SCSA WACE 2022-style practice: Calculator-free. Evaluate _0^π/2 sin^2 x\,dx exactly.

An even-power integral by power reduction. Use sin^2 x = 1 - cos 2x2. Then _0^π/2 sin^2 x\,dx = 12_0^π/2 (1 - cos 2x)\,dx = 12[x - sin 2x2]_0^π/2. At x = π2: π2 - sinπ2 = π2 - 0. At x = 0: 0 - 0 = 0. So the value is 12 × π2 = π4. Markers reward the power-reduction identity, integrating cos 2x to 12sin 2x, and the exact answer π4.

SCSA WACE 2024-style practice: Calculator-assumed. Find sin^3 x\,dx.

An odd-power integral by split-and-substitute. Write sin^3 x = sin x · sin^2 x = sin x (1 - cos^2 x). Let u = cos x, so du = -sin x\,dx, that is sin x\,dx = -du. The integral becomes (1 - u^2)(-du) = (u^2 - 1)\,du = u^33 - u + C = cos^3 x3 - cos x + C. Markers reward peeling off one sin x, converting the rest with sin^2 x = 1 - cos^2 x, the substitution u = cos x, and substituting back.

§-Syllabus dot point

WASpecialist MathematicsSyllabus dot point

How do trigonometric identities convert products and powers of sine and cosine into integrable forms?

Integrate trigonometric functions by first applying double-angle, Pythagorean and product identities

WACE Specialist Unit 4 trigonometric integration: using double-angle identities to integrate sine and cosine squared, the Pythagorean identity for odd powers, and product-to-sum identities, with a worked example.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

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

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What this dot point is asking
Even powers: the double-angle identities
Where the power-reduction identities come from
Odd powers: split and substitute
Higher even powers
Products of different angles
Choosing the route from the integrand

What this dot point is asking

SCSA wants you to choose the right identity to transform a trigonometric integrand into a sum of standard integrals.

Even powers: the double-angle identities

The integrals of $\sin^2 x$ and $\cos^2 x$ rely on lowering the power. From the double-angle formula for cosine,

Each right-hand side integrates termwise to a sum of a linear term and a $\sin 2x$ term. Higher even powers are reduced by applying these identities repeatedly.

Where the power-reduction identities come from

The power-reduction identities are rearrangements of the double-angle formula for cosine, $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ . Solving the first form for $\sin^2 x$ gives $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ , and solving the second for $\cos^2 x$ gives $\cos^2 x = \dfrac{1 + \cos 2x}{2}$ . Knowing the derivation means you can reconstruct the identities under exam pressure rather than relying on memory, and it links this dot point back to the double-angle work of Methods. The same double-angle relation, read in the other direction, also reverses the process when you need to express a $\cos 2x$ answer back in terms of $\cos^2 x$ .

Odd powers: split and substitute

For an odd power such as $\sin^3 x$ , peel off one factor of $\sin x$ and convert the remaining even power using $\sin^2 x = 1 - \cos^2 x$ . Then substitute $u = \cos x$ , since $du = -\sin x\,dx$ supplies the leftover factor. The same idea handles odd powers of cosine with $u = \sin x$ .

Higher even powers

A fourth power such as $\cos^4 x$ is reduced by applying the power-reduction identity twice. First, $\cos^4 x = (\cos^2 x)^2 = \left(\dfrac{1 + \cos 2x}{2}\right)^2 = \dfrac{1}{4}(1 + 2\cos 2x + \cos^2 2x)$ . The remaining $\cos^2 2x$ is itself reduced by $\cos^2 2x = \dfrac{1 + \cos 4x}{2}$ , leaving only constants and single cosine terms in $2x$ and $4x$ , all of which integrate directly. The pattern is that each application of the identity halves the highest power, so any even power of sine or cosine eventually flattens into a sum of cosines of multiple angles.

Products of different angles

For products like $\sin mx \cos nx$ , use the product-to-sum identities to rewrite as a sum of single trigonometric terms, each of which integrates directly. The relevant identities are $\sin A\cos B = \dfrac{1}{2}\big[\sin(A + B) + \sin(A - B)\big]$ , $\cos A\cos B = \dfrac{1}{2}\big[\cos(A + B) + \cos(A - B)\big]$ and $\sin A\sin B = \dfrac{1}{2}\big[\cos(A - B) - \cos(A + B)\big]$ . After rewriting, each term is a single sine or cosine of a multiple angle and integrates with the usual $\dfrac{1}{k}$ factor.

