IB Maths AA HLTopic 5 — CalculusPaper 1 & 2HL~10 min read
Integrating with Inverse Trigonometric Functions
The integrands look purely algebraic — fractions involving 1/(a²+x²) and 1/√(a²−x²) — yet the antiderivatives are inverse trig functions. The two general forms ARE in the formula booklet, so the technique is to spot the shape, match the form, and apply the formula. When the denominator isn’t already in standard form, two tools convert it: factoring out a² (when there’s a coefficient on x²) or completing the square (when the denominator is a 3-term quadratic).
📘 What you need to know
Two formula-booklet integrals: ∫ 1a²+x² dx = (1/a) arctan(x/a) + c; ∫ 1√(a²−x²) dx = arcsin(x/a) + c.
Special cases (a = 1, not in booklet but deducible): ∫ 1/(1+x²) dx = arctan x + c; ∫ 1/√(1−x²) dx = arcsin x + c.
The 1/a coefficient asymmetry: arctan form HAS a 1/a factor in front; arcsin form does NOT. The chain rule consumes the 1/a on differentiating arcsin(x/a) and produces an extra a for arctan — that’s the asymmetry.
For a coefficient on x² (e.g., 4 + 9x²), factor out the constant: 4 + 9x² = 4(1 + (3x/2)²), then apply with u = 3x/2.
For 3-term quadratics in the denominator, COMPLETE THE SQUARE first: x²+6x+13 = (x+3)²+4 lands you in standard form with a=2 and inner argument (x+3).
arccos antiderivative also exists: ∫ −1/√(1−x²) dx = arccos x + c. But unless a question specifically asks for arccos, ALWAYS default to arcsin — they only differ by a constant (since arcsin x + arccos x = π/2).
√(x²−a²) ≠ √(a²−x²): these are completely different functions. Only the second matches the arcsin antiderivative. The first leads to hyperbolic functions (out of syllabus).
+c on every indefinite integral. For definite integrals, use exact unit-circle values: arctan(1) = π/4, arcsin(1/2) = π/6, arctan(√3) = π/3, etc.
The two formula-booklet integrals
Formula booklet — inverse trig integrals (general case)
∫ 1a² + x² dx = 1a arctan(xa) + c
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∫ 1√(a² − x²) dx = arcsin(xa) + c
The two integrals look symmetric but they’re not — note the 1/a factor on arctan but NOT on arcsin. Visually, the area under the arctan integrand 1/(1+x²) accumulates as x grows, and the total accumulation across the whole real line is exactly π.
The shaded green region is the area under y = 1/(1+x²) between x = −1 and x = 1. By the FTC, that area equals the antiderivative arctan x evaluated at the bounds: arctan(1) − arctan(−1) = π/4 + π/4 = π/2. The integrand is algebraic but the antiderivative is inverse trig — that’s the surprise that makes these integrals worth memorising as a separate technique.
The 1/a asymmetry, geometrically: differentiate arcsin(x/a) — the chain rule gives (1/a) · 1/√(1−(x/a)²) = 1/√(a²−x²), so the 1/a “self-consumes.” Differentiate arctan(x/a) — you get (1/a) · 1/(1+(x/a)²) = a/(a²+x²), with an extra a on top, so the antiderivative needs a 1/a in FRONT to cancel it.
Converting non-standard denominators
Coefficient on x² (b ≠ 1)
4 + 9x² = 4(1 + 9x²4) = 4(1 + (3x2)²)
Factor out the a² (here 4) so the bracket starts with 1 — then the inner argument is 3x/2 instead of x.
3-term quadratic (complete the square)
x² + 6x + 13 = (x+3)² + 4 = (x+3)² + 2²
Complete the square to convert the 3-term denominator into the form (x+k)² + a² — then the inner argument is (x+k) and the standard arctan form applies.
🧭 Recipe — integrate any inverse-trig form
Identify the shape: does the denominator look like (a²+x²) [arctan] or √(a²−x²) [arcsin]? Order matters — a² must come FIRST in the arcsin form.
If b ≠ 1 (there’s a coefficient on x²), factor out a² to convert the bracket into 1 + (something)² or 1 − (something)². Identify the inner argument.
