP ($\int e^x\sin x$, $\int e^x\cos x$): apply IBP twice, keeping the same $u$ each time. Name the integral $I$, solve $2I = \ldots$:
- $\int e^x \sin x\,dx = \dfrac{e^x(\sin x - \cos x)}{2} + C$
- $\int e^x \cos x\,dx = \dfrac{e^x(\sin x + \cos x)}{2} + C$
Tabular method (polynomial $\times$ exp/trig): differentiate the polynomial column, integrate the exp/trig column, alternate $+/-$ signs, read diagonally.
Reduction formula: $n I_n = (n-1) I_{n-2}$ for $I_n = \int_0^{\pi/2} \cos^n x\,dx$.
§5 — Partial Fractions AHL 5.15
Decomposition forms
- Distinct linear: $\dfrac{1}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$
- Repeated linear: $\dfrac{1}{(x-a)^2} = \dfrac{A}{(x-a)^2} + \dfrac{B}{x-a}$
- Irreducible quadratic: $\dfrac{1}{(x-a)(x^2+b)} = \dfrac{A}{x-a} + \dfrac{Bx + C}{x^2 + b}$
Each term integrates to $\ln|\cdot|$ or $\arctan$.
5-step method
- Factorise the denominator.
- Write the decomposition with unknowns $A$, $B$, $C$.
- Multiply through by the denominator.
- Substitute strategic $x$-values or compare coefficients to solve for $A$, $B$, $C$.
- Integrate each term.
Complete the square for $\int \dfrac{1}{x^2 + bx + c}\,dx$: $(x+p)^2 + q^2 \to \dfrac{1}{q}\arctan\dfrac{x+p}{q}$.
§6 — Areas SL 5.11
Five area cases
| Case | Setup | Formula |
|---|---|---|
| 1 | $f(x) \geq 0$ on $[a,b]$ | $A = \int_a^b f\,dx$ |
| 2 | $f(x) \leq 0$ on $[a,b]$ | $A = -\int_a^b f\,dx$ |
| 3 | $f$ changes sign, root at $c$ | $A = \left|\int_a^c f\,dx\right| + \left|\int_c^b f\,dx\right|$ |
| 4 | Between curves $f \geq g$ on $[a,b]$ | $A = \int_a^b (f - g)\,dx$ |
| 5 | Curves cross inside $[a,b]$ | Find crossing, split, $\int |f-g|$ |
Area with $y$-axis: $A = \int_c^d g(y)\,dy$ where $x = g(y)$ (invert the curve).
Sign-change procedure
- Find all roots of $f$ on $[a,b]$.
- Sketch — identify sign on each interval.
- Split the integral at every root.
- Take $|\ldots|$ of each piece.
- Sum all pieces.
Even function: $\int_{-a}^a f = 2\int_0^a f$. Odd function: $\int_{-a}^a f = 0$.
§7 — Volumes of Revolution AHL 5.17
All four formulas
| Situation | Axis | Formula |
|---|---|---|
| Single curve $y = f(x)$ | $x$-axis | $V = \pi\int_a^b [f(x)]^2\,dx$ |
| Single curve (invert to $x = g(y)$) | $y$-axis | $V = \pi\int_c^d [g(y)]^2\,dy$ |
| Outer $f$, inner $g$ (washer) | $x$-axis | $V = \pi\int_a^b ([f]^2 - [g]^2)\,dx$ |
| Outer $g_1$, inner $g_2$ | $y$-axis | $V = \pi\int_c^d ([g_1]^2 - [g_2]^2)\,dy$ |
Worked examples
- $y = \sqrt{x}$, $x \in [0, 4]$, about $x$-axis: $V = \pi\int_0^4 x\,dx = 8\pi$.
- $y = \sin x$, $x \in [0, \pi]$, about $x$-axis: use $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ to get $V = \dfrac{\pi^2}{2}$.
