интеграл (2x-1) sin x/3 dx
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Ответил nafanya2014
1
Интегрирование по частям:
![\int{(2x-1)\cdot sin \frac{x}{3} } \, dx = \\ =[ u=2x-1, du=2dx; dv=sin \frac{x}{3}dx, v=-3cos \frac{x}{3}]= \\ =uv- {v} \, du= \\ \\ =-3(2x-1)cos \frac{x}{3}+6 \int {cos \frac{x}{3} } \, dx= \\ =-3(2x-1)cos \frac{x}{3}+ 18 sin \frac{x}{3}+C](https://tex.z-dn.net/?f=+%5Cint%7B%282x-1%29%5Ccdot+sin++%5Cfrac%7Bx%7D%7B3%7D+%7D+%5C%2C+dx+%3D+%5C%5C+%3D%5B+u%3D2x-1%2C+du%3D2dx%3B+++dv%3Dsin+%5Cfrac%7Bx%7D%7B3%7Ddx%2C+v%3D-3cos+%5Cfrac%7Bx%7D%7B3%7D%5D%3D+++%5C%5C+%3Duv-+%7Bv%7D+%5C%2C+du%3D+%5C%5C++%5C%5C+%3D-3%282x-1%29cos+%5Cfrac%7Bx%7D%7B3%7D%2B6+%5Cint+%7Bcos+%5Cfrac%7Bx%7D%7B3%7D+%7D+%5C%2C+dx%3D+%5C%5C+%3D-3%282x-1%29cos+%5Cfrac%7Bx%7D%7B3%7D%2B+18+sin+%5Cfrac%7Bx%7D%7B3%7D%2BC++++ "\int{(2x-1)\cdot sin \frac{x}{3} } \, dx = \\ =[ u=2x-1, du=2dx; dv=sin \frac{x}{3}dx, v=-3cos \frac{x}{3}]= \\ =uv- {v} \, du= \\ \\ =-3(2x-1)cos \frac{x}{3}+6 \int {cos \frac{x}{3} } \, dx= \\ =-3(2x-1)cos \frac{x}{3}+ 18 sin \frac{x}{3}+C")
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