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

$\begin{array}{l}g(x+\pi )={\int }_0^{\pi +x} {\cos }^{4}tdt\\ ={\int }_0^{\pi } {\cos }^{4}tdt+{\int }_{\pi }^{\pi +x} {\cos }^{4}tdt=g\left(\pi \right)+I_2.\end{array}$

Put $t=\pi +\theta$ and adjust the limits in $I_2$

$\begin{array}{l}\therefore I_2={\int }_0^x {\cos }^4\left(\pi +\theta \right)d\theta ={\int }_0^x {\cos }^{4}tdt=g\left(x\right)\\ \therefore g(\pi +x)=g\left(\pi \right)+g\left(x\right)\end{array}$