Question

Помогите решить интеграл: $\int\limits {sinx*sin\frac{x}{2} sin\frac{x}{3} dx}$
Ответ должен выйти: $\frac{3}{2} cos\frac{x}{6} - \frac{3}{10} cos\frac{5x}{6} -\frac{3}{14}cos\frac{7x}{6} +\frac{3}{32} cos\frac{11x}{6} +C$

Answer 1

Ответ:

Пошаговое объяснение:

$\int {\sin x*\sin\frac{x}{2}\sin\frac{x}{3} } \, dx =\int {\sin x*\frac{1}{2}(\cos(\frac{x}{2}-\frac{x}{3}) -\cos(\frac{x }{2}+\frac{x}{3}) )} \, dx =\\\frac{1}{2} \int {\sin x(\cos\frac{x}{6} -\cos\frac{5x}{6} )} \, dx =\frac{1}{2}\int {(\sin x\cos\frac{x}{6} -\sin x\cos\frac{5x}{6} )} \, dx =\\\frac{1}{2}\int {(\frac{1}{2}(\sin(x+\frac{x}{6})+\sin(x-\frac{x}{6}))-\frac{1}{2} (\sin(x+\frac{5x}{6}) +\sin(x-\frac{5x}{6}))) } \, dx =$

$\frac{1}{4}\int {(\sin\frac{7x}{6} +\sin\frac{5x}{6} -\sin\frac{11x}{6} -\sin\frac{x}{6} ) } \, dx =\frac{1}{4}*\frac{6}{7}\int {\sin\frac{7x}{6}* } \, \frac{7}{6} dx +\\\frac{1}{4}*\frac{6}{5}\int {\sin\frac{5x}{6}* } \, \frac{5}{6} dx-\frac{1}{4}*\frac{6}{11}\int {\sin\frac{11x}{6} *} \, \frac{11}{6} dx-\frac{1}{4}*6\int {\sin\frac{x}{6}* } \, \frac{1}{6} dx=$

$\frac{3}{14}\int {\sin\frac{7x}{6} } \, d(\frac{7x}{6})+\frac{3}{10}\int {\sin\frac{5x}{6} } \, d(\frac{5x}{6})-\frac{3}{22}\int {\sin\frac{11x}{6} } \, d(\frac{11x}{6})-\\\frac{3}{2}\int {\sin\frac{x}{6} } \, d(\frac{x}{6})=-\frac{3}{14}\cos\frac{7x}{6}-\frac{3}{10}\ cos\frac{5x}{6}+\frac{3}{22}\cos\frac{11x}{6}+\frac{3}{2}\cos\frac{x}{6}+C$