Question

Найдите интеграл, не используя метод подстановки

Answer 1

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

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