Papers 2 and 3. There is no rule for integrating \(\sin^2 x\), \(\cos^2 x\) or \(\tan^2 x\) directly. An identity turns each one into something you can integrate: a constant plus a cosine, or \(\sec^2 x\) minus 1.
Squares of sine and cosine are integrated by lowering the power with the double-angle formula: \(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\) and \(\cos^2 x = \dfrac{1}{2}(1 + \cos 2x)\). For \(\tan^2 x\) use \(\tan^2 x = \sec^2 x - 1\).
Fig. 1: y = sin²x = 1/2 - (1/2)cos 2x oscillates about the line y = 1/2; over 0 to π the cosine part integrates to zero, so the area is (1/2) × π = π/2
Section 01
Squares of sine and cosine: lower the power.
Rearrange \(\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1\) to express each square in terms of \(\cos 2x\). The square disappears, and \(\cos 2x\) is a standard integral.
\(\sin^2 x = \dfrac{1}{2} - \dfrac{1}{2}\cos 2x\), so \(\displaystyle\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{1}{4}\sin 2x + c\).
\(\cos^2 x = \dfrac{1}{2} + \dfrac{1}{2}\cos 2x\), so \(\displaystyle\int \cos^2 x\,dx = \dfrac{x}{2} + \dfrac{1}{4}\sin 2x + c\).
The angle doubles: \(\cos^2 3x = \dfrac{1}{2}(1 + \cos 6x)\) and \(\sin^2\dfrac{x}{2} = \dfrac{1}{2}(1 - \cos x)\).
Worked example Integrate sine squared
Find \(\displaystyle\int \sin^2 x\,dx\).
Use \(\cos 2x = 1 - 2\sin^2 x\), so \(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\).
\(\displaystyle\int \sin^2 x\,dx\) is NOT \(\dfrac{\sin^3 x}{3}\). The power rule only works on \(x^{n}\); differentiating \(\dfrac{\sin^3 x}{3}\) gives \(\sin^2 x\cos x\), not \(\sin^2 x\). And when the angle is \(2x\), the identity brings in \(\cos 4x\), not \(\cos 2x\).
Section 02
Tan squared and products: find the hidden identity.
Other integrands become standard once you spot the right identity. Ask: which identity turns this into \(\sec^2\), \(\sin\) or \(\cos\) of a single angle?
You see
Rewrite as
Integral
\(\tan^2 x\)
\(\sec^2 x - 1\)
\(\tan x - x + c\)
\(\sin x\cos x\)
\(\dfrac{1}{2}\sin 2x\)
\(-\dfrac{1}{4}\cos 2x + c\)
\(\dfrac{1}{1+\cos 2x}\)
\(\dfrac{1}{2\cos^2 x} = \dfrac{1}{2}\sec^2 x\)
\(\dfrac{1}{2}\tan x + c\)
Worked example Tan squared
Find the exact value of \(\displaystyle\int_0^{\pi/4} \tan^2 x\,dx\).
From \(1 + \tan^2 x = \sec^2 x\): \(\tan^2 x = \sec^2 x - 1\).
\(\tan^2 x = \sec^2 x - 1\), not \(\sec^2 x + 1\). Check with \(x = 0\): \(\tan^2 0 = 0\) and \(\sec^2 0 - 1 = 0\). The wrong sign turns \(1 - \dfrac{\pi}{4}\) into \(1 + \dfrac{\pi}{4}\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which expression is equal to \(\sin^2 x\)?
Quick check+10 XP
Find \(\displaystyle\int \tan^2 x\,dx\).
Quick check+10 XP
Which expression is equal to \(\cos^2 3x\)?
Quick check+10 XP
Find \(\displaystyle\int \sin x\cos x\,dx\).
Section 03
Expand first, then use the identities.
A bracket squared often contains \(\sin^2 x + \cos^2 x\), which is 1, and a cross term \(2\sin x\cos x\), which is \(\sin 2x\). Higher even powers are reduced by applying the double-angle formula twice.
Worked example A squared bracket
Find the exact value of \(\displaystyle\int_0^{\pi/2} (\sin x + \cos x)^2\,dx\).
Expand: \(\sin^2 x + 2\sin x\cos x + \cos^2 x\).
Use \(\sin^2 x + \cos^2 x = 1\) and \(2\sin x\cos x = \sin 2x\): the integrand is \(1 + \sin 2x\).