- $y = e^x$, $x \in [0, 2]$, about $x$-axis: $V = \pi\int_0^2 e^{2x}\,dx = \dfrac{\pi}{2}(e^4 - 1)$.
- Washer between $y = \sqrt{x}$ and $y = x^2$: $V = \pi\int_0^1 (x - x^4)\,dx = \dfrac{3\pi}{10}$.
Useful IBP result for $y$-axis rotations: $\int (\ln y)^2\,dy = y[(\ln y)^2 - 2\ln y + 2] + C$.
§8 — Kinematics (Integration View) SL 5.9
$v = \dfrac{ds}{dt}$, $a = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}$. $s = \int v\,dt$, $v = \int a\,dt$.
Displacement vs total distance
Displacement (signed): $\Delta s = \int_{t_1}^{t_2} v(t)\,dt$. Total distance (always non-negative): $d = \int_{t_1}^{t_2} |v(t)|\,dt$. Split at zeros of $v$ (direction changes).
Distance procedure
- Solve $v(t) = 0$ to find the times $t_1, t_2, \ldots$ at which the particle reverses.
- Check the sign of $v$ on each sub-interval.
- Integrate $v$ on each sub-interval.
- Sum the absolute values.
Speed $= |v|$ (always $\geq 0$). Particle at rest: $v = 0$.
nInt(abs(v(t)),t,a,b).§9 — Maclaurin Series & Integration AHL 5.19
Standard series (memorise)
- $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$
- $\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$
- $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$
- $\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$
Key substitution: replace $x$ with the relevant expression (e.g. $e^{-x^2}$: substitute $x \to -x^2$ in the series for $e^x$).
Integrating series term by term
Some integrals have no elementary closed form — e.g. $\int e^{-x^2}\,dx$, $\int \dfrac{\sin x}{x}\,dx$. Expand as a Maclaurin series and integrate each polynomial term separately:
- $\int e^{-x^2}\,dx \approx x - \dfrac{x^3}{3} + \dfrac{x^5}{10} - \cdots$
- $\dfrac{1}{1 + x^2} = 1 - x^2 + x^4 - \cdots \Rightarrow \arctan x = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots$
§10 — Exam Attack Plan All sections
When you see this in the question — reach for that:
| Trigger | Method |
|---|---|
| Numerator $=$ derivative of denominator | $\int \dfrac{g'}{g}\,dx = \ln|g|$ — inspection |
| Product: one factor is the derivative of the other | Inspection (reverse chain rule) |
| Power $\times$ composite function | Inspection or substitution |
| Product: two different function types | Integration by parts (LIATE) |
| $\int \ln x$, $\int \arctan x$, $\int \arcsin x$ | IBP with $u = \ln / \arctan / \arcsin$, $dv = dx$ |
| $\int e^x \sin x$ or $\int e^x \cos x$ | Cyclic IBP, name it $I$ |
| Rational function (poly / poly) | Partial fractions (long divide if needed) |
| $\int \dfrac{1}{x^2 + bx + c}\,dx$ | Complete the square $\to \arctan$ |
| "Use substitution $u = \ldots$" | Formal substitution, change limits if definite |
| Area "enclosed by" curves | Find intersections, set up $\int (f - g)$ |
| Area where $f$ changes sign | Split at roots, sum $|\ldots|$ of pieces |
| "Rotate about $x$-axis" | $V = \pi \int [f(x)]^2\,dx$ |
| "Rotate about $y$-axis" | Invert to $x = g(y)$, $V = \pi\int [g(y)]^2\,dy$ |
| "Total distance travelled" | $\int |v|\,dt$ — split at $v = 0$ |
| "Find displacement" | $\int v\,dt$ — signed, can be negative |
| Integral with no closed form | Maclaurin series, integrate term by term |
Top 5 marks lost — based on past papers
- Missing $|x|$ in $\ln$: writing $\ln x$ instead of $\ln|x|$ in $\int \dfrac{1}{x}\,dx$ loses a mark every time. Build $|x|$ into muscle memory.
