Trigonometric Substitution

When to Use

Use trigonometric substitution for integrals containing:

• $\sqrt{a^2 - x^2}$
• $\sqrt{a^2 + x^2}$
• $\sqrt{x^2 - a^2}$

The Three Cases

Case 1: $\sqrt{a^2 - x^2}$

Substitution: $x = a \sin(\theta)$, $dx = a \cos(\theta) \, d\theta$

Identity: $\sqrt{a^2 - x^2} = a \cos(\theta)$

Range: $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$

Case 2: $\sqrt{a^2 + x^2}$

Substitution: $x = a \tan(\theta)$, $dx = a \sec^2(\theta) \, d\theta$

Identity: $\sqrt{a^2 + x^2} = a \sec(\theta)$

Range: $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$

Case 3: $\sqrt{x^2 - a^2}$

Substitution: $x = a \sec(\theta)$, $dx = a \sec(\theta) \tan(\theta) \, d\theta$

Identity: $\sqrt{x^2 - a^2} = a \tan(\theta)$

Range: $0 \leq \theta < \frac{\pi}{2}$ or $\pi \leq \theta < \frac{3\pi}{2}$

Steps

1. Identify the form and make substitution
2. Replace $\sqrt{\phantom{x}}$ expression using identity
3. Substitute $dx$
4. Simplify and integrate
5. Use reference triangle to convert back to $x$

Examples

Example 1: $\sqrt{a^2 - x^2}$

$\int \sqrt{4 - x^2} \, dx$

Let $x = 2\sin(\theta)$, $dx = 2\cos(\theta) \, d\theta$

$\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos(\theta)$

$= \int 2\cos(\theta) \cdot 2\cos(\theta) \, d\theta = 4 \int \cos^2(\theta) \, d\theta$

$= 4 \int \frac{1 + \cos(2\theta)}{2} \, d\theta = 2\left[\theta + \frac{\sin(2\theta)}{2}\right] + C$

$= 2\theta + \sin(2\theta) + C = 2\theta + 2\sin(\theta)\cos(\theta) + C$

Convert back: $\sin(\theta) = \frac{x}{2}$, $\cos(\theta) = \frac{\sqrt{4-x^2}}{2}$

$= 2 \arcsin\left(\frac{x}{2}\right) + 2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4-x^2}}{2} + C$

$= 2 \arcsin\left(\frac{x}{2}\right) + \frac{x\sqrt{4-x^2}}{2} + C$

Example 2: $\sqrt{a^2 + x^2}$

$\int \frac{1}{\sqrt{x^2 + 9}} \, dx$

Let $x = 3\tan(\theta)$, $dx = 3\sec^2(\theta) \, d\theta$

$\sqrt{x^2 + 9} = 3\sec(\theta)$

$= \int \frac{3\sec^2(\theta)}{3\sec(\theta)} \, d\theta = \int \sec(\theta) \, d\theta$

$= \ln|\sec(\theta) + \tan(\theta)| + C$

Convert back: $\tan(\theta) = \frac{x}{3}$, $\sec(\theta) = \frac{\sqrt{x^2+9}}{3}$

$= \ln\left|\frac{\sqrt{x^2+9}}{3} + \frac{x}{3}\right| + C = \ln\left|\sqrt{x^2+9} + x\right| - \ln(3) + C$

$= \ln\left|\sqrt{x^2+9} + x\right| + C_1$

Example 3: $\sqrt{x^2 - a^2}$

$\int \frac{\sqrt{x^2 - 4}}{x} \, dx$

Let $x = 2\sec(\theta)$, $dx = 2\sec(\theta)\tan(\theta) \, d\theta$

$\sqrt{x^2 - 4} = 2\tan(\theta)$

$= \int \frac{2\tan(\theta)}{2\sec(\theta)} \cdot 2\sec(\theta)\tan(\theta) \, d\theta = 2 \int \tan^2(\theta) \, d\theta$

$= 2 \int (\sec^2(\theta) - 1) \, d\theta = 2[\tan(\theta) - \theta] + C$

Convert back: $\sec(\theta) = \frac{x}{2}$, $\tan(\theta) = \frac{\sqrt{x^2-4}}{2}$

$= 2 \cdot \frac{\sqrt{x^2-4}}{2} - 2 \text{arcsec}\left(\frac{x}{2}\right) + C$

$= \sqrt{x^2 - 4} - 2 \text{arcsec}\left(\frac{x}{2}\right) + C$

Example 4: Completing the square first

$\int \frac{1}{\sqrt{x^2 + 4x + 8}} \, dx$

Complete the square: $x^2 + 4x + 8 = (x + 2)^2 + 4$

Let $u = x + 2$, then $\sqrt{u^2 + 4}$

Let $u = 2\tan(\theta)$, $du = 2\sec^2(\theta) \, d\theta$

$= \int \frac{2\sec^2(\theta)}{2\sec(\theta)} \, d\theta = \int \sec(\theta) \, d\theta$

$= \ln|\sec(\theta) + \tan(\theta)| + C$

$= \ln\left|\frac{\sqrt{u^2+4}}{2} + \frac{u}{2}\right| + C$

$= \ln\left|\sqrt{(x+2)^2+4} + (x+2)\right| + C_1$