If the denominator is a 3-term quadratic (x² + px + q), COMPLETE THE SQUARE first to rewrite as (x+k)² + a². The inner argument shifts to (x+k).
Quote the formula-booklet integral with a read off your standard form. arctan picks up a 1/a in front; arcsin does NOT.
Handle the numerator: a constant numerator multiplies the whole antiderivative. For definite integrals, substitute the limits using exact values (arctan 1 = π/4, arcsin(1/2) = π/6, etc.).
Worked examples
WE 1
Definite integral matching the simplest arctan form
Evaluate ∫0111+x² dx, giving your answer in exact form.
Step 1 — identify the form (a = 1, so the simplest case)1/(1+x²) matches 1/(a²+x²) with a = 1∫ 1/(1+x²) dx = arctan x + c (the 1/a = 1 is invisible)Step 2 — apply the FTC with limits 0 and 1∫ from 0 to 1 = [arctan x] from 0 to 1 = arctan(1) − arctan(0)Step 3 — exact unit-circle valuesarctan(1) = π/4 (since tan(π/4) = 1)arctan(0) = 0Step 4 — combine = π/4 − 0 = π/4∫ from 0 to 1 of 1/(1+x²) dx = π4half the symmetric area in the SVG above — that one was from −1 to 1 giving π/2. This one is from 0 to 1, so naturally it gives π/4 (since 1/(1+x²) is symmetric about the y-axis).
WE 2
Indefinite integral matching the arcsin form (a ≠ 1)
Find ∫ 1√(16 − x²) dx.
Step 1 — identify the form1/√(16 – x²) matches 1/√(a² – x²) with a² = 16, so a = 4Step 2 — quote the formula-booklet integral∫ 1/√(a² – x²) dx = arcsin(x/a) + c NO 1/a factor in front — chain rule consumes itStep 3 — substitute a = 4∫ 1/√(16 – x²) dx = arcsin(x/4) + c∫ 1/√(16 − x²) dx = arcsin(x4) + cverify by differentiating: d/dx[arcsin(x/4)] = (1/4) · 1/√(1 – x²/16) = (1/4) · 4/√(16 – x²) = 1/√(16 – x²). ✓ the chain rule’s 1/4 cancels the 1/4 produced by the square root algebra.
WE 3
Constant numerator multiplying the arctan form
Find ∫ 325 + x² dx.
Step 1 — pull the constant 3 outside∫ 3/(25 + x²) dx = 3 ∫ 1/(25 + x²) dxStep 2 — identify a² = 25, so a = 5∫ 1/(a² + x²) dx = (1/a) arctan(x/a) + c∫ 1/(25 + x²) dx = (1/5) arctan(x/5) + cStep 3 — multiply by the 3 from outside∫ 3/(25 + x²) dx = 3 · (1/5) arctan(x/5) + c = (3/5) arctan(x/5) + c∫ 3/(25 + x²) dx = 35 arctan(x5) + ctwo coefficients to track: the 3 from the numerator and the 1/5 from the formula. Combine them at the end. Easy mark to lose if you forget either.
WE 4
Coefficient on x² — factor out a² first
Find ∫ 14 + 9x² dx.
Step 1 — the denominator isn’t yet in standard a² + x² form (b = 9, not 1)Factor 4 out so the bracket starts with 1:4 + 9x² = 4(1 + (9/4)x²) = 4(1 + (3x/2)²)Step 2 — rewrite the integral∫ 1/(4 + 9x²) dx = (1/4) ∫ 1/(1 + (3x/2)²) dxStep 3 — apply ∫ 1/(1+u²) du = arctan u, with u = 3x/2du/dx = 3/2, so dx = (2/3) du (“adjust and compensate”)(1/4) ∫ 1/(1 + u²) · (2/3) du = (1/4)(2/3) arctan(u) + c = (1/6) arctan(3x/2) + c∫ 1/(4 + 9x²) dx = 16 arctan(3x2) + cverify: d/dx[(1/6) arctan(3x/2)] = (1/6) · (3/2)/(1 + 9x²/4) = (1/4)/((4 + 9x²)/4) = 1/(4 + 9x²). ✓ the 1/4 from factoring AND the 2/3 from “adjust and compensate” combine to give 1/6 in front.
WE 5
3-term quadratic — complete the square
Find ∫ 1x² + 6x + 13 dx.
