du=dx,v=\frac{sin(2x)}{2}\mid=\\\\ =x*\frac{sin(2x)}{2}-\int\frac{sin(2x)}{2}dx=\mid t=2x \mid = x*\frac{sin(2x)}{2}-\frac{1}{2}*\int\frac{sint(t)}{2}dt=\\\\ =x*\frac{sin(2x)}{2}-\frac{1}{4}*(-cos(t))= (x*\frac{sin(2x)}{2}+\frac{cos(2x)}{4})\mid ^{\pi/2}_0=\\\\=\frac{\frac{\pi}{2}*sin(2*\frac{\pi}{2}) }{2}+\frac{cos(2*\frac{\pi}{2})}{4}-(\frac{0*sin(2*0)}{2}+\frac{cos(2*0)}{4})=\frac{\frac{\pi}{2}*0}{2}+\frac{cos(\pi)}{4}-(0+\frac{1}{4})=" alt="\int^{\frac{\pi}{2} }_0x*cos(2x)dx=\mid u=x,dv=cos(2x)=>du=dx,v=\frac{sin(2x)}{2}\mid=\\\\ =x*\frac{sin(2x)}{2}-\int\frac{sin(2x)}{2}dx=\mid t=2x \mid = x*\frac{sin(2x)}{2}-\frac{1}{2}*\int\frac{sint(t)}{2}dt=\\\\ =x*\frac{sin(2x)}{2}-\frac{1}{4}*(-cos(t))= (x*\frac{sin(2x)}{2}+\frac{cos(2x)}{4})\mid ^{\pi/2}_0=\\\\=\frac{\frac{\pi}{2}*sin(2*\frac{\pi}{2}) }{2}+\frac{cos(2*\frac{\pi}{2})}{4}-(\frac{0*sin(2*0)}{2}+\frac{cos(2*0)}{4})=\frac{\frac{\pi}{2}*0}{2}+\frac{cos(\pi)}{4}-(0+\frac{1}{4})=" align="absmiddle" class="latex-formula">