Вычислить ! подробное решение

[latex] \int\limits^{\pi/2}_0 {cos2x} \, dx = \frac{1}{2} sin2x\big|_0^{\pi/2}=\frac{1}{2}( sin(2*\pi/2)-sin(2*0))= \\ =\frac{1}{2} (sin\pi-sin0)= \frac{1}{2}sin\pi= \frac{1}{2}*0=0[/latex] [latex] \int\limits^{\pi}_{\pi/2} {sinx} \, dx =-cosx\big|_{\pi/2}^{\pi}=-(cos\pi-cos(\pi/2))=-(-1-0)= \\ =-(-1)=1[/latex]