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 

$$f(x+\pi)-f(\pi)=\int_0^{x+\pi}\cos^4t dt-\int_0^{\pi}\cos^4t dt=\int_{\pi}^{x+
\pi}\cos^4t dt$$
Putting $u=t-\pi, dt=du$ and $\cos t=\cos(u+\pi)=-\cos u$
$$\int_{\pi}^{x+\pi}\cos^4t dt=\int_0^x(-\cos u)^4du=f(x)$$
