IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~10 min read

Integrating with Reciprocal Trigonometric Functions

The four reciprocal-trig antiderivatives are direct reversals of the four reciprocal-trig derivatives — but unlike their differentiation cousins, these antiderivatives are NOT in the formula booklet. You deduce them by reading the derivative formulas backwards: d/dx[tan x] = sec²x means ∫sec²x dx = tan x + c, and similarly for the other three.

📘 What you need to know

The four core antiderivatives: ∫sec²x dx = tan x + c; ∫sec x tan x dx = sec x + c; ∫cosec²x dx = −cot x + c; ∫cosec x cot x dx = −cosec x + c.
NOT in the formula booklet — but the four derivative formulas ARE. Reading them backwards gives all four antiderivatives. That’s the whole game.
Mind the minuses: the cot and cosec antiderivatives have a leading negative sign. d/dx[cot x] = −cosec²x, so ∫cosec²x dx needs a minus to invert it.
Linear argument rule: ∫sec²(ax+b) dx = (1/a) tan(ax+b) + c. Same 1/a factor for all four.
“Adjust and compensate”: spot reverse chain rule. The inner coefficient a appears as 1/a outside the antiderivative.
+c on every indefinite integral — never optional. Drop it on Paper 1 and it costs marks.
Trig identities unlock more integrands: cot²x = cosec²x − 1 and tan²x = sec²x − 1 convert “non-standard” integrands into the standard four.
Domain reminder: sec and tan blow up at odd multiples of π/2; cosec and cot blow up at multiples of π. Definite integrals can’t span these singularities.

The four antiderivatives — read derivatives backwards

Four antiderivatives — derive by reversing the formula-booklet derivatives ∫sec²x dx = tan x + c | ∫sec x tan x dx = sec x + c

∫cosec²x dx = −cot x + c | ∫cosec x cot x dx = −cosec x + c

None of those four antiderivatives is given in the formula booklet. But the four derivatives they’re built from are — so the trick on Paper 1 is to quote the derivative, then invert it.

The shaded green region is the area under the curve y = cosec²x between x = π/6 and x = π/3. By the FTC, that area equals the antiderivative −cot x evaluated at those bounds: −cot(π/3) − (−cot(π/6)) = −1/√3 + √3 = 2√3/3 ≈ 1.155. The minus sign in the antiderivative isn’t decorative — it’s what makes the area come out positive.

Why the minus signs are there: differentiating cot x gives −cosec²x (a negative). So to reverse it we need ∫cosec²x dx = −cot x + c — the minus turns the −cosec² back into +cosec² on re-differentiation. Same story for cosec x cot x → −cosec x.

Linear arguments — adjust and compensate

When the argument is a linear function (ax+b) rather than just x, every antiderivative picks up a 1/a factor from reverse chain rule. This is the same “adjust and compensate” technique you used for ∫e^ax+b dx = (1/a) e^ax+b + c.

Linear arg — sec² and sec·tan

∫sec²(ax+b) dx = 1a tan(ax+b) + c

divide the antiderivative by the inner coefficient — same 1/a applies to ∫sec(ax+b)tan(ax+b) dx = (1/a)sec(ax+b) + c

Linear arg — cosec² and cosec·cot

∫cosec²(ax+b) dx = −1a cot(ax+b) + c

again 1/a, plus the minus sign from the base case — ∫cosec(ax+b)cot(ax+b) dx = −(1/a)cosec(ax+b) + c

🧭 Recipe — integrate any reciprocal-trig integrand

Identify the reciprocal-trig function and its argument: is it sec², sec·tan, cosec², or cosec·cot? Is the argument just x, linear (ax+b), or something that needs an identity?
Quote the matching derivative from the formula booklet, then read it backwards. d/dx[tan x] = sec²x → ∫sec²x dx = tan x + c. Check signs carefully — cot and cosec carry minus signs.
If linear argument (ax+b), divide by a (“adjust and compensate”). Verify by differentiating your candidate answer — it should reproduce the integrand.
If the integrand isn’t in standard form, apply trig identities to convert: cot²x = cosec²x − 1, tan²x = sec²x − 1, or rewrite using sin/cos.
Add “+c“ for an indefinite integral, or substitute the upper minus the lower limit for a definite integral. Keep exact form (π, √3, ln, etc.) on Paper 1.

