Integration of trigonometric functions: QCE Maths Methods Unit 4

Syllabus dot point

“Integrate trigonometric functions including $\sin(kx)$ , $\cos(kx)$ and $\sec^2(kx)$ , and apply the linear reverse-chain rule for integrals of the form $f(ax+b)$ ”

8 min readUpdated 7 Oct 2026

Quick answer

Integration of trigonometric functions in QCE Methods Unit 4 follows the standard antiderivative rules ( $\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C$ , $\int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C$ , $\int \sec^2(kx) \, dx = \frac{1}{k} \tan(kx) + C$ ) with the linear reverse-chain $\frac{1}{a}$ factor; Paper 1 evaluations require exact values at standard angles ( $\pi/6, \pi/4, \pi/3, \pi/2, \pi$ ) and the most-tested detail is including the $\frac{1}{k}$ factor in non-unit coefficient integrals.

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What this dot point is asking
Standard trigonometric antiderivatives
The linear reverse chain rule
Definite integrals with exact values
Area between trig curves
Applications in PSMT and EA
Exam-style questions

What this dot point is asking

QCAA wants you to integrate trigonometric functions of the form $\sin(kx)$ , $\cos(kx)$ and $\sec^2(kx)$ , evaluate definite integrals with exact values at standard angles, and apply trig integration in kinematics and area contexts. The dot point feeds the IA3 PSMT and the EA Paper 1 short response.

Standard trigonometric antiderivatives

The reverse of the Unit 3 trig derivatives, with the $\frac{1}{k}$ reverse-chain factor for non-unit coefficients.

Integrand	Antiderivative
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$
$\sin(kx)$	$-\frac{1}{k} \cos(kx) + C$
$\cos(kx)$	$\frac{1}{k} \sin(kx) + C$
$\sec^2(kx)$	$\frac{1}{k} \tan(kx) + C$

The $\frac{1}{k}$ factor is the reverse of the chain-rule factor that appears when differentiating $\sin(kx)$ to get $k \cos(kx)$ . Forgetting this factor is the single most common Paper 1 error in trig integration.

Sign check

The pattern for derivatives is:

$\frac{d}{dx} \sin x = +\cos x$ (no sign change).
$\frac{d}{dx} \cos x = -\sin x$ (sign change).

Reversing:

$\int \cos x \, dx = +\sin x$ (no sign change).
$\int \sin x \, dx = -\cos x$ (sign change).

The antiderivative of $\sin$ is $-\cos$ . The antiderivative of $\cos$ is $+\sin$ . These two patterns generate every trig sign error you'll see.

The linear reverse chain rule

For an integrand of the form $f(ax + b)$ where $a, b$ are constants:

\int f(ax + b) \, dx = \frac{1}{a} F(ax + b) + C

where $F$ is an antiderivative of $f$ .

Examples:

$\int \sin(3x + 1) \, dx = -\frac{1}{3} \cos(3x + 1) + C$ .
$\int \cos(2x - 5) \, dx = \frac{1}{2} \sin(2x - 5) + C$ .
$\int e^{4x + 2} \, dx = \frac{1}{4} e^{4x + 2} + C$ .

The $\frac{1}{a}$ corrects for the chain rule factor that would appear when differentiating the antiderivative.

Definite integrals with exact values

For definite integrals of trig functions evaluated at standard angles, the QCAA Paper 1 expects exact-value answers (in terms of $\pi$ , fractions, surds).

Standard exact values:

$\theta$	$\sin \theta$	$\cos \theta$	$\tan \theta$
$0$	$0$	$1$	$0$
$\pi/6$	$1/2$	$\sqrt{3}/2$	$1/\sqrt{3}$
$\pi/4$	$\sqrt{2}/2$	$\sqrt{2}/2$	$1$
$\pi/3$	$\sqrt{3}/2$	$1/2$	$\sqrt{3}$
$\pi/2$	$1$	$0$	undefined
$\pi$	$0$	$-1$	$0$
$3\pi/2$	$-1$	$0$	undefined
$2\pi$	$0$	$1$	$0$

The Paper 1 expects fluency with these. Substituting $\pi = 3.14159$ and computing decimals loses marks on a Paper 1 exact-value question.

Worked example

Three definite integrals of trigonometric functions, each requiring exact-value evaluation at standard angles.

Step 1: Integrate $\sec^2 x$ on $[0, \pi/4]$

The antiderivative of $\sec^2 x$ is $\tan x$ (this is the reverse of the differentiation rule). There is no chain rule factor needed here because the argument is simply $x$ .

\int_{0}^{\pi/4} \sec^2(x) \, dx = [\tan x]_{0}^{\pi/4} = \tan\!\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1

Step 2: Integrate $\cos(3x)$ on $[0, \pi/6]$ , including the reverse-chain factor

The antiderivative of $\cos(kx)$ is $\frac{1}{k}\sin(kx)$ . Here $k = 3$ , so the factor $\frac{1}{3}$ appears in front. This corrects for the factor of $3$ that the chain rule would produce on differentiation.

\int_{0}^{\pi/6} \cos(3x) \, dx = \left[\frac{1}{3}\sin(3x)\right]_{0}^{\pi/6} = \frac{1}{3}\sin\!\left(\frac{\pi}{2}\right) - \frac{1}{3}\sin(0) = \frac{1}{3}(1) - 0 = \frac{1}{3}

Step 3: Integrate a mixed integrand on $[0, \pi/2]$

For $\int_{0}^{\pi/2} [2\sin x + \cos(2x)] \, dx$ , we antidifferentiate each term. The antiderivative of $2\sin x$ is $-2\cos x$ (the sign changes), and the antiderivative of $\cos(2x)$ includes the $\frac{1}{2}$ reverse-chain factor.

