Question

помогите пожалуйста решить​

Answer 1

1.

1)

$\int \frac{\cos(x)}{2\sin(x) + 1} dx\\d\sin(x) = \cos(x)dx\\\int \frac{d\sin(x)}{2\sin(x) + 1} = \frac12 \int \frac{d(2\sin(x) + 1)}{2\sin(x) + 1} = \frac12 \ln(2\sin(x) + 1) + C$

2)

$\int\limits^{\pi}_{0} x^3 \sin(x) dx$

$\int x^3 \sin(x) dx = \\\fbox{ u = x^3, dv = \sin(x)dx \Rightarrow du = 3x^2dx, v = -\cos(x) }\\= -x^3\cos(x) + \int 3x^2 \cos(x) dx = \\\fbox{ u = x^2, dv = \cos(x)dx \Rightarrow du = 2xdx, v = \sin(x) }\\-x^3\cos(x) + 3(x^2\sin(x) - \int 2x\sin(x)dx) = \\\fbox{ u = x, dv = \sin(x)dx \Rightarrow du = dx, v = -\cos(x) }\\-x^3\cos(x) + 3(x^2\sin(x) - 2(-x\cos(x) + \int \cos(x)dx)) =\\-x^3\cos(x) + 3(x^2\sin(x) - 2(-x\cos(x) + \sin(x))) + C = \\-x^3\cos(x) + 3x^2\sin(x) + 6x\cos(x) - 6\sin(x) + C$

$\int\limits_{0}^{\pi} x^3 \sin(x) dx = (-x^3\cos(x) + 3x^2\sin(x) + 6x\cos(x) - 6\sin(x))\Big|^{\pi}_{0} =\\= -\pi^3 \cdot (-1) + 3\pi^3 \cdot 0 + 6\pi \cdot (-1) - 6\cdot 0 - 0 + 0 + 0 - 0 = \pi^3 - 6\pi$

2. Не понимаю что в скобках написано

1)

$2xy' = 1 - x^2\\dy = \frac12(\frac{1}{x} - x)dx\\\int dy = \frac12\int (\frac{1}{x} - x)dx\\y = \frac12 (\ln(x) - \frac12 x^2) + c_0$

2)

$y'' + 2y' = 0\\y' = u(y) \Rightarrow y'' = u'y' = u'u\\u'u + 2u = 0\\u' = -2\\\int du = -2\int dy\\u = -2y\\y' = -2y\\\int \frac{dy}{y} = -2\int dx\\\ln(y) = -2x + c_0\\y = e^{-2x + c_0} + c_1$

3.

$y' - y \mathrm{tg}(x) = \frac{1}{\cos(x)}, y(0) = 0\\\cos(x) dy - y \sin(x)dx = dx\\\cos(x) dy + y d\cos(x) = dx\\\int d(y\cos(x)) = \int dx\\y \cos(x) = x + c_0\\y = \frac{x + c_0}{\cos(x)}\\y(0) = 0 = \frac{0 + c_0}{1} \Rightarrow c_0 = 0$

$y = \frac{x}{\cos(x)}$

4. На рисунке должна быть перевёрнутая парабола, от которой прямая $y = 0$ отсекает кусочек, который мы ищем

$y = 4x - x^2 = 0 \Rightarrow x_0 = 0, x_1 = 4$

$S = \int\limits_{0}^{4} (4x - x^2) dx = (2x^2 - \frac{x^3}{3})\Big|_{0}^{4} = 32 - 64/3 = \frac{32}{3}$