- Signed area instead of geometric area: "Find the area" always means take $|\ldots|$ and split at roots. Computing $\int_a^b f\,dx$ directly when $f$ dips below the axis loses every accuracy mark.
- Wrong constant in inspection: $\int 6x(x^2 + 1)^4\,dx \neq \dfrac{(x^2+1)^5}{5} \times 6x$. Always isolate the constant before applying the pattern.
- Cyclic IBP — swapping $u$: choosing $u = e^x$ in round 1, then $u = \sin x$ in round 2, gives the trivial $0 = 0$. Keep the same $u$ throughout.
- Volume with wrong variable: rotating about the $y$-axis means integrating with respect to $y$. Writing $\pi \int [f(x)]^2\,dx$ for a $y$-axis rotation scores zero marks for method.
Worked Example — IB-Style Integration by Parts (Cyclic)
Question (HL Paper 1 style — 8 marks)
Use integration by parts twice to show that $\int e^{2x}\sin(3x)\,dx = \dfrac{e^{2x}}{13}\bigl(2\sin 3x - 3\cos 3x\bigr) + C$.
Solution
- Let $I = \int e^{2x}\sin(3x)\,dx$. Apply IBP with $u = \sin(3x)$, $dv = e^{2x}\,dx$ (or equivalently $u = e^{2x}$, $dv = \sin(3x)\,dx$ — choice is fixed for both rounds). Take $u = e^{2x}$, $dv = \sin(3x)\,dx$, so $du = 2e^{2x}\,dx$ and $v = -\tfrac{1}{3}\cos(3x)$. Then $I = -\tfrac{1}{3}e^{2x}\cos(3x) + \tfrac{2}{3}\int e^{2x}\cos(3x)\,dx$. (M1)(A1)
- Apply IBP again to $J = \int e^{2x}\cos(3x)\,dx$, keeping $u = e^{2x}$, $dv = \cos(3x)\,dx$, so $du = 2e^{2x}\,dx$ and $v = \tfrac{1}{3}\sin(3x)$. Then $J = \tfrac{1}{3}e^{2x}\sin(3x) - \tfrac{2}{3}\int e^{2x}\sin(3x)\,dx = \tfrac{1}{3}e^{2x}\sin(3x) - \tfrac{2}{3}I$. (M1)(A1)
- Substitute back into the expression for $I$: $I = -\tfrac{1}{3}e^{2x}\cos(3x) + \tfrac{2}{3}\!\left[\tfrac{1}{3}e^{2x}\sin(3x) - \tfrac{2}{3}I\right]$. (M1)
- Expand: $I = -\tfrac{1}{3}e^{2x}\cos(3x) + \tfrac{2}{9}e^{2x}\sin(3x) - \tfrac{4}{9}I$. Collect $I$ on the LHS: $\tfrac{13}{9}I = e^{2x}\!\left(\tfrac{2}{9}\sin 3x - \tfrac{1}{3}\cos 3x\right)$. (A1)
- Multiply both sides by $\tfrac{9}{13}$: $I = \dfrac{e^{2x}}{13}\bigl(2\sin 3x - 3\cos 3x\bigr) + C$. (A1)(AG)
Examiner's note: The fatal error in cyclic IBP is swapping the choice of $u$ in round 2. If you chose $u = e^{2x}$ in round 1, you must keep $u = e^{2x}$ in round 2 — otherwise the second IBP undoes the first and you get $I = I$ (no progress). Also remember to write "$+ C$" on indefinite integrals, even on "show that" questions.
Common Student Questions
How do I decide which integration technique to use?
What does LIATE mean and how do I apply it for integration by parts?
Why is "find the area" different from computing $\int f\,dx$ directly?
What's the difference between rotating about the $x$-axis and the $y$-axis?
When do I need partial fractions instead of just substitution?
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