Trigonometric substitution is a technique for integrating expressions involving square roots of quadratic forms: $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, and $\sqrt{x^2 - a^2}$. By substituting $x$ with a trigonometric function, the square root simplifies nicely using Pythagorean identities.
🎯 In this section you will learn
- The three main trigonometric substitutions and when to use each
- How to simplify $\sqrt{a^2 - x^2}$ using $x = a\sin\theta$
- How to simplify $\sqrt{a^2 + x^2}$ using $x = a\tan\theta$
- How to simplify $\sqrt{x^2 - a^2}$ using $x = a\sec\theta$
- How to convert back to $x$ after integrating
📌 The Three Cases
| Expression | Substitution | Identity | $dx$ |
|---|
| $\sqrt{a^2 - x^2}$ | $x = a\sin\theta$, $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ | $1 - \sin^2\theta = \cos^2\theta$ | $dx = a\cos\theta\, d\theta$ |
| $\sqrt{a^2 + x^2}$ | $x = a\tan\theta$, $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ | $1 + \tan^2\theta = \sec^2\theta$ | $dx = a\sec^2\theta\, d\theta$ |
| $\sqrt{x^2 - a^2}$ | $x = a\sec\theta$, $0 \le \theta < \frac{\pi}{2}$ or $\pi \le \theta < \frac{3\pi}{2}$ | $\sec^2\theta - 1 = \tan^2\theta$ | $dx = a\sec\theta\tan\theta\, d\theta$ |
💡 Key Identity
The substitution is chosen so that the square root becomes a perfect square:
- $\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\sqrt{1-\sin^2\theta} = a|\cos\theta|$
- $\sqrt{a^2 + x^2} = \sqrt{a^2 + a^2\tan^2\theta} = a\sqrt{1+\tan^2\theta} = a|\sec\theta|$
- $\sqrt{x^2 - a^2} = \sqrt{a^2\sec^2\theta - a^2} = a\sqrt{\sec^2\theta - 1} = a|\tan\theta|$
Find $\displaystyle \int \frac{1}{\sqrt{4 - x^2}} dx$.
①
Identify $a$ and substitution
$a = 2$, $x = 2\sin\theta$, $dx = 2\cos\theta\, d\theta$
②
Simplify $\sqrt{4 - x^2}$
$\sqrt{4 - 4\sin^2\theta} = \sqrt{4(1-\sin^2\theta)} = 2\cos\theta$ (for $\theta$ where $\cos\theta \ge 0$)
③
Rewrite the integral
$\int \frac{1}{2\cos\theta} \cdot 2\cos\theta\, d\theta = \int 1\, d\theta = \theta + C$
④
Convert back to $x$
From $x = 2\sin\theta$, $\theta = \arcsin\left(\frac{x}{2}\right)$
$\int \frac{1}{\sqrt{4-x^2}} dx = \arcsin\left(\frac{x}{2}\right) + C$
Find $\displaystyle \int \frac{1}{x^2\sqrt{x^2 + 4}} dx$.
①
Identify $a$ and substitution
$a = 2$, $x = 2\tan\theta$, $dx = 2\sec^2\theta\, d\theta$
②
Simplify $\sqrt{x^2 + 4}$
$\sqrt{4\tan^2\theta + 4} = \sqrt{4(\tan^2\theta+1)} = 2\sec\theta$
③
Rewrite the integral
$\int \frac{1}{4\tan^2\theta \cdot 2\sec\theta} \cdot 2\sec^2\theta\, d\theta = \int \frac{\sec\theta}{4\tan^2\theta} d\theta = \frac{1}{4} \int \frac{\cos\theta}{\sin^2\theta} d\theta$
④
Integrate (let $u = \sin\theta$)
$\frac{1}{4} \int u^{-2} du = -\frac{1}{4} u^{-1} + C = -\frac{1}{4\sin\theta} + C$
⑤
Convert back to $x$
$\tan\theta = \frac{x}{2}$, so $\sin\theta = \frac{x}{\sqrt{x^2+4}}$
$\int \frac{1}{x^2\sqrt{x^2+4}} dx = -\frac{\sqrt{x^2+4}}{4x} + C$
Find $\displaystyle \int \frac{1}{\sqrt{x^2 - 9}} dx$ for $x > 3$.
①
Identify $a$ and substitution
$a = 3$, $x = 3\sec\theta$, $dx = 3\sec\theta\tan\theta\, d\theta$
②
Simplify $\sqrt{x^2 - 9}$
$\sqrt{9\sec^2\theta - 9} = \sqrt{9(\sec^2\theta-1)} = 3\tan\theta$ (for $\theta$ where $\tan\theta \ge 0$)
③
Rewrite the integral
$\int \frac{1}{3\tan\theta} \cdot 3\sec\theta\tan\theta\, d\theta = \int \sec\theta\, d\theta = \ln|\sec\theta + \tan\theta| + C$
④
Convert back to $x$
$\sec\theta = \frac{x}{3}$, $\tan\theta = \frac{\sqrt{x^2-9}}{3}$
$\int \frac{1}{\sqrt{x^2-9}} dx = \ln\left|\frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right| + C = \ln\left|x + \sqrt{x^2-9}\right| + C'$
⚠️ Common Mistakes
- Wrong substitution: Make sure you match the form ($\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, or $\sqrt{x^2-a^2}$) with the correct substitution.
- Forgetting $dx$: When you substitute $x$, you must also replace $dx$ with the appropriate $d\theta$ expression.
- Not converting back to $x$: Always express the final answer in terms of the original variable.
- Ignoring domain restrictions: The substitution must be invertible on the interval of integration.
🔍 Key Takeaways
- Use $x = a\sin\theta$ for $\sqrt{a^2 - x^2}$.
- Use $x = a\tan\theta$ for $\sqrt{a^2 + x^2}$.
- Use $x = a\sec\theta$ for $\sqrt{x^2 - a^2}$.
- After substituting, simplify using Pythagorean identities.
- Convert back to $x$ using a right triangle or inverse trig functions.
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