[tex]\frac{1}{\pi} \int_{0}^{2\pi } (2x+3) cos kx dx= \frac{1}{k\pi} \int_{0}^{2\pi } (2x+3) cos kx d kx= \frac{1}{k\pi} \int_{0}^{2\pi } (2x+3) d sin kx= \frac{1}{k\pi} [(2x+3)sin kx|_{0}^{2\pi}-\int_{0}^{ 2\pi} sin kxd(2x+3)]=[/tex]
[tex]= \frac{1}{k\pi} [(2x+3)sin kx|_{0}^{2\pi}-2\int_{0}^{ 2\pi} sin kxdx]= \frac{1}{k\pi} [(2x+3)sin kx|_{0}^{2\pi}-\frac{2}{k} \int_{0}^{ 2\pi} sin kxdkx]=[/tex]
[tex]=\frac{1}{k\pi} [(2x+3)sin kx|_{0}^{2\pi}+\frac{2}{k} cos kx|_{0}^{2\pi}]=...[/tex]
Остава само да се приложи формулата на Нютон-Лайбниц