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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.

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What this dot point is asking

QCAA wants you to integrate trigonometric functions of the form sin⁑(kx)\sin(kx), cos⁑(kx)\cos(kx) and sec⁑2(kx)\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 1k\frac{1}{k} reverse-chain factor for non-unit coefficients.

Integrand Antiderivative
IMATH_5 IMATH_6
IMATH_7 IMATH_8
IMATH_9 IMATH_10
IMATH_11 IMATH_12
IMATH_13 IMATH_14
IMATH_15 IMATH_16

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

Sign check

The pattern for derivatives is:

  • IMATH_20 (no sign change).
  • IMATH_21 (sign change).

Reversing:

  • IMATH_22 (no sign change).
  • IMATH_23 (sign change).

The antiderivative of sin⁑\sin is βˆ’cos⁑-\cos. The antiderivative of cos⁑\cos is +sin⁑+\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)f(ax + b) where a,ba, b are constants:

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

where FF is an antiderivative of ff.

Examples:

  • IMATH_32 .
  • IMATH_33 .
  • IMATH_34 .

The 1a\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 Ο€\pi, fractions, surds).

Standard exact values:

IMATH_37 IMATH_38 IMATH_39 IMATH_40
IMATH_41 IMATH_42 IMATH_43 IMATH_44
IMATH_45 IMATH_46 IMATH_47 IMATH_48
IMATH_49 IMATH_50 IMATH_51 IMATH_52
IMATH_53 IMATH_54 IMATH_55 IMATH_56
IMATH_57 IMATH_58 IMATH_59 undefined
IMATH_60 IMATH_61 IMATH_62 IMATH_63
IMATH_64 IMATH_65 IMATH_66 undefined
IMATH_67 IMATH_68 IMATH_69 IMATH_70

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

Worked examples

Example 1. Simple definite integral

∫0Ο€/4sec⁑2(x) dx=[tan⁑x]0Ο€/4=tan⁑(Ο€/4)βˆ’tan⁑(0)=1βˆ’0=1\int_{0}^{\pi/4} \sec^2(x) \, dx = [\tan x]_{0}^{\pi/4} = \tan(\pi/4) - \tan(0) = 1 - 0 = 1.

Example 2. With reverse-chain factor

∫0Ο€/6cos⁑(3x) dx=[13sin⁑(3x)]0Ο€/6=13sin⁑(Ο€/2)βˆ’13sin⁑(0)=13(1)βˆ’0=13\int_{0}^{\pi/6} \cos(3x) \, dx = \left[ \frac{1}{3} \sin(3x) \right]_{0}^{\pi/6} = \frac{1}{3} \sin(\pi/2) - \frac{1}{3} \sin(0) = \frac{1}{3}(1) - 0 = \frac{1}{3}.

Example 3. Mixed integrand

∫0Ο€/2[2sin⁑x+cos⁑(2x)] dx\int_{0}^{\pi/2} [2 \sin x + \cos(2x)] \, dx.

Antiderivative. βˆ’2cos⁑x+12sin⁑(2x)-2 \cos x + \frac{1}{2} \sin(2x).

Evaluate at x=Ο€/2x = \pi/2. βˆ’2cos⁑(Ο€/2)+12sin⁑(Ο€)=0+0=0-2 \cos(\pi/2) + \frac{1}{2} \sin(\pi) = 0 + 0 = 0.

Evaluate at x=0x = 0. βˆ’2cos⁑(0)+12sin⁑(0)=βˆ’2+0=βˆ’2-2 \cos(0) + \frac{1}{2} \sin(0) = -2 + 0 = -2.

Subtract. 0βˆ’(βˆ’2)=20 - (-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⁑xy = \sin x and y=cos⁑xy = \cos x on [0,Ο€/4][0, \pi/4].

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

Area =∫0Ο€/4(cos⁑xβˆ’sin⁑x) dx=[sin⁑x+cos⁑x]0Ο€/4=(sin⁑(Ο€/4)+cos⁑(Ο€/4))βˆ’(0+1)=(2/2+2/2)βˆ’1=2βˆ’1= \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 1k\frac{1}{k} factor.

Common errors

Missing the 1k\frac{1}{k} factor. ∫sin⁑(2x) dx=βˆ’12cos⁑(2x)+C\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C, not βˆ’cos⁑(2x)+C-\cos(2x) + C. This is the single most common error.

Wrong sign on cos⁑\cos antiderivative. ∫sin⁑x=βˆ’cos⁑x\int \sin x = -\cos x (sign change); ∫cos⁑x=+sin⁑x\int \cos x = +\sin x (no sign change).

Using degrees instead of radians. QCAA Methods Paper 1 uses radians. sin⁑(Ο€/2)=1\sin(\pi/2) = 1; sin⁑(90)\sin(90) in calculator mode-degrees is the same value, but the formula expects radians.

Substituting Ο€\pi as 3.14 in Paper 1. Paper 1 expects exact values.

Forgetting the constant of integration. Indefinite integrals always include CC. 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.

In one sentence

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

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2024 QCAA-style P13 marksEvaluate $\int_{0}^{\pi/2} (3 \cos(2x) - 2 \sin(x)) \, dx$ exactly.
Show worked answer β†’

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

∫3cos⁑(2x) dx=3β‹…12sin⁑(2x)=32sin⁑(2x)\int 3 \cos(2x) \, dx = 3 \cdot \frac{1}{2} \sin(2x) = \frac{3}{2} \sin(2x).

βˆ«βˆ’2sin⁑(x) dx=2cos⁑(x)\int -2 \sin(x) \, dx = 2 \cos(x).

Combined antiderivative: F(x)=32sin⁑(2x)+2cos⁑(x)F(x) = \frac{3}{2} \sin(2x) + 2 \cos(x).

Apply the fundamental theorem.

F(Ο€/2)βˆ’F(0)=[32sin⁑(Ο€)+2cos⁑(Ο€/2)]βˆ’[32sin⁑(0)+2cos⁑(0)]=(0+0)βˆ’(0+2)=βˆ’2F(\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 1/21/2 factor inside sin⁑(2x)\sin(2x), the correct sign on cos⁑\cos antiderivative (it's βˆ’(βˆ’sin⁑)=+cos⁑-(-\sin) = +\cos - wait, antiderivative of βˆ’2sin⁑x-2\sin x is +2cos⁑x+2\cos x here, careful), exact-value evaluation at Ο€\pi and Ο€/2\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 $t = 0$ to $t = \pi$ seconds.
Show worked answer β†’

(a) Position function. Antidifferentiate velocity.

x(t)=∫5sin⁑(2t) dt=5β‹…βˆ’12cos⁑(2t)+C=βˆ’52cos⁑(2t)+Cx(t) = \int 5 \sin(2t) \, dt = 5 \cdot -\frac{1}{2} \cos(2t) + C = -\frac{5}{2} \cos(2t) + C.

Apply x(0)=0x(0) = 0: βˆ’52β‹…1+C=0-\frac{5}{2} \cdot 1 + C = 0, so C=52C = \frac{5}{2}.

Therefore x(t)=52(1βˆ’cos⁑(2t))x(t) = \frac{5}{2} (1 - \cos(2t)) m.

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

Displacement=x(Ο€)βˆ’x(0)=52(1βˆ’cos⁑(2Ο€))βˆ’0=52(1βˆ’1)=0\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=Ο€t = \pi (one full period of sin⁑(2t)\sin(2t)).

Markers reward the antiderivative with βˆ’12-\frac{1}{2} factor, applying initial condition to find CC, and the displacement calculation showing the particle returns to start.

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