QLDSpecialist MathematicsSyllabus dot point

Topic 1: Integration and applications of integration

Evaluate integrals using integration by parts and integrate trigonometric expressions using identities such as double-angle and Pythagorean identities and products of sines and cosines

A focused answer to the QCE Specialist Mathematics Unit 4 dot point on integration by parts and trigonometric integrals. Covers the parts formula and the LIATE choice, repeated parts, and reducing powers of sine and cosine with double-angle and Pythagorean identities, with a verified worked example and the wrong-choice trap.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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What this dot point is asking

QCAA wants you to integrate products using integration by parts and to integrate powers and products of trigonometric functions by first rewriting them with identities. These techniques extend the substitution and partial-fraction methods, and the assessable skill is choosing the right approach and carrying out the algebra cleanly.

The answer

Integration by parts

Integration by parts reverses the product rule. For functions $u$ and $v$ ,

\int u \, \frac{dv}{dx}\, dx = uv - \int v\, \frac{du}{dx}\, dx.

You split the integrand into a part to differentiate ( $u$ ) and a part to integrate ( $dv$ ). A good choice makes the new integral simpler than the original.

Choosing $u$ with LIATE

The order Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential (LIATE) is a reliable guide: pick $u$ to be whichever factor appears first in that list, since differentiating it tends to simplify. For $\int x\cos x\, dx$ , the algebraic $x$ comes before the trigonometric $\cos x$ , so $u = x$ and $dv = \cos x\, dx$ .

Repeated parts

Sometimes you apply parts twice. For $\int x^2 e^x\, dx$ , the first application leaves $\int 2x e^x\, dx$ , which needs parts again. Each pass lowers the power of the algebraic factor until it disappears.

Trigonometric integrals: even powers

To integrate even powers of sine or cosine, use the power-reduction forms of the double-angle identity:

\cos^2 x = \frac{1 + \cos 2x}{2}, \qquad \sin^2 x = \frac{1 - \cos 2x}{2}.

These turn a squared trigonometric function into a linear one that integrates directly.

Trigonometric integrals: odd powers

For an odd power such as $\int \sin^3 x\, dx$ , peel off one factor and convert the rest with the Pythagorean identity $\sin^2 x = 1 - \cos^2 x$ , then substitute $u = \cos x$ . The single peeled factor becomes the $du$ .

Products of sine and cosine

For $\int \sin mx \cos nx\, dx$ and similar, the product-to-sum identities rewrite the product as a sum of single trigonometric terms, each integrating directly. The goal throughout is to reach an integrand that is a sum of standard forms.

Worked example

Evaluate $\displaystyle\int x\cos x\, dx$ .

Choose $u$ and $dv$ by LIATE. Algebraic before trigonometric, so

u = x, \quad \frac{dv}{dx} = \cos x \ \Rightarrow\ \frac{du}{dx} = 1, \quad v = \sin x.

Apply the parts formula.

\int x\cos x\, dx = uv - \int v\, \frac{du}{dx}\, dx = x\sin x - \int \sin x \, dx.

Integrate the remaining term.

\int x\cos x\, dx = x\sin x - (-\cos x) + C = x\sin x + \cos x + C.

Check by differentiating. $\dfrac{d}{dx}\big(x\sin x + \cos x\big) = \sin x + x\cos x - \sin x = x\cos x.$ This matches the integrand, confirming the result.

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.

2022 QCAA7 marksThe function f(x) passes through the origin. The gradient function of f(x) is defined as g(x) = e^x * arcsin(e^x). Determine f(x).

Show worked answer →

Since g(x) = f'(x), f(x) = integral of e^x * arcsin(e^x) dx.

Substitute to simplify [2 marks]: let w = e^x, so dw = e^x dx. Then f = integral of arcsin(w) dw.

Integrate by parts [2 marks]: take u = arcsin(w) (so du = 1/sqrt(1 - w^2) dw) and dv = dw (so v = w):
integral of arcsin(w) dw = w * arcsin(w) - integral of w / sqrt(1 - w^2) dw.

The remaining integral by substitution [1 mark]: integral of w / sqrt(1 - w^2) dw = -sqrt(1 - w^2).
So f = w * arcsin(w) + sqrt(1 - w^2) + c.

Return to x [1 mark]: f(x) = e^x * arcsin(e^x) + sqrt(1 - e^(2x)) + c.

Apply f(0) = 0 [1 mark]: e^0 * arcsin(1) + sqrt(1 - 1) + c = pi/2 + c = 0, so c = -pi/2.
Therefore f(x) = e^x * arcsin(e^x) + sqrt(1 - e^(2x)) - pi/2.