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 integrals related to circles: the equation $x^2 + y^2 = a^2$ gives $y = \sqrt{a^2 - x^2}$, the upper semicircle. Direct substitution cannot remove the square root because the derivative of the inside is $-2x$, and there is no extra $x$ factor in the integrand to absorb it. Instead, a trigonometric substitution eliminates the square root by exploiting the identity $1 - \sin^2\theta = \cos^2\theta$.
Quick Reference
Substitution: $x = a\sin\theta$, $dx = a\cos\theta\,d\theta$, $\theta \in [-\pi/2,\, \pi/2]$.
Key simplification: $\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\sqrt{1-\sin^2\theta} = a|\cos\theta| = a\cos\theta$ (since $\cos\theta \geq 0$ on $[-\pi/2, \pi/2]$).
Back-substitution: From $\sin\theta = x/a$, draw the reference triangle: opposite $= x$, hypotenuse $= a$, adjacent $= \sqrt{a^2 - x^2}$. Read off $\cos\theta = \sqrt{a^2-x^2}/a$ and $\tan\theta = x/\sqrt{a^2-x^2}$.
Key Concepts
1. Why This Substitution Works
The goal is to eliminate $\sqrt{a^2 - x^2}$. Substituting $x = a\sin\theta$ gives: $$a^2 - x^2 = a^2 - a^2\sin^2\theta = a^2(1 - \sin^2\theta) = a^2\cos^2\theta.$$
So $\sqrt{a^2 - x^2} = a\cos\theta$ (taking positive root since $\theta \in [-\pi/2, \pi/2]$). The square root vanishes.
2. The Reference Triangle
After integrating in $\theta$, you must return to the original variable $x$. The relationship $\sin\theta = x/a$ defines a right triangle:
- Opposite = $x$
- Hypotenuse = $a$
- Adjacent = $\sqrt{a^2 - x^2}$ (by Pythagorean theorem)
Every trig function of $\theta$ can be read from this triangle: $$\cos\theta = \frac{\sqrt{a^2-x^2}}{a}, \qquad \tan\theta = \frac{x}{\sqrt{a^2-x^2}}.$$
3. The Restricted Domain
Restricting $\theta \in [-\pi/2, \pi/2]$ ensures that $\sin\theta$ is one-to-one (so the substitution can be inverted) and that $\cos\theta \geq 0$ (so $|\cos\theta| = \cos\theta$, avoiding absolute values). For definite integrals, you can either convert the limits from $x$ to $\theta$ at the start, or back-substitute and use the original $x$-limits.
“$\sqrt{a^2\cos^2\theta} = a\cos\theta$ always, without the domain restriction.” Without restricting $\theta$, $\sqrt{a^2\cos^2\theta} = a|\cos\theta|$. The absolute value simplifies to $a\cos\theta$ only because we restrict to $\theta \in [-\pi/2, \pi/2]$, where $\cos\theta \geq 0$. Using the substitution outside this range introduces sign errors.
Worked Example
Evaluate $\displaystyle\int \frac{x^2}{\sqrt{9-x^2}}\,dx$.
Step 1 -- Identify. $a = 3$, form is $\sqrt{a^2 - x^2}$.
Step 2 -- Substitute. $x = 3\sin\theta$, $dx = 3\cos\theta\,d\theta$, $\sqrt{9-x^2} = 3\cos\theta$.
Step 3 -- Rewrite.
