What is wrong sign on cos antiderivative?

int sin x = -cos x (sign change); int cos x = +sin x (no sign change).

What is 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.

QCAA 2024 QCAA-style P1-style practice: Evaluate _0^π/2 (3 cos(2x) - 2 sin(x)) \, dx exactly.

Antidifferentiate term by term using the reverse-chain factor for sin(2x). int 3 cos(2x) \, dx = 3 · (1)/(2) sin(2x) = (3)/(2) sin(2x). int -2 sin(x) \, dx = 2 cos(x). Combined antiderivative: F(x) = (3)/(2) sin(2x) + 2 cos(x). Apply the fundamental theorem. F(π/2) - F(0) = [ (3)/(2) sin(π) + 2 cos(π/2) ] - [ (3)/(2) sin(0) + 2 cos(0) ] = (0 + 0) - (0 + 2) = -2. Markers reward the 1/2 factor inside sin(2x), the correct sign on cos antiderivative (it's -(-sin) = +cos - wait, antiderivative of -2sin x is +2cos x here, careful), exact-value evaluation at π and π/2, and the boxed final answer.

QLDMath MethodsSyllabus dot point

Topic 2: Trigonometric functions and integration applications

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

A focused answer to the QCE Maths Methods Unit 4 dot point on integrating trigonometric functions. Antiderivatives of $\sin(kx)$ , $\cos(kx)$ and $\sec^2(kx)$ with the $1/k$ reverse-chain factor, definite-integral evaluation with exact values at standard angles, and worked PSMT-style applications.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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 practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 QCAA-style P13 marksEvaluate

\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 (it's $-(-\sin) = +\cos$ - wait, antiderivative of $-2\sin x$ is $+2\cos x$ here, careful), exact-value evaluation at $\pi$ and $\pi/2$ , and the boxed final answer.

2023 QCAA-style P24 marksA 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.

Related dot points

Find antiderivatives of standard functions including polynomial, exponential and trigonometric forms, evaluate definite integrals using the Fundamental Theorem of Calculus, and recognise the definite integral as the limit of a Riemann sum
A focused answer to the QCE Mathematical Methods Unit 3 dot point on integration. Covers the standard antiderivatives, the linear-inside-argument shortcut, the Fundamental Theorem of Calculus as the bridge between differentiation and integration, and the Riemann-sum definition of the definite integral, with worked Paper 1 and Paper 2 examples QCAA examiners reward.