If $g\left(x\right)={\int }_0^x \cos 4tdt$,, then $g(x+\pi )$ equals

$\begin{array}{l}g(x+\pi )={\int }_0^{x+\pi } \cos 4tdt\\ ={\int }_0^{\pi } \cos 4tdt+{\int }_{\pi }^{x+\pi } \cos 4tdt\\ =g\left(\pi \right)+I\end{array}$

where $I={\int }_{\pi }^{x+\pi } \cos 4tdt,$ Put $t=\pi +\theta$ so that

$I={\int }_0^x \cos (4\pi +4\theta )d\theta ={\int }_0^x \cos 4\theta d\theta =g\left(x\right)$

So $g(x+\pi )=g\left(\pi \right)+g\left(x\right)$ but

$\begin{array}{l}g\left(\pi \right)={\int }_0^{\pi } \cos 4tdt=\frac{1}{4}\left({\left.\sin 4t\right|}_0=0.\right.\end{array}$

$\therefore g(x+\pi )=g\left(x\right)-g\left(\pi \right)$ also.