Worked examples

WE 1

Basic sec² with linear argument

Find ∫6 sec²(3x) dx.

Step 1 — identify the form and the inner coefficient integrand is k · sec²(ax+b) with k = 6, a = 3 Step 2 — quote the derivative backwards d/dx[tan x] = sec²x ⟹ ∫sec²x dx = tan x + c Step 3 — adjust and compensate for the inner coefficient 3 ∫sec²(3x) dx = (1/3) tan(3x) + c Step 4 — pull the outer 6 through ∫6 sec²(3x) dx = 6 · (1/3) tan(3x) + c = 2 tan(3x) + c ∫6 sec²(3x) dx = 2 tan(3x) + c verify by differentiating: d/dx[2 tan(3x)] = 2 · 3 sec²(3x) = 6 sec²(3x). ✓ the outer 6 and inner 3 combined to give the (1/3) · 6 = 2 coefficient.

WE 2

sec·tan with linear argument including a constant shift

Find ∫sec(4x − π/3) tan(4x − π/3) dx.

Step 1 — identify the form sec(ax+b) tan(ax+b) with a = 4, b = -π/3 Step 2 — derivative backwards d/dx[sec x] = sec x tan x ⟹ ∫sec x tan x dx = sec x + c Step 3 — adjust by 1/a for the linear argument ∫sec(4x – π/3) tan(4x – π/3) dx = (1/4) sec(4x – π/3) + c = 14 sec(4x − π/3) + c the constant shift -π/3 doesn’t affect the 1/a factor — only the coefficient of x matters for reverse chain rule. Verify: d/dx[(1/4)sec(4x – π/3)] = (1/4) · 4 sec(4x – π/3) tan(4x – π/3) = the integrand. ✓

WE 3

Definite integral with cosec² — exact answer

Evaluate ∫_π/6^π/3 cosec²x dx, giving your answer in exact form.

Step 1 — derivative backwards (mind the sign) d/dx[cot x] = -cosec²x ⟹ ∫cosec²x dx = -cot x + c Step 2 — apply the FTC with the antiderivative -cot x ∫ from π/6 to π/3 of cosec²x dx = [-cot x] from π/6 to π/3 = -cot(π/3) – (-cot(π/6)) = -cot(π/3) + cot(π/6) Step 3 — use exact unit-circle values cot(π/3) = cos(π/3)/sin(π/3) = (1/2)/(√3/2) = 1/√3 = √3/3 cot(π/6) = cos(π/6)/sin(π/6) = (√3/2)/(1/2) = √3 Step 4 — combine = -√3/3 + √3 = -√3/3 + 3√3/3 = 2√3/3 ∫ cosec²x dx (π/6 to π/3) = 2√33 the area under cosec²x between these two limits is exactly 2√3/3 ≈ 1.155 — the value matches the geometric shaded area in the SVG above.

WE 4

cosec·cot with linear argument and outer coefficient

Find ∫3 cosec(2x) cot(2x) dx.

Step 1 — derivative backwards (mind the sign) d/dx[cosec x] = -cosec x cot x ⟹ ∫cosec x cot x dx = -cosec x + c Step 2 — adjust by 1/a = 1/2 for the linear argument 2x ∫cosec(2x) cot(2x) dx = -(1/2) cosec(2x) + c Step 3 — pull the outer 3 through ∫3 cosec(2x) cot(2x) dx = 3 · -(1/2) cosec(2x) + c = -(3/2) cosec(2x) + c = −32 cosec(2x) + c two sources of minus signs to track: (a) the base antiderivative has a minus, (b) the inner coefficient gives 1/2, not -1/2. Combine carefully — the final answer is negative because the antiderivative of cosec·cot is negative.

WE 5

Initial value problem — find the curve from its derivative

A curve has gradient function dy/dx = 8 sec²(2x) and passes through the point (π/8, 1). Find an expression for y in terms of x.