$$\int\frac{9\sin^2\theta}{3\cos\theta}\cdot 3\cos\theta\,d\theta = \int 9\sin^2\theta\,d\theta.$$
Step 4 -- Integrate. Use $\sin^2\theta = \dfrac{1-\cos(2\theta)}{2}$:
$$9\int\frac{1-\cos(2\theta)}{2}\,d\theta = \frac{9}{2}\left[\theta - \frac{\sin(2\theta)}{2}\right] + C = \frac{9\theta}{2} - \frac{9\sin(2\theta)}{4} + C.$$
Step 5 -- Back-substitute. Use $\sin(2\theta) = 2\sin\theta\cos\theta = 2\cdot\dfrac{x}{3}\cdot\dfrac{\sqrt{9-x^2}}{3} = \dfrac{2x\sqrt{9-x^2}}{9}$ and $\theta = \arcsin(x/3)$:
$$= \frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) - \frac{9}{4}\cdot\frac{2x\sqrt{9-x^2}}{9} + C = \frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) - \frac{x\sqrt{9-x^2}}{2} + C.$$
Boxed answer: $\dfrac{9}{2}\arcsin\!\left(\dfrac{x}{3}\right) - \dfrac{x\sqrt{9-x^2}}{2} + C$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting $dx = a\cos\theta\,d\theta$ | Substituting for $x$ and $\sqrt{a^2-x^2}$ but leaving $dx$ unchanged | Every instance of $x$ and $dx$ must be replaced by expressions in $\theta$ and $d\theta$ |
| Skipping the reference triangle | Leaving the answer in $\theta$ | Draw the triangle with opposite $=x$, hypotenuse $=a$ and read off $\cos\theta$, $\tan\theta$ to back-substitute |
| Taking $\sqrt{a^2\cos^2\theta} = a\cos\theta$ outside the restricted domain | Using the substitution for $\theta > \pi/2$ | The simplification $|\cos\theta| = \cos\theta$ holds only for $\theta\in[-\pi/2,\pi/2]$ |
Leveled Practice
Level 1 -- Setup
Problem 1. For the integral $\displaystyle\int \frac{1}{\sqrt{4-x^2}}\,dx$, write down the substitution, compute $dx$ and $\sqrt{4-x^2}$ in terms of $\theta$.
Show answer
$a = 2$, $x = 2\sin\theta$, $dx = 2\cos\theta\,d\theta$, $\sqrt{4-x^2} = 2\cos\theta$.
$\displaystyle\int\frac{2\cos\theta\,d\theta}{2\cos\theta} = \int d\theta = \theta + C = \arcsin(x/2) + C$.
Level 2 -- Full Computation
Problem 2. Evaluate $\displaystyle\int_0^{3/2}\frac{x^3}{\sqrt{9-x^2}}\,dx$.
Show answer
$x = 3\sin\theta$, $dx = 3\cos\theta\,d\theta$, $\sqrt{9-x^2} = 3\cos\theta$.
New limits: $x=0\Rightarrow\theta=0$; $x=3/2\Rightarrow\sin\theta=1/2\Rightarrow\theta=\pi/6$.
$\displaystyle\int_0^{\pi/6}\frac{27\sin^3\theta}{3\cos\theta}\cdot 3\cos\theta\,d\theta = 27\int_0^{\pi/6}\sin^3\theta\,d\theta$.
$\sin^3\theta = \sin\theta(1-\cos^2\theta)$. Let $u=\cos\theta$, $du=-\sin\theta\,d\theta$:
$27\int_1^{\sqrt{3}/2}(-(1-u^2))\,du = 27\int_{\sqrt{3}/2}^1(1-u^2)\,du = 27\left[u-\frac{u^3}{3}\right]_{\sqrt{3}/2}^1$
$= 27\left[\left(1-\frac{1}{3}\right) - \left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{8}\right)\right] = 27\left[\frac{2}{3} - \frac{3\sqrt{3}}{8}\right] = 18 - \frac{81\sqrt{3}}{8}$.
Mastery Checklist
Mental Model
The substitution $x = a\sin\theta$ is parametric: it traces the circle $x^2 + y^2 = a^2$ via the angle $\theta$. The expression $\sqrt{a^2-x^2}$ is the $y$-coordinate of a point on that circle, and it simplifies to $a\cos\theta$ because the Pythagorean theorem for the circle gives $y^2 = a^2 - x^2 = a^2\cos^2\theta$.
Connections
Looking back
- Trig integrals (Section 7.2): After substitution, the integral typically becomes a trig integral.
- Inverse trig functions: $\theta = \arcsin(x/a)$ is needed for back-substitution.
Looking ahead
- Trig sub: $\sqrt{a^2+x^2}$ (Section 7.3): Uses $x = a\tan\theta$ and the identity $1 + \tan^2\theta = \sec^2\theta$.
- Trig sub: $\sqrt{x^2-a^2}$ (Section 7.3): Uses $x = a\sec\theta$ and the identity $\sec^2\theta - 1 = \tan^2\theta$.
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