received_965847035002459.jpeg

# Lecture 14: October 23 ## Integration of Rational Functions ### Partial Fraction Decomposition * We want to integrate $\int \frac{P(x)}{Q(x)} dx$ where $P(x)$ and $Q(x)$ are polynomials. * Assume that the degree of $P(x)$ is less than the degree of $Q(x)$. If not, do long division. $$\...

# Lecture 14: October 23 ## Integration of Rational Functions ### Partial Fraction Decomposition * We want to integrate $\int \frac{P(x)}{Q(x)} dx$ where $P(x)$ and $Q(x)$ are polynomials. * Assume that the degree of $P(x)$ is less than the degree of $Q(x)$. If not, do long division. $$\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$$ where the degree of $R(x)$ is less than the degree of $Q(x)$. * Factor $Q(x)$ completely into linear and irreducible quadratic factors. $$Q(x) = (ax + b)^n \dots (cx^2 + dx + e)^m \dots$$ where $cx^2 + dx + e$ cannot be factored further, i.e., $d^2 - 4ce < 0$. * Then we can write $\frac{P(x)}{Q(x)}$ as a sum of partial fractions: $$\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \dots + \frac{A_n}{(ax + b)^n} + \dots + \frac{B_1 x + C_1}{cx^2 + dx + e} + \frac{B_2 x + C_2}{(cx^2 + dx + e)^2} + \dots + \frac{B_m x + C_m}{(cx^2 + dx + e)^m} + \dots$$ * **Example:** Find $\int \frac{1}{x^2 - a^2} dx$. $$\frac{1}{x^2 - a^2} = \frac{1}{(x - a)(x + a)} = \frac{A}{x - a} + \frac{B}{x + a} = \frac{A(x + a) + B(x - a)}{(x - a)(x + a)}$$ $$A(x + a) + B(x - a) = 1$$ Set $x = a$: $2aA = 1$, $A = \frac{1}{2a}$. Set $x = -a$: $-2aB = 1$, $B = -\frac{1}{2a}$. $$\int \frac{1}{x^2 - a^2} dx = \int \frac{1}{2a} \left( \frac{1}{x - a} - \frac{1}{x + a} \right) dx = \frac{1}{2a} (\ln |x - a| - \ln |x + a|) + C = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C$$ * **Example:** $\int \frac{x^2 + 2x - 1}{2x^3 + 3x^2 - 2x} dx$ $$2x^3 + 3x^2 - 2x = x(2x^2 + 3x - 2) = x(2x - 1)(x + 2)$$ $$\frac{x^2 + 2x - 1}{2x^3 + 3x^2 - 2x} = \frac{A}{x} + \frac{B}{2x - 1} + \frac{C}{x + 2} = \frac{A(2x - 1)(x + 2) + Bx(x + 2) + Cx(2x - 1)}{x(2x - 1)(x + 2)}$$ $$A(2x - 1)(x + 2) + Bx(x + 2) + Cx(2x - 1) = x^2 + 2x - 1$$ Set $x = 0$: $-2A = -1$, $A = \frac{1}{2}$. Set $x = \frac{1}{2}$: $\frac{1}{2} \left( \frac{1}{2} + 2 \right) B = \frac{1}{4} + 1 - 1 = \frac{1}{4}$, $\frac{5}{4} B = \frac{1}{4}$, $B = \frac{1}{5}$. Set $x = -2$: $C(-2)(-5) = 4 - 4 - 1 = -1$, $10C = -1$, $C = -\frac{1}{10}$. $$\int \frac{x^2 + 2x - 1}{2x^3 + 3x^2 - 2x} dx = \int \left( \frac{1/2}{x} + \frac{1/5}{2x - 1} - \frac{1/10}{x + 2} \right) dx = \frac{1}{2} \ln |x| + \frac{1}{10} \ln |2x - 1| - \frac{1}{10} \ln |x + 2| + C$$ ## Trigonometric Substitution ### When to use Integrals containing $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, $\sqrt{x^2 - a^2}$. ### Substitution * For $\sqrt{a^2 - x^2}$, let $x = a \sin \theta$, $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$. Then $dx = a \cos \theta d\theta$ and $\sqrt{a^2 - x^2} = a \cos \theta$. * For $\sqrt{a^2 + x^2}$, let $x = a \tan \theta$, $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. Then $dx = a \sec^2 \theta d\theta$ and $\sqrt{a^2 + x^2} = a \sec \theta$. * For $\sqrt{x^2 - a^2}$, let $x = a \sec \theta$, $0 \le \theta < \frac{\pi}{2}$ or $\pi \le \theta < \frac{3\pi}{2}$. Then $dx = a \sec \theta \tan \theta d\theta$ and $\sqrt{x^2 - a^2} = a \tan \theta$. ### Example $$\int \frac{x^3}{\sqrt{1 - x^2}} dx$$ Let $x = \sin \theta$, $dx = \cos \theta d\theta$, $\sqrt{1 - x^2} = \cos \theta$. $$\int \frac{\sin^3 \theta}{\cos \theta} \cos \theta d\theta = \int \sin^3 \theta d\theta = \int \sin \theta (1 - \cos^2 \theta) d\theta = \int (\sin \theta - \sin \theta \cos^2 \theta) d\theta$$ $$= -\cos \theta + \frac{\cos^3 \theta}{3} + C = -\sqrt{1 - x^2} + \frac{(1 - x^2)^{3/2}}{3} + C$$