26.4 Further Trigonometric Integrals
AHL 5.16
This section is for HL students. It builds on 26.3.
- how to integrate \(\sec^2 x\) and \(\sec^2(ax + b)\)
- how to integrate \(f'(x)\sec^2\big[f(x)\big]\), \(f'(x)\sec\big[f(x)\big]\tan\big[f(x)\big]\) and similar products by inspection
- how to integrate \(\tan^2 x\) and \(\cot^2 x\) using a Pythagorean identity
The integrals of \(\sec^2 x\), \(\sec x\tan x\), \(\operatorname{cosec}^2 x\) and \(\operatorname{cosec} x\cot x\) come from the HL derivatives in AHL 5.15. Angles are always in radians.
Secant squared
Each rule is a derivative read backwards: \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\tan x) = \sec^2 x\).
Integral of secant squared Not in the formula booklet
\[ \int \sec^2 x\,\mathrm{d}x = \tan x + C. \]
With a linear expression inside (\(a \neq 0\)), divide by \(a\):
\[ \int \sec^2(ax + b)\,\mathrm{d}x = \frac{1}{a}\tan(ax + b) + C. \]
- \(C\): the constant of integration
- \(a\): the coefficient of \(x\) (\(a \neq 0\))
- \(b\): the constant term
\(\sec^2 x\) appears in the booklet only as the derivative of \(\tan x\) (HL 5.15).
Example 1 Find \(\displaystyle\int \frac{5 - 3\cos^2 x}{2\cos^2 x}\,\mathrm{d}x\).
Split the fraction: \(\dfrac{5}{2\cos^2 x} - \dfrac{3\cos^2 x}{2\cos^2 x} = \dfrac{5}{2}\sec^2 x - \dfrac{3}{2}\).
\[ \int \left(\frac{5}{2}\sec^2 x - \frac{3}{2}\right)\mathrm{d}x = \frac{5}{2}\tan x - \frac{3}{2}x + C \]
Further integrals by inspection
In 26.3 we read the derivatives of \(\sin\big[f(x)\big]\) and \(\cos\big[f(x)\big]\) backwards. The HL derivatives give four more patterns.
Further trigonometric integrals by inspection Not in the formula booklet
| Integral | Result |
|---|---|
| \(\int f'(x)\sec^2\big[f(x)\big]\,\mathrm{d}x\) | \(\tan\big[f(x)\big] + C\) |
| \(\int f'(x)\sec\big[f(x)\big]\tan\big[f(x)\big]\,\mathrm{d}x\) | \(\sec\big[f(x)\big] + C\) |
| \(\int f'(x)\operatorname{cosec}\big[f(x)\big]\cot\big[f(x)\big]\,\mathrm{d}x\) | \(-\operatorname{cosec}\big[f(x)\big] + C\) |
| \(\int f'(x)\operatorname{cosec}^2\big[f(x)\big]\,\mathrm{d}x\) | \(-\cot\big[f(x)\big] + C\) |
- \(f(x)\): a function of \(x\) (the inside function)
- \(f'(x)\): the derivative of \(f(x)\)
- \(C\): the constant of integration
The HL booklet lists the derivatives of \(\tan x\), \(\sec x\), \(\operatorname{cosec} x\) and \(\cot x\) (5.15).
Example 2 Find \(\displaystyle\int x^3\sec^2\left(x^4\right)\mathrm{d}x\).
\(f(x) = x^4\), \(f'(x) = 4x^3\), so \(x^3 = \dfrac{1}{4}f'(x)\):
\[ \int x^3\sec^2\left(x^4\right)\mathrm{d}x = \frac{1}{4}\tan\left(x^4\right) + C. \]
Squares of tangent and cotangent
For \(\tan^2(ax + b)\) and \(\cot^2(ax + b)\) use the Pythagorean identities
\[ \tan^2\theta = \sec^2\theta - 1, \qquad \cot^2\theta = \operatorname{cosec}^2\theta - 1, \]
because \(\sec^2\) and \(\operatorname{cosec}^2\) can be integrated directly.
\(1 + \tan^2\theta = \sec^2\theta\) and \(1 + \cot^2\theta = \operatorname{cosec}^2\theta\) are in the HL formula booklet (AHL 3.9).
Example 3 Find \(\displaystyle\int \tan^2(3x - \pi)\,\mathrm{d}x\).
There is no direct rule for \(\tan^2\). Use \(1 + \tan^2\theta = \sec^2\theta\), so \(\tan^2\theta = \sec^2\theta - 1\):
\[ \int \left[\sec^2(3x - \pi) - 1\right]\mathrm{d}x = \frac{1}{3}\tan(3x - \pi) - x + C. \]
(Since \(\tan\) has period \(\pi\), this is also \(\dfrac{1}{3}\tan 3x - x + C\).)
Example 4 Find \(\displaystyle\int \cot^2\left(\frac{x}{2} + \frac{\pi}{3}\right)\mathrm{d}x\).
\(\cot^2\theta = \operatorname{cosec}^2\theta - 1\), and \(\int \operatorname{cosec}^2(ax + b)\,\mathrm{d}x = -\dfrac{1}{a}\cot(ax + b) + C\). Here \(a = \dfrac{1}{2}\):
\[ \int \left[\operatorname{cosec}^2\left(\frac{x}{2} + \frac{\pi}{3}\right) - 1\right]\mathrm{d}x = -2\cot\left(\frac{x}{2} + \frac{\pi}{3}\right) - x + C. \]
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\int \tan^2 x\,\mathrm{d}x = \dfrac{\tan^3 x}{3}\) | \(\tan x - x + C\) | Use \(\tan^2 x = \sec^2 x - 1\) first. |
| \(\int \operatorname{cosec}^2 x\,\mathrm{d}x = \cot x\) | \(-\cot x + C\) | \(\frac{\mathrm{d}}{\mathrm{d}x}\cot x = -\operatorname{cosec}^2 x\), so the sign changes. |
Exercises
1 Find \(\displaystyle\int \tan^2 2x\,\mathrm{d}x\).
\(\displaystyle\int (\sec^2 2x - 1)\,\mathrm{d}x = \frac{1}{2}\tan 2x - x + C\).
2 Find \(\displaystyle\int x\operatorname{cosec}\left(x^2\right)\cot\left(x^2\right)\mathrm{d}x\).
\(f(x) = x^2\), \(f'(x) = 2x\), so \(x = \dfrac{1}{2}f'(x)\):
\[ \int x\operatorname{cosec}\left(x^2\right)\cot\left(x^2\right)\mathrm{d}x = -\frac{1}{2}\operatorname{cosec}\left(x^2\right) + C. \]