Let $\text{⨍}\left(\mathrm{x}\right)\text{=}\left|\begin{array}{ccc}2{\cos }^2\mathrm{x}& \sin \text{ 2x}& -\sin \mathrm{x}\\ \sin \text{ 2x}& 2 {\sin }^2\mathrm{x}& \cos \text{ x}\\ \sin \text{ x}& -\cos \mathrm{x}& 0\end{array}\right|$ Then the value of ${\int }_0^{\mathrm{\pi }/2}\left[\text{⨍}\left(\mathrm{x}\right)\text{+⨍'}\left(\mathrm{x}\right)\right]\mathrm{dx}$

Applying ${\mathrm{C}}_1\to {\mathrm{C}}_1-2 {\mathrm{sinxC}}_3 \mathrm{and} {\mathrm{C}}_2\to {\mathrm{C}}_2+2{\mathrm{cosxC}}_3, \mathrm{we} \mathrm{get}$

$\text{⨍}\left(\mathrm{x}\right)\text{=}\left|\begin{array}{ccc}2& 0& -\sin \mathrm{x}\\ 0& 2& \cos \text{ x}\\ \sin \text{x}& -\cos \mathrm{x}& 0\end{array}\right|{\text{=2 cos}}^2{\text{x+2 sin}}^2\text{x=2}$

$\therefore \text{⨍'}\left(\mathrm{x}\right)\text{=0}$

$\therefore {\int }_0^{\mathrm{\pi }/2}\left[\text{⨍}\left(\mathrm{x}\right)\text{+⨍'}\left(\mathrm{x}\right)\right]\mathrm{dx}={\int }_0^{\mathrm{\pi }/2}2\mathrm{dx}={\left(2x\right)}_0^{\mathrm{\pi }/2}=2\left(\frac{\mathrm{\pi }}{2}\right)=\mathrm{\pi }$