\text{Antiderivative:} \quad -2\cos x + \frac{1}{2}\sin(2x)

Evaluate at the upper limit $x = \pi/2$ , using $\cos(\pi/2) = 0$ and $\sin(\pi) = 0$ :

-2\cos\!\left(\frac{\pi}{2}\right) + \frac{1}{2}\sin(\pi) = -2(0) + \frac{1}{2}(0) = 0

Evaluate at the lower limit $x = 0$ , using $\cos(0) = 1$ and $\sin(0) = 0$ :

-2\cos(0) + \frac{1}{2}\sin(0) = -2(1) + 0 = -2

Subtract lower from upper to get the definite integral.

Final answer: $0 - (-2) = 2$ .

Area between trig curves

Trig curves often appear in area problems. The general procedure (top minus bottom on the interval between intersection points) is the same as in the area-between-curves dot point.

Example. Area enclosed between $y = \sin x$ and $y = \cos x$ on $[0, \pi/4]$ .

At $x = 0$ : $\sin 0 = 0$ , $\cos 0 = 1$ . So $\cos x > \sin x$ on $[0, \pi/4)$ .

Area $= \int_{0}^{\pi/4} (\cos x - \sin x) \, dx = [\sin x + \cos x]_{0}^{\pi/4} = (\sin(\pi/4) + \cos(\pi/4)) - (0 + 1) = (\sqrt{2}/2 + \sqrt{2}/2) - 1 = \sqrt{2} - 1$ .

Applications in PSMT and EA

The QCAA Methods PSMT often involves modelling periodic phenomena (tides, biological cycles, oscillating systems). Integration of the model gives accumulated quantities (total flow, energy, average values).

The EA Paper 1 short response routinely includes trig integration questions involving the $\frac{1}{k}$ factor.

Common errors

Missing the $\frac{1}{k}$ factor: $\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C$ , not $-\cos(2x) + C$ . This is the single most common error.
Wrong sign on $\cos$ antiderivative: $\int \sin x = -\cos x$ (sign change); $\int \cos x = +\sin x$ (no sign change).
Using degrees instead of radians: QCAA Methods Paper 1 uses radians. $\sin(\pi/2) = 1$ ; $\sin(90)$ in calculator mode-degrees is the same value, but the formula expects radians.
Substituting $\pi$ as 3.14 in Paper 1: Paper 1 expects exact values.
Forgetting the constant of integration: Indefinite integrals always include $C$ . Definite integrals do not (it cancels in the subtraction).
Sign-cancellation forgotten when integrand changes sign on the interval: For total area, split at zeros and take absolute values; for signed integral, do not.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks

Evaluate

\int_{0}^{\pi/2} (3 \cos(2x) - 2 \sin(x)) \, dx

exactly.

Show worked answer →

Antidifferentiate term by term using the reverse-chain factor for $\sin(2x)$ .

$\int 3 \cos(2x) \, dx = 3 \cdot \frac{1}{2} \sin(2x) = \frac{3}{2} \sin(2x)$ .

$\int -2 \sin(x) \, dx = 2 \cos(x)$ .

Combined antiderivative: $F(x) = \frac{3}{2} \sin(2x) + 2 \cos(x)$ .

Apply the fundamental theorem.

$F(\pi/2) - F(0) = \left[ \frac{3}{2} \sin(\pi) + 2 \cos(\pi/2) \right] - \left[ \frac{3}{2} \sin(0) + 2 \cos(0) \right] = (0 + 0) - (0 + 2) = -2$ .

Markers reward the $1/2$ factor inside $\sin(2x)$ , the correct sign on $\cos$ antiderivative (the antiderivative of $-2\sin x$ is $+2\cos x$ , because the sign flips twice), exact-value evaluation at $\pi$ and $\pi/2$ , and the boxed final answer.

Practice4 marks

A particle moves with velocity

v(t) = 5 \sin(2t)

m/s, where

t

is in seconds. (a) Find the position function

x(t)

given

x(0) = 0

. (b) Find the displacement of the particle from

t = 0

t = \pi

seconds.

Show worked answer →

(a) Position function. Antidifferentiate velocity.

$x(t) = \int 5 \sin(2t) \, dt = 5 \cdot -\frac{1}{2} \cos(2t) + C = -\frac{5}{2} \cos(2t) + C$ .

Apply $x(0) = 0$ : $-\frac{5}{2} \cdot 1 + C = 0$ , so $C = \frac{5}{2}$ .

Therefore $x(t) = \frac{5}{2} (1 - \cos(2t))$ m.

(b) Displacement from $t = 0$ to $t = \pi$ .

$\text{Displacement} = x(\pi) - x(0) = \frac{5}{2}(1 - \cos(2\pi)) - 0 = \frac{5}{2}(1 - 1) = 0$ m.

The particle returns to its starting position at $t = \pi$ (one full period of $\sin(2t)$ ).

Markers reward the antiderivative with $-\frac{1}{2}$ factor, applying initial condition to find $C$ , and the displacement calculation showing the particle returns to start.

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA