Question

Ребята, СРОЧНО!!!Найти интеграл sin^2x*sin3xdx

Answer 1

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