Question

Evaluate the definite integral ${\int }_0^{\frac{\pi }{2}}\cos 2xdx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\cos 2xdx$
$\Rightarrow \int \cos 2xdx=\left(\frac{\sin 2x}{2}\right)=F\left(x\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(\frac{\pi }{2}\right)-F\left(0\right)$
$=\frac{1}{2}[\sin 2\left(\frac{\pi }{2}\right)-\sin 0]$
$=\frac{1}{2}[\sin \pi -\sin 0]=\frac{1}{2}[0-0]=0$