Integration by parts and trigonometric integrals in QCE Specialist Mathematics Unit 4

Syllabus dot point

“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”

QCESpecialist MathematicsUnit 4: Further calculus, and…8 min readUpdated 2 Jun 2026

Quick answer

Integration by parts is $\int u\, dv = uv - \int v\, du$ ; choose $u$ by LIATE so the remaining integral is simpler, applying parts repeatedly if needed. For trigonometric integrals, reduce even powers with $\cos^2 x = \frac{1+\cos 2x}{2}$ and $\sin^2 x = \frac{1-\cos 2x}{2}$ , handle odd powers by peeling one factor and using $\sin^2 x = 1 - \cos^2 x$ before substituting, and convert products of sines and cosines with product-to-sum identities.

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What this dot point is asking
The answer
Exam-style questions

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. The identities are $\sin A\cos B = \tfrac{1}{2}[\sin(A + B) + \sin(A - B)]$ , $\cos A\cos B = \tfrac{1}{2}[\cos(A - B) + \cos(A + B)]$ and $\sin A\sin B = \tfrac{1}{2}[\cos(A - B) - \cos(A + B)]$ . Each turns a product into a sum of cosines or sines whose antiderivatives are immediate.

The cycling case of repeated parts

A special situation arises with integrands like $\int e^x \sin x\,dx$ , where applying parts twice returns a multiple of the original integral. Setting $I = \int e^x\sin x\,dx$ , two passes give $I = e^x\sin x - e^x\cos x - I$ , and solving algebraically yields $I = \tfrac{1}{2}e^x(\sin x - \cos x) + C$ . Recognising the cycle and solving for $I$ rather than integrating endlessly is the key insight, and QCAA expects the algebraic rearrangement to be shown explicitly.

Definite integrals by parts

For a definite integral the formula carries the limits on every term: $\displaystyle\int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du$ . Evaluate the boundary term $[uv]_a^b$ at both limits and keep the limits on the remaining integral. A frequent error is to apply the limits only at the end; instead the $uv$ product is evaluated at the limits as soon as it appears, which keeps the bookkeeping clean and avoids losing a boundary contribution.

Choosing between techniques

When an integrand is a product, ask first whether a simple substitution works (one factor is a constant multiple of the derivative of the other); if so, substitution is faster than parts. Reserve integration by parts for genuine products of unrelated functions, such as a polynomial times an exponential or a logarithm standing alone (with $dv = dx$ ). For powers and products of trigonometric functions, reach for identities before either substitution or parts, because reducing the power first usually leaves a standard form.

Worked example

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

This is a product of an algebraic function and a trigonometric function. Neither is the derivative of the other, so substitution will not help. Integration by parts is the right technique.

Step 1: Choose $u$ and $dv$ using LIATE

The LIATE rule puts Algebraic before Trigonometric, so we let $u = x$ (algebraic) and $dv = \cos x\, dx$ (trigonometric). Differentiating and integrating to find $du$ and $v$ :

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

Choosing $u = x$ means the new integral will contain $v\,du = \sin x \cdot 1\,dx$ , which is simpler than the original.

Step 2: Apply the parts formula

Substitute into $\displaystyle\int u\,dv = uv - \int v\,du$ :

\int x\cos x\, dx = x\sin x - \int \sin x \, dx.

Step 3: Integrate the remaining term

The remaining integral is a standard form:

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

Step 4: Check by differentiating

Differentiate the result to verify it reproduces the original integrand:

\frac{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 answer.

Final answer: $\displaystyle\int x\cos x\, dx = x\sin x + \cos x + C$ .

Common trap

The most common error is choosing $u$ and $dv$ the wrong way round, for example taking $u = \cos x$ in $\int x\cos x\, dx$ , which makes the new integral harder rather than easier. Use LIATE. A second slip is trying to integrate $\sin^2 x$ or $\cos^2 x$ directly; you must first apply the power-reduction identity, since there is no simple antiderivative of the squared form on its own.

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.

Practice7 marks

Paper 2 (complex unfamiliar). The function

f

passes through the origin and has gradient

f'(x) = e^x \arcsin(e^x)

. Determine

f(x)

Show worked answer →

$f(x) = \displaystyle\int e^x \arcsin(e^x)\,dx$ . Let $w = e^x$ , so $dw = e^x\,dx$ , giving $\displaystyle\int \arcsin(w)\,dw$ .

Integrate by parts with $u = \arcsin w$ ( $du = \dfrac{dw}{\sqrt{1 - w^2}}$ ) and $dv = dw$ ( $v = w$ ): $\displaystyle\int \arcsin w\,dw = w\arcsin w - \int \dfrac{w}{\sqrt{1 - w^2}}\,dw = w\arcsin w + \sqrt{1 - w^2}.$

Return to $x$ : $f(x) = e^x \arcsin(e^x) + \sqrt{1 - e^{2x}} + c$ . Apply $f(0) = 0$ : $\arcsin 1 + 0 + c = \tfrac{\pi}{2} + c = 0$ , so $c = -\tfrac{\pi}{2}$ . Thus $f(x) = e^x\arcsin(e^x) + \sqrt{1 - e^{2x}} - \tfrac{\pi}{2}.$

Markers reward the substitution, the parts step, the secondary substitution for the remaining integral, and applying the initial condition.

Practice4 marks

Paper 1 (technique). Determine

\displaystyle\int x e^{2x}\,dx