Choosing the route from the integrand

Diagnosing the integrand quickly is the key skill. Scan the powers of sine and cosine present. If the highest power is even and there is no convenient derivative pairing, reach for power reduction. If there is an odd power, split off one factor of that function and substitute, letting the leftover factor become $du$ . If the integrand is a product of trig functions of different angles, use the product-to-sum identities. A mixed integrand such as $\sin^2 x\cos^3 x$ combines methods: peel off one $\cos x$ for the substitution $u = \sin x$ and convert the remaining $\cos^2 x = 1 - \sin^2 x$ , turning the whole thing into a polynomial in $u$ . Identifying the route before integrating avoids the dead ends that cost time in the exam.

Worked example

Evaluate $\displaystyle\int \cos^2 x\,dx$ .

Step 1: Apply the power-reduction identity

The integrand $\cos^2 x$ is an even power of cosine, and there is no paired derivative present to use a substitution. The power-reduction identity converts the squared cosine into a linear expression in $\cos 2x$ , which can be integrated term by term.

\cos^2 x = \frac{1 + \cos 2x}{2}

Step 2: Split and integrate each term

Substituting the identity rewrites the integral as two standard integrals. The constant $\tfrac{1}{2}$ is distributed across each, and each term is then integrated separately using the rule $\int \cos(ax)\,dx = \tfrac{1}{a}\sin(ax)$ .

\int \cos^2 x\,dx = \int \frac{1 + \cos 2x}{2}\,dx = \frac{1}{2}\int 1\,dx + \frac{1}{2}\int \cos 2x\,dx

The first integral gives $\tfrac{1}{2}x$ . For the second, $\int \cos 2x\,dx = \tfrac{1}{2}\sin 2x$ , so $\tfrac{1}{2} \times \tfrac{1}{2}\sin 2x = \tfrac{1}{4}\sin 2x$ .

Step 3: Combine and write the final answer

Adding the two results and including the constant of integration gives the complete antiderivative. The factor of $\tfrac{1}{2}$ on $\cos 2x$ compensates for the inner derivative of $2x$ .

\int \cos^2 x\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C

Final answer: $\dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$ .

Exam-style practice questions

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

WACE 20225 marksCalculator-free. Evaluate

\displaystyle\int_0^{\pi/2} \sin^2 x\,dx

exactly.

Show worked answer →

An even-power integral by power reduction.

Use $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ . Then $\displaystyle\int_0^{\pi/2} \sin^2 x\,dx = \dfrac{1}{2}\displaystyle\int_0^{\pi/2} (1 - \cos 2x)\,dx = \dfrac{1}{2}\left[x - \dfrac{\sin 2x}{2}\right]_0^{\pi/2}$ .

At $x = \dfrac{\pi}{2}$ : $\dfrac{\pi}{2} - \dfrac{\sin\pi}{2} = \dfrac{\pi}{2} - 0$ . At $x = 0$ : $0 - 0 = 0$ . So the value is $\dfrac{1}{2} \times \dfrac{\pi}{2} = \dfrac{\pi}{4}$ .

Markers reward the power-reduction identity, integrating $\cos 2x$ to $\dfrac{1}{2}\sin 2x$ , and the exact answer $\dfrac{\pi}{4}$ .

WACE 20246 marksCalculator-assumed. Find

\displaystyle\int \sin^3 x\,dx

Show worked answer →

An odd-power integral by split-and-substitute.

Write $\sin^3 x = \sin x \cdot \sin^2 x = \sin x (1 - \cos^2 x)$ . Let $u = \cos x$ , so $du = -\sin x\,dx$ , that is $\sin x\,dx = -du$ .

The integral becomes $\displaystyle\int (1 - u^2)(-du) = \displaystyle\int (u^2 - 1)\,du = \dfrac{u^3}{3} - u + C = \dfrac{\cos^3 x}{3} - \cos x + C$ .

Markers reward peeling off one $\sin x$ , converting the rest with $\sin^2 x = 1 - \cos^2 x$ , the substitution $u = \cos x$ , and substituting back.