Question

А) y'=2x(y-1)^2 б) интеграл п/12 по п/18 cos3x dx как решить?

Answer 1

$1)\; \; y'=2x(y-1)^2\\\\ \frac{dy}{dx}=2x(y-1)^2\\\\\int \frac{dy}{(y-1)^2}=\int 2x\, dx\\\\-\frac{1}{y-1}=x^2+C\\\\2)\; \; \int\limits^{\frac{\pi}{12}}_{\frac{\pi}{18}} {cos3x} \, dx = \frac{1}{3} \; \int\limits^{\frac{\pi}{12}}_{\frac{\pi}{18}} {cos3x} \cdot \underbrace {3dx}_{d(3x)}= \frac{1}{3}\cdot sin3x\Big |_{\frac{\pi}{18}} ^{\frac{\pi}{12}}=\\\\= \frac{1}{3}\cdot (sin\frac{\pi }{4}-sin\frac{\pi }{6} )= \frac{1}{3}\cdot (\frac{\sqrt2}{2}-\frac{1}{2})= \frac{\sqrt2-1}{6}$