Integration using trigonometric identities (WACE Year 12 Mathematics Specialist Unit 4)

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.