Step 1 — integrate to find y up to constant y = ∫8 sec²(2x) dx = 8 · (1/2) tan(2x) + c y = 4 tan(2x) + c Step 2 — substitute the point (π/8, 1) to find c 1 = 4 tan(2 · π/8) + c 1 = 4 tan(π/4) + c 1 = 4 · 1 + c (since tan(π/4) = 1) c = 1 – 4 = -3 Step 3 — write the final expression y = 4 tan(2x) − 3 always integrate FIRST, THEN substitute the point — never the other way round. The constant of integration c is exactly what the given point pins down. Verify by checking: d/dx[4 tan(2x) – 3] = 4 · 2 sec²(2x) = 8 sec²(2x). ✓

WE 6

Non-standard integrand — use a trig identity first

Evaluate ∫_π/4^π/3 cot²x dx, giving your answer in exact form.

Step 1 — cot²x is NOT one of the standard four — convert using an identity Pythagorean identity: 1 + cot²x = cosec²x ⟹ cot²x = cosec²x – 1 Step 2 — rewrite and split the integral ∫cot²x dx = ∫(cosec²x – 1) dx = ∫cosec²x dx – ∫1 dx = -cot x – x + c (antiderivative form) Step 3 — apply the limits π/4 and π/3 ∫ from π/4 to π/3 = [-cot x – x] from π/4 to π/3 = (-cot(π/3) – π/3) – (-cot(π/4) – π/4) = (-√3/3 – π/3) – (-1 – π/4) = -√3/3 – π/3 + 1 + π/4 Step 4 — combine the π terms (π/4 – π/3 = -π/12) = 1 – √3/3 – π/12 ∫ cot²x dx (π/4 to π/3) = 1 − √33 − π12 cot²x and tan²x always need an identity first — they’re not in the “four standard antiderivatives” but become trivial once you swap them for cosec²x – 1 or sec²x – 1. Numerically: 1 – 0.577 – 0.262 ≈ 0.161, which is small and positive — makes sense, since cot²x is small on (π/4, π/3).

💡 Top tips

Read derivative formulas backwards — the four antiderivatives aren’t in the formula booklet, but the four derivatives are. d/dx[tan x] = sec²x reverses to ∫sec²x dx = tan x + c.
Always check by differentiating your answer. If d/dx[your antiderivative] equals the original integrand, you’ve got it right — including the sign and the 1/a.
Spot trig identities early: cot²x, tan²x, sin²x, cos²x all need rewriting before any reciprocal-trig formula applies. cot²x = cosec²x − 1 is the most common.
Keep exact form on Paper 1: π/3, √3, √2/2, and so on. Don’t switch to decimals unless the question explicitly asks for them.
For initial-value problems, integrate FIRST (keeping +c), THEN substitute the given point. Reversing this order is the most common error in this topic.

⚠ Common mistakes

Dropping the minus sign on ∫cosec²x dx = −cot x + c or ∫cosec x cot x dx = −cosec x + c. The minus is essential — drop it and the answer differentiates to −cosec²x or −cosec x cot x.
Forgetting the 1/a factor for linear arguments: ∫sec²(3x) dx = (1/3)tan(3x) + c, NOT tan(3x) + c.
Forgetting “+c“ on indefinite integrals — costs a mark every time, even when the rest is perfect.
Trying to integrate cot²x or tan²x directly — they’re not in the standard four. You MUST use cot²x = cosec²x − 1 (or the tan version) first.
Substituting limits before integrating on initial-value problems. Always integrate to find the family y = F(x) + c first, then use the boundary point to nail down c.

Up next: Integrating with Inverse Trigonometric Functions. The reverse direction of arcsin, arccos, arctan derivatives turns out to give SURPRISING antiderivatives — purely algebraic expressions like ∫ 1/√(1−x²) dx = arcsin x + c and ∫ 1/(1+x²) dx = arctan x + c. Same “read derivative backwards” trick, but the integrands look algebraic and the antiderivatives are inverse trig. Plus completing the square unlocks denominators like 5 − x² + 4x.

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