Prove that $f(x)=\int_{0}^x \cos^4 t\, dt\implies f(x+\pi)=f(x)+f(\pi)$ How to prove that if $f(x)=\int_{0}^x \cos^4 t\, dt$ then $f(x+\pi)$ equals to $f(x)+f(\pi)$
I thought to first integrate $\cos^4 t$ but this might be the problem of Definite Integral 

This is a general statement about periodic functions: If $f(x)$ is periodic with period $\pi$ then $\int_{x}^{x+\pi}f(s)ds$ is independent of x.
