\( \int_{0}^{\frac{\pi}{2}}(2 \pi+2 \cos 2 x) d x \)
we split this into sum of integrals i.e
\( \int_{0}^{\frac{\pi}{2}} 2 \pi+\int_{0}^{\frac{\pi}{2}} 2 \cos 2 x d x|2 \pi x|_{10}^{+}+\left.2 \cdot \frac{\sin 2 x}{2}\right|_{0} ^{\frac{\pi}{2}} \)
We subtract the lower limit from the upper limit \(2 \pi(\pi / 2)-2 \pi(0)+\sin 2(\pi / 2)-\sin 2(0)\)
\(\pi^{2}-0+\sin \pi-0=\pi^{2}+\sin \pi \)
but \(\sin \pi=0\).
\(\pi^{2}+\sin \pi=\pi^{2}\)