[Equações Diferenciais Ordinárias] Aula 3: revisão de integrais

Integrais simples
a)
$$\int x^8 dx = (\dfrac{1}{9} x^9) + C = \dfrac{x^9}{9} + C$$
b)
$$\int \dfrac{1}{x^3} dx = \int x^{-3} dx = (\dfrac{1}{-2} x^{-2}) = \dfrac{-1}{2x^2} + C$$
c)
$$\int \sqrt[3]{x^2} dx = \int x^{2/3} dx = (\dfrac{3}{5} x^{5/3}) = \dfrac{3 \sqrt[3]{x^5}}{5} + C$$
d)
$$\int (5x^3 + 2 \cos{x}) dx = 5 \int x^3 dx + 2 \int \cos{x} dx = 5 (\dfrac{1}{4}) x^4 + 2 \sin{x} = \\ \dfrac{5}{4} x^4 + 2 \sin{x} + C$$
e)
$$\int (8t^2 + 6e^t + \dfrac{1}{t})dt = 8 \int t^2 dt + 6 \int e^t dt + \int \dfrac{1}{t} dt = 8 (\dfrac{1}{3} t^3) + 6e^t + \ln{t} = \\ \dfrac{8}{3} t^3 + 6e^t + \ln{t} + C$$
f)
$$\int \dfrac{(x-1)^2}{x^2} dx = \int \dfrac{(x^2 - 2x + 1)}{x^2} dx = \int \dfrac{x^2}{x^2} dx - 2 \int \dfrac{x}{x^2} dx + \int \dfrac{1}{x^2} dx = \\ \int 1 dx - 2 \int \dfrac{1}{x} dx + \int x^{-2} dx = x - 2 \ln{x} + (-x^{-1}) = x - 2 \ln{x} - \dfrac{1}{x} + C$$
g)
$$\int \dfrac{\sin{x}}{\tan{x}} dx = \int \dfrac{\sin{x}}{\dfrac{\sin{x}}{\cos{x}}} dx = \int \dfrac{\sin{x} \cos{x}}{\sin{x}} = \int \cos{x} dx = \sin{x} + C$$

Integrais por substituição
a)
$$\int (x^2 + 3x)^3 (2x + 3)dx \\ u = x^2 + 3x, du = (2x + 3)dx \\ \int u^3 du = \dfrac{1}{4} u^4 = \dfrac{1}{4} (x^2 + 3x)^4 + C$$
b)
$$\int \dfrac{5x}{(x^2 + 1)} dx \\ u = x^2 + 1, du = (2x)dx \\ \int \dfrac{5x}{u} \dfrac{du}{2x} = \dfrac{5}{2} \int \dfrac{1}{u} du = \dfrac{5}{2} \ln{|u|} = \dfrac{5}{2} \ln{|(x^2 + 1)|} + C$$
c)
$$\int \dfrac{x^2}{\sqrt[3]{1 - 2x^3}} dx = \int x^2 ((1 - 2x^3)^{-1/3} dx \\ u = 1 - 2x^3, du = (-6x^2)dx \\ \int x^2 u^{-1/3} \dfrac{du}{-6x^2} = \dfrac{-1}{6} \int u^{-1/3} du = \\ \dfrac{-1}{6} (\dfrac{3}{2} u^{2/3}) = - \dfrac{3}{12} (1 - 2x^3)^{2/3} = - \dfrac{1}{4} \sqrt[3]{(1 - 2x^3)^2} + C$$
d)
$$\int 2e^{-x} dx \\ u = -x, du = (-1)dx \\ 2 \int e^u \dfrac{du}{-1} = -2e^u = -2e^{-x} + C$$
e)
$$\int \sin{3x} dx \\ u = 3x, du = 3dx \\ \sin{u} \dfrac{du}{3} = - \dfrac{1}{3} \cos{3x} + C$$
f)
$$\int 2x \cos{x^2} dx \\ u = x^2, du = (2x)dx \\ \int \cos{u} du = \sin{u} = \sin{x^2} + C$$
g)
$$\int \sin{x} \cos{x} dx \\ u = \sin{x}, du = \cos{x} dx \\ \int u du = \dfrac{1}{2} u^2 = \dfrac{1}{2} \sin^2 {x} + C$$

