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

$g(x+\pi )={\int }_0^{x+\pi }{\cos }^{4}tdt$ Split the integral:$g(x+\pi )={\int }_0^{\pi }{\cos }^{4}tdt+{\int }_{\pi }^{x+\pi }{\cos }^{4}tdt$ The first term is $g(\pi )$.In the second term, let $u=t-\pi$, then $du=dt$. When $t=\pi ,u=0$. When $t=x+\pi ,u=x$.${\int }_0^x{\cos }^4(u+\pi )du={\int }_0^x(-\cos u{)}^{4}du={\int }_0^x{\cos }^{4}udu=g(x)$.Thus, $g(x+\pi )=g(\pi )+g(x)$.