Question

интеграл sin2x-sin8xdx=

Answer 1

$\\int (sin2x-sin8x), dx=int sin2x, dx-int sin8x, dx=\\=-frac{1}{2}cos2x+frac{1}{8}cos8x+C\\Formyla:; ; int sin(ax+b), dx=-frac{1}{a}cdot cos(ax+b)+C$