Avaliação 2 - Cálculo 1 2021 2

De 3 A = \frac{2}{5} De 1 B = -\frac{2}{5} De 2 D = \frac{4}{5} + C - 1 C = \frac{1}{5} Então \int \frac{x + 2}{x(x^2 + 2x + 5)} \, dx = \frac{2}{5} \int \frac{dx}{x} + \frac{1}{5} \int \frac{-1x - 2}{x^2 + 2x + 5} \, dx Resolvendo, \frac{1}{5} \int \frac{-1x - 2}{x^2 + 2x + 5} \, dx = \frac{1}{5} \left( \frac{3}{x^2 + 2x + 5} - \frac{2x + 2}{x^2 + 2x + 5} \right) \, dx = -\frac{1}{5} \int \frac{2x + 2}{x^2 + 2x + 5} \, dx + \frac{3}{5} \int \frac{dx}{x^2 + 2x + 5} Substituições: u = x^2 + 2x + 5 \rightarrow du = (2x + 2)dx e \, \text{note que} \, x^2 + 2x + 5 = (x + 1)^2 + 4 v \rightarrow v = x + 1 \rightarrow dv = dx = -\frac{1}{5} \int \frac{du}{u} + \frac{3}{5} \int \frac{dv}{v^2 - 4} = -\frac{1}{5} \int \frac{du}{u} + \frac{3}{20} \int \frac{dv}{\left( \frac{v^2}{4} + 1 \right)} Cálculo 1 1) Integrais a) \int \sin \left( \frac{x}{2} \right) \, dx Substituição: u = \frac{x}{2} \rightarrow du = \frac{dx}{2} \rightarrow 2du = dx \int \sin \left( \frac{x}{2} \right) \, dx = 2 \int \sin u \, du = -2 \cos u + C \int \sin \left( \frac{x}{2} \right) \bigg| dx = -2 \cos \left( \frac{x}{2} \right) + C b) \int \left( \frac{1}{3} \cos 3x - \frac{1}{7} \sin 7x \right) \, dx Substituições: u = 3x \rightarrow du = 3dx \rightarrow \frac{du}{3} = dx v = 7x \rightarrow dv = 7dx \rightarrow \frac{dv}{7} = dx \int \left( \frac{1}{3} \cos 3x - \frac{1}{7} \sin 7x \right) \, dx = \frac{1}{3} \int \cos 3x \, dx - \frac{1}{7} \int \sin 7x \, dx = \frac{1}{9} \int \cos u \, du - \frac{1}{49} \int \sin v \, dv = \frac{1}{9} \sin u + \frac{1}{49} \cos v + C \int \left( \frac{1}{3} \cos 3x - \frac{1}{7} \sin 7x \right) \, dx = \frac{1}{9} \sin 3x + \frac{1}{49} \cos 7x + C c) \int_0^1 xe^{-x^2} \, dx Substituição: u = -x^2 \rightarrow du = -2xdx -\frac{du}{2} = xdx x = 0 \rightarrow u = 0 x = 1 \rightarrow u = -1 \int_0^1 xe^{-x^2} \, dx = -\frac{1}{2} \int_0^{-1} e^u \, du = \frac{1}{2} \int_{-1}^0 e^u \, du = \frac{1}{2} e^u \bigg|_{-1}^{0} = \frac{1}{2} (1 - e^{-1}) \int_0^1 xe^{-x^2} \, dx = \frac{e - 1}{2e} d) \int \frac{x + 2}{x(x^2 + 2x + 5)} \, dx = \int \frac{x + 2}{x(x^2 + 2x + 5)} \, dx Funções parciais \frac{x + 2}{x(x^2 + 2x + 5)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2x + 5} x + 2 = A(x^2 + 2x + 5) + Bx + Cx x - 2 = x^2 (A + B) + x(2A + C) + 5A comparando os termos: 1) 0 = A + B 2) 1 = 2A + C 3) 2 = 5A Substituições. t = \frac{v}{2} \rightarrow dt = \frac{dv}{2} = -\frac{1}{5} \int \frac{du}{u} + \frac{3}{10} \int \frac{dt}{t^2 + 1} = -\frac{1}{5} \ln |u| + \frac{3}{10} \operatorname{arctg}\, t + C = -\frac{1}{5} \ln (x^2 + 2x + 5) + \frac{3}{10} \operatorname{arctg}\left(x + \frac{1}{2}\right) + C então, \int \frac{x+2}{x^2 + 2x + 5} = \frac{2}{5} \ln x - \frac{1}{5} (x^2 + 2x + 5) + \frac{3}{10} \operatorname{arctg}\left(x + \frac{1}{2}\right) + C b) \int \frac{\cos^4 x}{\sin^4 x} \, dx = \int \operatorname{cotg}^4 x \, dx usando a fórmula de redução: \int \operatorname{cotg}^n x \, dx = -\frac{1}{3} \operatorname{cotg}^3 x - \int \operatorname{cotg}^2 x \, dx = -\frac{1}{3} \operatorname{cotg}^3 x - \int (\operatorname{cosec}^2 x - 1) dx \int \frac{\cos^4 x}{\sin^4 x} \, dx = -\frac{1}{3} \operatorname{cotg}^3 x - \operatorname{cotg} x - x + C c) \int (x + e^x) \, dx = \int x dx + \int e^x \, dx = \frac{x^2}{2} + e^x + C d) \int \cos(3x) \cos(7x) \, dx = \frac{1}{2} \int (\cos(-4x) + \cos(10x)) dx = \frac{1}{2} \int \cos 4x \, dx + \frac{1}{2} \int \cos 10x \, dx Substituições: u = 4x \rightarrow \frac{du}{4} = dx \quad \text{e} \quad v = 10x \rightarrow \frac{dv}{10} = dx = \frac{1}{5} \int \cos u \, du + \frac{1}{20} \int \cos v \, dv = \frac{1}{5} \sin u + \frac{1}{20} \sin v + C \int \cos 3x \cos 7x \, dx = \frac{1}{5} \sin 4x + \frac{1}{20} \sin 10x + C h) \int_0^1 e^{-x} \cos 2x \, dx Por partes: u = \cos 2x \quad du = -2 \sin 2x \, dx dv = e^{-x} dx \quad v = -e^{-x} \int_0^1 e^{-x} \cos 2x \, dx = -e^{-x} \cos 2x \Big|_0^1 - 2 \int_0^1 e^{-x} \sin 2x \, dx = \frac{1 - \cos 2}{e} - 2 \int_0^1 e^{-x} \sin 2x \, dx Por partes: u = \sin 2x \quad du = 2 \cos 2x \, dx dv = e^{-x} dx \quad v = -e^{-x} \int_0^1 e^{-x} \cos 2x \, dx = \frac{1 - \cos 2}{e} + 2 e^{-x} \sin 2x \Big|_0^1 - 4 \int_0^1 e^{-x} \cos 2x \, dx 5 \int_0^1 e^{-x} \cos 2x \, dx = \frac{1 - \cos 2}{e} + \frac{2 \sin 2}{e} - 2 \int_0^1 e^{-x} \cos 2x \, dx = \frac{2 \sin 2 - \cos 2 - 1}{5e} 1) \int x^2 \sqrt{x-1}dx \quad Substituicao: u = \sqrt{x-1} \quad du = \frac{dx}{2\sqrt{x-1}} nota que: \quad u = \sqrt{x-1} \rightarrow u^2 = x-1 \rightarrow x = u^2 + 1 \rightarrow x = u^2 + 1 \rightarrow x^2 = (u^2 + 1)^2 entao \int x^2 \sqrt{x-1}dx = \int(u^2 + 1)^2 u^2 du = 2 \int(u^6 + 2u^4 + u^2)du = \frac{2}{7}u^7 + \frac{4}{5}u^5 + \frac{2}{3}u^3 + C \int x^2 \sqrt{x-1}dx = \frac{2}{7} (x-1)^{7/2} + \frac{4}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2} + C 2) \int (x^2 + 5x + 4)dx = \frac{x^3}{3} + \frac{5}{2}x^2 + 2x + C 2) Dem. \int \cos^n x dx = \int \cos x \cos^{n-1} x dx Por partes: u = \cos^{n-1} x \qquad dv = \cos x dx du = -(n-1)\cos^{n-2} x \sin x dx \qquad v = \sin x \int \cos^n x dx = \cos^{n-1} x \sin x + (n-1)\int \cos^{n-2} x \sin^2 x dx como \sin^2 x = 1 - \cos^2 x \int \cos^n x dx = \cos^{n-1} x \sin x + (n-1)\int \cos^{n-2} x (1-\cos^2 x) dx = \cos^{n-1} x \sin x + (n-1)\int \cos^{n-2} x dx - (n-1)\int \cos^n x dx n \int \cos^n x dx = \cos^{n-1} x \sin x + (n-1)\int \cos^{n-2} x dx \int \cos^n x dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x dx com \quad n \geq 2 3) \int t^n e^{-5t} dt Por partes: u = t^n \qquad dv = e^{-5t} dt du = nt^{n-1} dx \qquad v = -\frac{1}{5} e^{-5t} \int t^n e^{-5t} dt = -\frac{1}{5} t^n e^{-5t} + \frac{n}{5} \int t^{n-1} e^{-5t} dt 4) Temos que f''(x)-f(x)=0, ou seja: f''(x)=f(x) para que g(x)=C, onde C=cte, g'(x)=0 \rightarrow e^x[f'(x)-f(x)]=C \rightarrow g'(x) = e^x f'(x) + c x f''(x) - e^x f(x) - cx f'(x) g'(x) = e^x f''(x) - e^x f(x) = e^x [f(x)-f(x)] = f(x) g'(x)=0