hasil dari integral (batas atas phi, batas bawah 0) cos 2x-phi dx

[tex] \int\limits^\pi_0 {cos(2x - \pi)} \, dx \\ = \frac{1}{2} sin(2x - \pi) ]^\pi_0\\ = \frac{1}{2} sin(2(\pi) - \pi) - \frac{1}{2} sin(2(0) - \pi)\\ = \frac{1}{2}(sin(\pi)-sin(-\pi))\\ = \frac{1}{2}(sin(\pi)+sin(\pi))\\ = \frac{1}{2}(2 \ sin(\pi))\\ = sin(\pi) = 0 [/tex]