Trigonometric identities for integration
Master trigonometric identities for integration for Year 12 Specialist Mathematics in Queensland (QCAA). These are the identities that let you integrate squares and products of trig functions — the power-reduction, double-angle and Pythagorean identities.
You will learn to power-reduce \(\sin^2 x\) and \(\cos^2 x\), rewrite \(\sin x\cos x\) as \(\tfrac12\sin 2x\), turn \(\tan^2 x\) into \(\sec^2 x-1\), integrate odd powers by substitution, and evaluate definite integrals as exact values — core integration techniques that support areas and volumes later in Unit 4.
Theory
Trigonometric identities for integration turn integrals that have no direct antiderivative into ones you can integrate on sight, in Year 12 Specialist Mathematics (QCAA, Queensland). The power-reduction identities \(\sin^2 x=\tfrac12(1-\cos 2x)\) and \(\cos^2 x=\tfrac12(1+\cos 2x)\), the double-angle form \(\sin x\cos x=\tfrac12\sin 2x\) and the Pythagorean identities \(1+\tan^2 x=\sec^2 x\), \(\cot^2 x+1=\operatorname{cosec}^2 x\) rewrite squares and products before you integrate. This page shows which identity to reach for and how to finish each type.
Powers and products of \(\sin\) and \(\cos\) have no antiderivative in their given form, so you first rewrite them with an identity that turns the awkward term into something you can integrate directly. The identity is chosen by the shape of the integrand.
The power-reduction identities handle a single square. From the double-angle form of \(\cos 2x\) you get \(\sin^2 x=\tfrac12(1-\cos 2x)\) and \(\cos^2 x=\tfrac12(1+\cos 2x)\); each replaces a square by a constant plus a single cosine, which integrates term by term. The same works with a multiple angle, e.g. \(\cos^2 3x=\tfrac12(1+\cos 6x)\).
The double-angle identity \(\sin 2x=2\sin x\cos x\) rearranges to \(\sin x\cos x=\tfrac12\sin 2x\), so a product of one sine and one cosine becomes a single sine. Squaring it turns \(\sin^2 x\cos^2 x\) into \(\tfrac14\sin^2 2x\), which is then power-reduced again.
The Pythagorean identities \(1+\tan^2 x=\sec^2 x\) and \(\cot^2 x+1=\operatorname{cosec}^2 x\) deal with \(\tan^2\) and \(\cot^2\), because \(\displaystyle\int\sec^2 x\,dx=\tan x+c\) and \(\displaystyle\int\operatorname{cosec}^2 x\,dx=-\cot x+c\) are standard results. An odd power such as \(\sin^3 x\) is split as \(\sin^2 x\cdot\sin x=(1-\cos^2 x)\sin x\), then a substitution \(u=\cos x\) finishes it.
The power-reduction identities remove a square from \(\sin\) or \(\cos\):
The double-angle identity turns a product into a single sine:
The Pythagorean identities, with their standard integrals:
Integrating a power or product of trig functions
- Classify the integrand: a single square \(\sin^2\) or \(\cos^2\), a product \(\sin x\cos x\), a \(\tan^2\)/\(\cot^2\) term, or an odd power such as \(\sin^3 x\).
- Rewrite with the matching identity: power-reduce a square to \(\tfrac12(1\mp\cos 2x)\); write \(\sin x\cos x=\tfrac12\sin 2x\); replace \(\tan^2 x\) with \(\sec^2 x-1\) (or \(\cot^2 x\) with \(\operatorname{cosec}^2 x-1\)); split an odd power and set \(u=\cos x\) or \(u=\sin x\).
- Integrate term by term: remember \(\displaystyle\int\cos kx\,dx=\tfrac1k\sin kx\) and \(\displaystyle\int\sec^2 x\,dx=\tan x\); keep the \(\tfrac12\) (or \(\tfrac14\)) factor out the front.
- Finish: add \(+c\) for an indefinite integral, or substitute the limits and simplify to an exact value (a multiple of \(\pi\), a surd, or a plain number) for a definite integral.
Replace \(\cos^2 x\) with its power-reduction identity:
| \(8\cos^2 x\) | \(=\) | \(8\cdot\dfrac{1}{2}(1+\cos 2x)\) |
| \(=\) | \(4(1+\cos 2x)\) |
Integrate term by term, using \(\displaystyle\int\cos 2x\,dx=\tfrac12\sin 2x\):
| \(\int 8\cos^2 x\,dx\) | \(=\) | \(4\int (1+\cos 2x)\,dx\) |
| \(=\) | \(4\left(x+\dfrac{1}{2}\sin 2x\right)+c\) | |
| \(=\) | \(4x+2\sin 2x+c\) |
\(\displaystyle\int 8\cos^2 x\,dx=4x+2\sin 2x+c\).
