Tutorials on Advanced Math and Computer Science Concepts

# Trigonometric Substituion

Due to how triangles and circles interact with trigonometric functions, there are a number of substitutions we can make using trig functions that are helpful. There are three main substitutions we will discuss that are helpful.

1. $\sqrt{a^2-x^2} = \sqrt{a^2 - (a*sin(\theta))^2}, when -\frac{\pi}{2} \le \theta \ge \frac{\pi}{2}$
2. $\sqrt{a^2+x^2} = \sqrt{a^2+(a*tan(\theta))^2}, when - \frac{\pi}{2} < \theta < \frac{\pi}{2}$
3. $\sqrt{x^2-a^2} = \sqrt{(a*sec(\theta))^2 - a^2}, when 0 \le \theta < \frac{pi}{2}$ or $\pi \le \theta < \frac{3\pi}{2}$

Example: Evaluate $\int \frac{\sqrt{9-x^2}}{x^2} dx$

We can let $x = 3sin(\theta)$, meaning that $dx = 3cos(\theta) d\theta$, and $\sqrt{9 - x^2} = \sqrt{9-9sin^2(\theta) = \sqrt{9cos^2(\theta)} = 3 cos(\theta)$

From here, our integral becomes easy to solve.

$\int \frac{\sqrt{9-x^2}}{x^2} dx = \int \frac{3cos(\theta)}{9 sin^2(\theta)} 3cos(\theta) d\theta$
$\int \frac{cos^2(\theta)}{sin^2(\theta)} d\theta = \int cot^2(\theta) d\theta = -cot(\theta) - \theta + C$