Trig Sub: √(a²+x²)
Trig Sub: $\sqrt{a^2 + x^2}$
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 expression $\sqrt{a^2 + x^2}$ appears in arc-length integrals, in the formula for the length of a catenary, and in electrostatics. Unlike $\sqrt{a^2 - x^2}$, this expression grows without bound as $x \to \infty$, so it cannot be eliminated by a sine substitution. The identity $1 + \tan^2\theta = \sec^2\theta$ provides the right tool: substituting $x = a\tan\theta$ turns $a^2 + x^2$ into $a^2\sec^2\theta$.
Quick Reference
Substitution: $x = a\tan\theta$, $dx = a\sec^2\theta\,d\theta$, $\theta \in (-\pi/2,\, \pi/2)$.
Key simplification: $\sqrt{a^2 + x^2} = \sqrt{a^2(1+\tan^2\theta)} = a|\sec\theta| = a\sec\theta$ (since $\sec\theta > 0$ on $(-\pi/2,\pi/2)$).
Reference triangle: $\tan\theta = x/a$, so opposite $= x$, adjacent $= a$, hypotenuse $= \sqrt{a^2+x^2}$.
$$\cos\theta = \frac{a}{\sqrt{a^2+x^2}}, \qquad \sin\theta = \frac{x}{\sqrt{a^2+x^2}}, \qquad \sec\theta = \frac{\sqrt{a^2+x^2}}{a}.$$
Key Concepts
1. Why This Substitution Works
With $x = a\tan\theta$: $$a^2 + x^2 = a^2 + a^2\tan^2\theta = a^2(1 + \tan^2\theta) = a^2\sec^2\theta.$$
So $\sqrt{a^2 + x^2} = a\sec\theta$ (positive, since $\sec\theta > 0$ on the restricted domain). The integral then involves powers of $\sec\theta$, which are handled by the techniques of Section 7.2.
2. The Reference Triangle
The substitution $\tan\theta = x/a$ corresponds to a right triangle with:
- Opposite = $x$
- Adjacent = $a$
- Hypotenuse = $\sqrt{a^2 + x^2}$
Read all trig functions of $\theta$ from this triangle for back-substitution.
“$\sqrt{a^2+x^2}$ should use $x = a\sin\theta$ just like $\sqrt{a^2-x^2}$.” The sign inside the square root determines the identity to use. For $a^2 - x^2$, use $\sin^2 + \cos^2 = 1$ (subtract); for $a^2 + x^2$, use $1 + \tan^2 = \sec^2$ (add). Matching the sign inside to the correct identity is the whole principle.
Worked Example
Evaluate $\displaystyle\int \frac{1}{\sqrt{x^2+4}}\,dx$.
Step 1. $a = 2$, $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$, $\sqrt{x^2+4} = 2\sec\theta$.
Step 2 -- Rewrite.
$$\int\frac{2\sec^2\theta\,d\theta}{2\sec\theta} = \int\sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C.$$
Step 3 -- Back-substitute. $\sec\theta = \dfrac{\sqrt{x^2+4}}{2}$, $\tan\theta = \dfrac{x}{2}$:
$$= \ln\!\left|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\right| + C = \ln\!\left|\frac{\sqrt{x^2+4}+x}{2}\right| + C = \ln\!\left|\sqrt{x^2+4}+x\right| + C_1$$
(absorbing $-\ln 2$ into the constant).
Boxed answer: $\ln\!\left|\sqrt{x^2+4}+x\right| + C$ (also written $\sinh^{-1}(x/2) + C$).
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $x = a\sin\theta$ for $\sqrt{a^2+x^2}$ | Substituting $x = 2\sin\theta$ in $\int\frac{1}{\sqrt{x^2+4}}\,dx$ | The sum $a^2+x^2$ requires the tangent identity; use $x = a\tan\theta$ |
| Forgetting $dx = a\sec^2\theta\,d\theta$ | Missing the $\sec^2\theta$ factor after substituting $x$ | Differentiate $x = a\tan\theta$ to get $dx = a\sec^2\theta\,d\theta$ |
| Using $|\sec\theta|$ instead of $\sec\theta$ | Writing $\sqrt{a^2\sec^2\theta} = a|\sec\theta|$ without simplifying | On $(-\pi/2,\pi/2)$, $\sec\theta > 0$, so $|\sec\theta| = \sec\theta$ |
Leveled Practice
Level 1 -- Setup
Problem 1. For $\displaystyle\int\frac{x^2}{\sqrt{x^2+9}}\,dx$, write down the substitution, find $dx$ and $\sqrt{x^2+9}$ in terms of $\theta$.
Show answer
$a = 3$, $x = 3\tan\theta$, $dx = 3\sec^2\theta\,d\theta$, $\sqrt{x^2+9} = 3\sec\theta$.
$\displaystyle\int\frac{9\tan^2\theta\cdot 3\sec^2\theta\,d\theta}{3\sec\theta} = 9\int\tan^2\theta\sec\theta\,d\theta$.
(This integral requires $\int\sec^3\theta\,d\theta$ -- a good review of the sec/tan techniques.)
Level 2 -- Full Computation
Problem 2. Evaluate $\displaystyle\int_0^1\frac{1}{(1+x^2)^{3/2}}\,dx$.
Show answer
$a = 1$, $x = \tan\theta$, $dx = \sec^2\theta\,d\theta$, $(1+x^2)^{3/2} = \sec^3\theta$.
Bounds: $x=0\Rightarrow\theta=0$; $x=1\Rightarrow\theta=\pi/4$.
$\displaystyle\int_0^{\pi/4}\frac{\sec^2\theta\,d\theta}{\sec^3\theta} = \int_0^{\pi/4}\cos\theta\,d\theta = \sin\theta\Big|_0^{\pi/4} = \frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2}$.
Mastery Checklist
Mental Model
Three forms, three identities, three substitutions:
| Form | Identity used | Substitution |
|---|---|---|
| $\sqrt{a^2-x^2}$ | $1 - \sin^2\theta = \cos^2\theta$ | $x = a\sin\theta$ |
| $\sqrt{a^2+x^2}$ | $1 + \tan^2\theta = \sec^2\theta$ | $x = a\tan\theta$ |
| $\sqrt{x^2-a^2}$ | $\sec^2\theta - 1 = \tan^2\theta$ | $x = a\sec\theta$ |
Match the sign inside the square root to the right row.
Connections
Looking back
- Trig sub: $\sqrt{a^2-x^2}$ (Section 7.3): The first case; this lesson extends the method to the sum form.
- Trig integrals (Section 7.2): Powers of $\sec\theta$ resulting from this substitution require the techniques there.
Looking ahead
- Trig sub: $\sqrt{x^2-a^2}$ (Section 7.3): The third and final case.
Back to Techniques of Integration | Next: Trig Sub -- sqrt(x^2-a^2)