Question

Ребята, СРОЧНО!!!

Найти интеграл sin^2x*sin3xdx

Answer 1

$sin ^{2}x= frac{1-cos2x}{2} \ int {sin ^{2}xcdot sin3x } , dx = int {frac{1-cos2x}{2} cdot sin3x } , dx= \ ={frac{1}{2}int sin3x } , dx- {frac{1}{2}int cos2xcdot sin3x } , dx= \ =frac{1}{2cdot3}(-cos3x)-{frac{1}{2}int {(frac{1}{2}sin(3x+2x)+frac{1}{2}sin(3x-2x))dx= \ =$
$=frac{1}{2cdot3}(-cos3x)-{frac{1}{4}int {sin5xdx-frac{1}{4}int sinxdx =$
$=frac{1}{2cdot3}(-cos3x)- frac{1}{4cdot5}(-cos5x)frac{1}{4}(-cosx)+C=$
$=-frac{cos3x}{6}+ frac{cos5x}{20}+frac{cosx}{4}+C$