Question

Найдите интеграл, не используя метод подстановки

Answer 1

$\int sinx*sin5x*cos\frac{2}{3}x* dx = | sina*sinb = \frac{1}{2}(cos(a-b) - cos(a+b))| = \int \frac{1}{2}(cos4x - cos6x)*cos\frac{2}{3}x *dx = \frac{1}{2}\int (cos4x*cos\frac{2}{3}x - cos6x*cos\frac{2}{3}x) dx = | cosa*cosb = \frac{1}{2}(cos(a+b) + cos(a-b))| = \frac{1}{2}\int (\frac{1}{2}(cos\frac{14}{3}x + cos\frac{10}{3}x) - \frac{1}{2}(cos\frac{20}{3}x + cos\frac{16}{3}x))dx = -\frac{1}{4}\int(cos\frac{20}{3}x+ cos\frac{16}{3}x - cos\frac{14}{3}x -cos\frac{10}{3}x)dx =$

$= | \int cos(ax)dx = \frac{1}{a}sin(ax) + c| = -\frac{1}{4}(\frac{3}{20}sin\frac{20}{3}x + \frac{3}{16}sin\frac{16}{3}x - \frac{3}{14}sin\frac{14}{3}x - \frac{3}{10}sin\frac{10}{3}x)+c = -\frac{3}{8}(\frac{1}{10}sin\frac{20}{3}x + \frac{1}{8}sin\frac{16}{3}x - \frac{1}{7}sin\frac{14}{3}x - \frac{1}{5}sin\frac{10}{3}x)+c\\ Answer: -\frac{3}{8}(\frac{1}{10}sin\frac{20}{3}x + \frac{1}{8}sin\frac{16}{3}x - \frac{1}{7}sin\frac{14}{3}x - \frac{1}{5}sin\frac{10}{3}x)+c$