Integrais por partes
a)
$$\int x \ln{x} dx \\ u = \ln{x}, du = \dfrac{1}{x}, dv = (x)dx, v = \dfrac{x^2}{2} \\  \int x \ln{x} dx  = \dfrac{x^2}{2} \ln{x} - \int \dfrac{1}{2} x^2 \dfrac{1}{x} dx = \dfrac{x^2}{2} \ln{x} - \dfrac{1}{2} \int x dx = \dfrac{x^2}{2} \ln{x} - \dfrac{1}{2} (\dfrac{1}{2} x^2) = \\ \dfrac{x^2}{2} \ln{x} - \dfrac{x^2}{4} = \dfrac{x^2}{2} (\ln{x} - \dfrac{1}{2})$$
b)
$$\int x \sin{x} dx \\ u = x, du = 1, dv = \sin{x} dx, v = - \cos{x} \\  \int x \sin{x} dx = x (- \cos{x}) - \int (- \cos{x})*1 dx = \\ -x \cos{x} + \int \cos{x} dx = -x \cos{x} + \sin{x} + C$$
c)
$$\int xe^{2x} dx \\ u = x, du = 1, dv = e^{2x} dx, v = \dfrac{e^{2x}}{2} \\ \int xe^{2x} dx = x (\dfrac{e^{2x}}{2}) - \int \dfrac{e^{2x}}{2} dx = \dfrac{xe^{2x}}{2} - \dfrac{1}{2} \dfrac{e^{2x}}{2} = \dfrac{xe^{2x}}{2} - \dfrac{e^{2x}}{4} = \dfrac{e^{2x}}{2} (x - \dfrac{1}{4}) + C$$

d) Integral cíclica (vou sempre pegar os números de Euler como u e as trigonométricas como v)
$$\int e^{x} \cos{x} dx = e^{x} \sin{x} - \int e^{x} \sin{x} dx = e^{x} \sin{x} - [e^{x} (- \cos{x}) - \int (- \cos{x}) e^{x} dx] = \\ e^{x} \sin{x} - [-e^x \cos{x} + \int e^{x} \cos{x} dx] = e^{x} \sin{x} + e^x \cos{x} - \int e^{x} \cos{x} dx \\ 2 \int e^{x} \cos{x} dx =  e^{x} \sin{x} + e^x \cos{x} \\ \int e^{x} \cos{x} dx = \dfrac{1}{2} (e^{x} \sin{x} + e^{x} \cos{x}) = \dfrac{e^x}{2} (\sin{x} + \cos{x}) + C$$
e)
$$\int x^2 e^{-x} dx \\ u = x^2, du = 2x, dv = e^{-x} dx, v = -e^{-x} \\  \int x^2 e^{-x} dx = x^2 (-e^{-x}) - \int (-e^{-x}) 2x dx = -x^2 e^{-x} + \int e^{-x} 2x dx \\ u = 2x, du = 2, dv = e^{-x} dx, v = -e^{-x} \\ -x^2 e^{-x} + [2x (-e^{-x}) - \int (-e^{-x}) 2 dx] = -x^2 e^{-x} + [-2xe^{-x} + 2 \int e^{-x} dx] = \\ -x^2 e^{-x} + [-2xe^{-x} - 2e^{-x}] = -x^2 e^{-x} - 2xe^{-x} - 2e^{-x} = \\ -e^{-x} (x^2 + 2x + 2)$$


É isso aí, galera. Acho que agora não tem mais como enrolar e vamos voltar a ver EDOs, mas só vendo pra saber. Boa noite a todos.