Given:
[tex]\int ^{\frac{\pi}{2}}_0|4\sin x-4\cos (2x)|dx[/tex]
First, eliminate absolute
This gives
[tex]\begin{gathered} \int ^{\frac{\pi}{2}}_0|4\sin x-4\cos (2x)|dx \\ =\int ^{\frac{\pi}{6}}_0-4\sin x+4\cos (2x)dx+\int ^{\frac{\pi}{2}}_{\frac{\pi}{6}}4\sin x-4\cos (2x)dx \end{gathered}[/tex]
Integrating each part gives
[tex]\begin{gathered} \int ^{\frac{\pi}{6}}_0-4\sin x+4\cos (2x)dx+\int ^{\frac{\pi}{2}}_{\frac{\pi}{6}}4\sin x-4\cos (2x)dx \\ =4\cos x+2\sin (2x)|^{\frac{\pi}{6}^{}_{}}_0-4\cos x-2\sin (2x)|^{\frac{\pi}{2}}_{\frac{\pi}{6}} \end{gathered}[/tex]
Simplifying further gives
[tex]\begin{gathered} 4\cos x+2\sin (2x)|^{\frac{\pi}{6}^{}_{}}_0 \\ =\mleft\lbrace(4\cos (\frac{\pi}{6})+2\sin (\frac{2\pi}{6})-(4\cos 0+2\sin 0)\mright\rbrace \\ =(2\sqrt[]{3}+\sqrt[]{3}-(4+0) \\ =3\sqrt[]{3}-4 \end{gathered}[/tex]
Also
[tex]\begin{gathered} -4\cos x-2\sin (2x)|^{\frac{\pi}{2}}_{\frac{\pi}{6}} \\ =-4\cos (\frac{\pi}{2})-2\sin (\frac{2\pi}{2})-(-4\cos (\frac{\pi}{6})-2\sin (\frac{2\pi}{6}) \\ =0+0-(-2\sqrt[]{3}-\sqrt[]{3)} \\ =2\sqrt[]{3}+\sqrt[]{3} \\ =3\sqrt[]{3} \end{gathered}[/tex]
Hence the solution is
[tex]\begin{gathered} \int ^{\frac{\pi}{6}}_0-4\sin x+4\cos (2x)dx+\int ^{\frac{\pi}{2}}_{\frac{\pi}{6}}4\sin x-4\cos (2x)dx \\ =3\sqrt[]{3}-4+3\sqrt[]{3} \\ =6\sqrt[]{3}-4 \end{gathered}[/tex]
Therefore, the answer is
[tex]6\sqrt[]{3}-4[/tex]