Step 1 — the denominator is a 3-term quadratic — COMPLETE THE SQUAREx² + 6x + 13 = (x² + 6x + 9) + 4 (since 13 = 9 + 4) = (x + 3)² + 4 = (x + 3)² + 2²Step 2 — rewrite the integral in standard form with a = 2 and inner argument (x + 3)∫ 1/(x² + 6x + 13) dx = ∫ 1/((x + 3)² + 4) dxStep 3 — apply ∫ 1/(a² + u²) du = (1/a) arctan(u/a), where u = x + 3du/dx = 1, so dx = du (no adjust needed since x + 3 has gradient 1)∫ 1/((x + 3)² + 4) dx = (1/2) arctan((x + 3)/2) + c∫ 1/(x² + 6x + 13) dx = 12 arctan(x + 32) + csince the derivative of (x + 3) is 1 (same as derivative of x), there’s no reverse-chain-rule coefficient — everything passes through cleanly. The 1/2 is just the 1/a from the standard formula.
WE 6
Definite arctan integral with a ≠ 1 — exact form
Evaluate ∫0√323 + x² dx, giving your answer in exact form.
Step 1 — identify a² = 3, so a = √3, and pull constant 2 outside∫ 2/(3 + x²) dx = 2 · (1/√3) arctan(x/√3) + c = (2/√3) arctan(x/√3) + cStep 2 — apply FTC with limits 0 and √3[(2/√3) arctan(x/√3)] from 0 to √3 = (2/√3) [arctan(√3/√3) − arctan(0/√3)] = (2/√3) [arctan(1) − arctan(0)]Step 3 — exact valuesarctan(1) = π/4, arctan(0) = 0 = (2/√3) · π/4 = π/(2√3) = π√3/6 (rationalise)∫ from 0 to √3 of 2/(3 + x²) dx = π√36the limits √3 and the a = √3 conveniently match: x/a = √3/√3 = 1 at the upper limit, which gives the nice arctan(1) = π/4. Examiners often choose limits that produce clean angles like π/4, π/6, or π/3 — when you see √3 or √2 in the limits, that’s the hint.
💡 Top tips
Quote the formula booklet for the two general forms: ∫1/(a²+x²) dx = (1/a) arctan(x/a) + c and ∫1/√(a²−x²) dx = arcsin(x/a) + c. The 1/a appears only on the arctan side.
Always check by differentiating your answer — for these antiderivatives it’s the only way to be sure you have the 1/a coefficient in the right place.
If the bracket isn’t 1 + (something)² or 1 − (something)², do the algebra FIRST: factor out a² or complete the square. Only then quote the formula.
Default to arcsin over arccos for ∫1/√(a²−x²) dx unless the question specifically calls for arccos. They differ only by a constant, so either is technically correct, but arcsin is the standard.
⚠ Common mistakes
Forgetting the 1/a on arctan: ∫1/(25+x²) dx = (1/5) arctan(x/5) + c, NOT arctan(x/5) + c.
Incorrectly adding 1/a to arcsin: ∫1/√(16−x²) dx = arcsin(x/4) + c, NOT (1/4) arcsin(x/4) + c. The chain rule consumed the 1/4 — putting it back doubles up.
Confusing the two forms: a denominator with a square root (a²−x²) is the arcsin form; a denominator without a square root (a²+x²) is the arctan form. Squared difference is arcsin, squared sum is arctan.
Treating √(x²−a²) like √(a²−x²) — these are different functions. Only the second is in your toolkit (it gives arcsin). The first leads outside the syllabus.
Skipping “factor out a²” or “complete the square” when the denominator isn’t already in standard form. ∫1/(4+9x²) dx ≠ (1/2) arctan(x/2) + c — the 9x² needs handling first.
Up next: Integrating with Exponential & Logarithmic Functions. The natural cases ∫ex dx = ex + c and ∫1/x dx = ln|x| + c you’ve met before — but the general cases ∫ax dx = (1/ln a) ax + c and the reverse-chain-rule shortcut ∫ f′(x)/f(x) dx = ln|f(x)| + c open up a much wider class of integrands. Same theme as this chapter: read the corresponding derivative formula backwards, then handle the bookkeeping.
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