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 

We have 
$$f(x+\pi)=f(x)+\int_x^{x+\pi} \cos^4 t\,dt.$$ 
Let $G(x)=\int_x^{x+\pi} \cos^4 t\,dt$. 
By the Fundamental Theorem of Calculus, $G'(x)=\cos^4(x+\pi)-\cos^4(x)=0$. Thus $G(x)$ is a constant function.  Set $x=0$.