Use \(1+\tan^2\theta=\sec^2\theta\) with \(\theta=3x\), so \(\tan^2 3x=\sec^2 3x-1\):
| \(\tan^2 3x\) | \(=\) | \(\sec^2 3x-1\) |
Integrate, using \(\displaystyle\int\sec^2 3x\,dx=\tfrac13\tan 3x\):
| \(\int \tan^2 3x\,dx\) | \(=\) | \(\int (\sec^2 3x-1)\,dx\) |
| \(=\) | \(\dfrac{1}{3}\tan 3x-x+c\) |
\(\displaystyle\int \tan^2 3x\,dx=\dfrac{1}{3}\tan 3x-x+c\).
Rewrite the product with the double-angle identity:
| \(\sin x\cos x\) | \(=\) | \(\dfrac{1}{2}\sin 2x\) |
Integrate, using \(\displaystyle\int\sin 2x\,dx=-\tfrac12\cos 2x\):
| \(\int \sin x\cos x\,dx\) | \(=\) | \(\dfrac{1}{2}\left(-\dfrac{1}{2}\cos 2x\right)\) |
| \(=\) | \(-\dfrac{1}{4}\cos 2x\) |
Substitute the limits \(0\) and \(\dfrac{\pi}{2}\):
| \(\int_0^{\pi/2} \sin x\cos x\,dx\) | \(=\) | \(\left[-\dfrac{1}{4}\cos 2x\right]_0^{\pi/2}\) |
| \(=\) | \(-\dfrac{1}{4}\cos\pi+\dfrac{1}{4}\cos 0\) | |
| \(=\) | \(-\dfrac{1}{4}(-1)+\dfrac{1}{4}(1)\) | |
| \(=\) | \(\dfrac{1}{2}\) |
\(\displaystyle\int_0^{\pi/2} \sin x\cos x\,dx=\dfrac{1}{2}\).
Split off one \(\sin x\) and use \(\sin^2 x=1-\cos^2 x\):
| \(\sin^3 x\) | \(=\) | \((1-\cos^2 x)\sin x\) |
Substitute \(u=\cos x\), so \(du=-\sin x\,dx\):
| \(\int \sin^3 x\,dx\) | \(=\) | \(-\int (1-u^2)\,du\) |
| \(=\) | \(-u+\dfrac{u^3}{3}+c\) | |
| \(=\) | \(-\cos x+\dfrac{1}{3}\cos^3 x+c\) |
Substitute the limits \(0\) and \(\dfrac{\pi}{2}\):
| \(\int_0^{\pi/2} \sin^3 x\,dx\) | \(=\) | \(\left[-\cos x+\dfrac{1}{3}\cos^3 x\right]_0^{\pi/2}\) |
| \(=\) | \((0+0)-\left(-1+\dfrac{1}{3}\right)\) | |
| \(=\) | \(\dfrac{2}{3}\) |
\(\displaystyle\int_0^{\pi/2} \sin^3 x\,dx=\dfrac{2}{3}\).
Common pitfalls
Frequently asked questions
How do you integrate sin squared x?
Power-reduce first: \(\sin^2 x=\tfrac12(1-\cos 2x)\), so \(\displaystyle\int\sin^2 x\,dx=\tfrac12 x-\tfrac14\sin 2x+c\).
How do you integrate cos squared x?
Use \(\cos^2 x=\tfrac12(1+\cos 2x)\), which gives \(\displaystyle\int\cos^2 x\,dx=\tfrac12 x+\tfrac14\sin 2x+c\).
What is the integral of sec squared x?
It is a standard result: \(\displaystyle\int\sec^2 x\,dx=\tan x+c\), because \(\dfrac{d}{dx}(\tan x)=\sec^2 x\).
How do you integrate tan squared x?
Replace it using the Pythagorean identity: \(\tan^2 x=\sec^2 x-1\), so \(\displaystyle\int\tan^2 x\,dx=\tan x-x+c\).
How do you integrate sin x cos x?
Use the double-angle identity \(\sin x\cos x=\tfrac12\sin 2x\), giving \(\displaystyle\int\sin x\cos x\,dx=-\tfrac14\cos 2x+c\).
How do you integrate an odd power like sin cubed x?
Split off one factor: \(\sin^3 x=(1-\cos^2 x)\sin x\), then substitute \(u=\cos x\) to get \(-\cos x+\tfrac13\cos^3 x+c\).