Welcome to Professor Baker's Math Class!
Tonight's lesson focuses on Sections 7.2 and 7.3: Trigonometric Integrals and Trigonometric Substitution. We will explore techniques to solve integrals involving trigonometric functions. Let's dive in!
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Section 7.2: Trigonometric Integrals
This section uses trigonometric identities to integrate combinations of trigonometric functions. We'll start with powers of sine and cosine.
Strategy for Evaluating $\int \sin^m(x) \cos^n(x) dx$
- Case 1: If $n$ is odd ($n = 2k + 1$), save one cosine factor and use $\cos^2(x) = 1 - \sin^2(x)$ to express the remaining factors in terms of sine. Then, substitute $u = \sin(x)$. $$\int \sin^m(x) \cos^{2k+1}(x) dx = \int \sin^m(x) (\cos^2(x))^k \cos(x) dx = \int \sin^m(x) (1 - \sin^2(x))^k \cos(x) dx$$
- Case 2: If $m$ is odd ($m = 2k + 1$), save one sine factor and use $\sin^2(x) = 1 - \cos^2(x)$ to express the remaining factors in terms of cosine. Then, substitute $u = \cos(x)$. $$\int \sin^{2k+1}(x) \cos^n(x) dx = \int (\sin^2(x))^k \cos^n(x) \sin(x) dx = \int (1 - \cos^2(x))^k \cos^n(x) \sin(x) dx$$
- Case 3: If both $m$ and $n$ are even, use the half-angle identities: $$\sin^2(x) = \frac{1}{2}(1 - \cos(2x)), \quad \cos^2(x) = \frac{1}{2}(1 + \cos(2x))$$
It is sometimes helpful to use the identity: $\sin(x) \cos(x) = \frac{1}{2} \sin(2x)$
Example
Let's evaluate the following integral: $\int \sin^5(x) \cos^2(x) dx$
Since the power of sine is odd, we can use the strategy outlined in Case 2. We separate a sine factor and rewrite the remaining sine factors in terms of cosine. Let $u = \cos(x)$, so $du = -\sin(x) dx$.
$\int \sin^5(x) \cos^2(x) dx = \int \sin^4(x) \cos^2(x) \sin(x) dx = \int (1 - \cos^2(x))^2 \cos^2(x) \sin(x) dx$
$= \int (1-u^2)^2 u^2 (-du) = -\int (1 - 2u^2 + u^4)u^2 du = -\int (u^2 - 2u^4 + u^6) du$
$= -[\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7}] + C = -\frac{\cos^3(x)}{3} + \frac{2\cos^5(x)}{5} - \frac{\cos^7(x)}{7} + C$
Section 7.3: Trigonometric Substitution
Trigonometric substitution involves using trigonometric functions to substitute for expressions involving square roots, particularly when dealing with integrals of the form $\int \sqrt{a^2 - x^2} dx$, $\int \sqrt{a^2 + x^2} dx$, or $\int \sqrt{x^2 - a^2} dx$.
Table of Trigonometric Substitutions
| Expression | Substitution | Identity |
|---|---|---|
| $\sqrt{a^2 - x^2}$ | $x = a \sin(\theta)$, $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ | $1 - \sin^2(\theta) = \cos^2(\theta)$ |
| $\sqrt{a^2 + x^2}$ | $x = a \tan(\theta)$, $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ | $1 + \tan^2(\theta) = \sec^2(\theta)$ |
| $\sqrt{x^2 - a^2}$ | $x = a \sec(\theta)$, $0 \le \theta < \frac{\pi}{2}$ or $\pi \le \theta < \frac{3\pi}{2}$ | $\sec^2(\theta) - 1 = \tan^2(\theta)$ |
Example
Evaluate $\int \frac{\sqrt{9 - x^2}}{x^2} dx$.
Let $x = 3\sin(\theta)$, then $dx = 3\cos(\theta) d\theta$. Substituting into the integral, we get:
$\int \frac{\sqrt{9 - (3\sin(\theta))^2}}{(3\sin(\theta))^2} 3\cos(\theta) d\theta = \int \frac{\sqrt{9 - 9\sin^2(\theta))}}{9\sin^2(\theta)} 3\cos(\theta) d\theta$
$= \int \frac{3\cos(\theta)}{9\sin^2(\theta)} 3\cos(\theta) d\theta = \int \frac{\cos^2(\theta)}{\sin^2(\theta)} d\theta = \int \cot^2(\theta) d\theta = \int (\csc^2(\theta) - 1) d\theta$
$= -\cot(\theta) - \theta + C$
Since $x = 3\sin(\theta)$, we have $\sin(\theta) = \frac{x}{3}$, and $\theta = \arcsin(\frac{x}{3})$. Using a right triangle, we can find that $\cot(\theta) = \frac{\sqrt{9 - x^2}}{x}$.
Therefore, the integral evaluates to $ -\frac{\sqrt{9 - x^2}}{x} - \arcsin(\frac{x}{3}) + C$
Keep practicing, and you'll master these techniques! See you in the next class!