$\int \dfrac{1-\cos 2x}{2} dx =\dfrac{1}{2}x -\dfrac{1}{4}\sin 2x +C$ tự thay cận
$\int
_0^{\frac{\pi}{2}} \dfrac{1-\cos 2x}{2} dx =
\bigg (\dfrac{1}{2}x -\dfrac{1}{4}\sin 2x
\bigg ) \bigg |_0^{\frac{\pi}{2}}$$
=\dfrac{\pi}{4}-\dfrac{1}{2}\sin \pi =\dfrac
{\pi}{4}$