Trig Sub: √(x²-a²)
Trig Sub: $\sqrt{x^2 - a^2}$
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 3.3: “Trigonometric Substitution” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/3-3-trigonometric-substitution |
| Textbook used in class | Stewart, Calculus, Section 7.3: “Trigonometric Substitution” |
Opening Scenario
The third and final trig substitution case handles $\sqrt{x^2 - a^2}$, which appears in arc-length problems for hyperbolas and in gravitational fields. The identity $\sec^2\theta - 1 = \tan^2\theta$ does the work: substituting $x = a\sec\theta$ turns $x^2 - a^2$ into $a^2\tan^2\theta$, eliminating the square root.
Quick Reference
Substitution: $x = a\sec\theta$, $dx = a\sec\theta\tan\theta\,d\theta$.
Domain restriction: $\theta \in [0, \pi/2)$ when $x \geq a > 0$; $\theta \in (\pi/2, \pi]$ when $x \leq -a$.
Key simplification: $\sqrt{x^2-a^2} = \sqrt{a^2\sec^2\theta - a^2} = a\sqrt{\sec^2\theta-1} = a|\tan\theta|$.
On $[0,\pi/2)$: $\tan\theta \geq 0$, so $|\tan\theta| = \tan\theta$ and $\sqrt{x^2-a^2} = a\tan\theta$.
Reference triangle: $\sec\theta = x/a$, so hypotenuse $= x$, adjacent $= a$, opposite $= \sqrt{x^2-a^2}$.
Key Concepts
1. Matching the Form
For $\sqrt{x^2 - a^2}$, the variable $x$ is under the square root in the form $x^2 - a^2$. The identity that makes this vanish is:
$$\sec^2\theta - 1 = \tan^2\theta \quad \Longrightarrow \quad x^2 - a^2 = a^2\sec^2\theta - a^2 = a^2\tan^2\theta.$$
This only works when $x \geq a$ (or $x \leq -a$), since $\sqrt{x^2 - a^2}$ requires $|x| \geq a$.
2. The Reference Triangle
With $\sec\theta = x/a$ (i.e., $\cos\theta = a/x$), draw the right triangle:
- Hypotenuse = $x$
- Adjacent = $a$
- Opposite = $\sqrt{x^2 - a^2}$
Then $\tan\theta = \dfrac{\sqrt{x^2-a^2}}{a}$ and $\sin\theta = \dfrac{\sqrt{x^2-a^2}}{x}$.
3. The Three-Case Summary
| Form | Substitution | Identity used |
|---|---|---|
| $\sqrt{a^2-x^2}$ | $x = a\sin\theta$ | $\cos^2 = 1 - \sin^2$ |
| $\sqrt{a^2+x^2}$ | $x = a\tan\theta$ | $\sec^2 = 1 + \tan^2$ |
| $\sqrt{x^2-a^2}$ | $x = a\sec\theta$ | $\tan^2 = \sec^2 - 1$ |
“For $\sqrt{x^2-a^2}$, use $x = a\tan\theta$ since it involves tangent.” The form $\sqrt{x^2-a^2}$ requires $x = a\sec\theta$, not $x = a\tan\theta$. A tangent substitution turns $a^2 + a^2\tan^2\theta = a^2\sec^2\theta$ (the sum form). Only the secant substitution produces the difference $a^2\sec^2\theta - a^2 = a^2\tan^2\theta$.
Worked Example
Evaluate $\displaystyle\int\frac{\sqrt{x^2-9}}{x}\,dx$ for $x > 3$.
Step 1. $a = 3$, $x = 3\sec\theta$, $dx = 3\sec\theta\tan\theta\,d\theta$, $\sqrt{x^2-9} = 3\tan\theta$ (for $\theta\in[0,\pi/2)$).
Step 2 -- Rewrite.
$$\int\frac{3\tan\theta}{3\sec\theta}\cdot 3\sec\theta\tan\theta\,d\theta = 3\int\tan^2\theta\,d\theta = 3\int(\sec^2\theta-1)\,d\theta.$$
Step 3 -- Integrate.
$3(\tan\theta - \theta) + C$.
Step 4 -- Back-substitute. $\tan\theta = \dfrac{\sqrt{x^2-9}}{3}$, $\theta = \text{arcsec}(x/3)$:
$$= 3\cdot\frac{\sqrt{x^2-9}}{3} - 3\,\text{arcsec}\!\left(\frac{x}{3}\right) + C = \sqrt{x^2-9} - 3\,\text{arcsec}\!\left(\frac{x}{3}\right) + C.$$
Boxed answer: $\sqrt{x^2-9} - 3\arccos\!\left(\dfrac{3}{x}\right) + C$ (since $\text{arcsec}(x/3) = \arccos(3/x)$).
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Confusing the secant case with the tangent case | Setting $x = 3\tan\theta$ for $\sqrt{x^2-9}$ | The secant substitution creates $\sec^2\theta - 1 = \tan^2\theta$; use $x = 3\sec\theta$ |
| Forgetting $dx = a\sec\theta\tan\theta\,d\theta$ | Missing the $\sec\theta\tan\theta$ factor in $dx$ | Differentiate $x = a\sec\theta$: $dx = a\sec\theta\tan\theta\,d\theta$ |
| Taking $|\tan\theta| = \tan\theta$ without checking | Applying this simplification for $x < -a$ | For $x < -a$, $\theta \in (\pi/2,\pi]$ where $\tan\theta < 0$; need $|\tan\theta| = -\tan\theta$ |
Leveled Practice
Level 1 -- Setup
Problem 1. For $\displaystyle\int\frac{x^2}{\sqrt{x^2-1}}\,dx$ ($x > 1$), identify the substitution, find $dx$ and $\sqrt{x^2-1}$ in terms of $\theta$.
Show answer
$a = 1$, $x = \sec\theta$, $dx = \sec\theta\tan\theta\,d\theta$, $\sqrt{x^2-1} = \tan\theta$.
$\displaystyle\int\frac{\sec^2\theta}{\tan\theta}\cdot\sec\theta\tan\theta\,d\theta = \int\sec^3\theta\,d\theta$.
(Use the formula $\int\sec^3\theta\,d\theta = \frac{1}{2}\sec\theta\tan\theta + \frac{1}{2}\ln|\sec\theta+\tan\theta| + C$.)
Mastery Checklist
Mental Model
Think of the three forms geometrically:
- $\sqrt{a^2-x^2}$: the height of a point on a circle of radius $a$. Parametrize by $x = a\sin\theta$.
- $\sqrt{a^2+x^2}$: the hypotenuse of a right triangle with legs $a$ and $x$. Parametrize by $x = a\tan\theta$ so the hypotenuse is $a\sec\theta$.
- $\sqrt{x^2-a^2}$: one leg of a right triangle with hypotenuse $x$ and other leg $a$. Parametrize by $x = a\sec\theta$ so this leg is $a\tan\theta$.
Each substitution matches a specific geometric configuration.
Connections
Looking back
- Trig sub: $\sqrt{a^2-x^2}$ and $\sqrt{a^2+x^2}$ (Section 7.3): The first two cases; this is the third.
- Trig integrals (Section 7.2): The substitution typically produces integrals involving $\sec\theta$ and $\tan\theta$.
Looking ahead
- Partial fractions (Section 7.4): A different technique for rational functions, not requiring a geometric substitution.
Back to Techniques of Integration | Next: Partial Fractions -- Distinct Linear Factors