Prove that $f(x)=\int_{0}^x \cos^4 t\, dt\implies f(x+\pi)=f(x)+f(\pi)$

Question

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

Answer 1

$$f(x+\pi)-f(\pi)=\int_0^{x+\pi}\cos^4t dt-\int_0^{\pi}\cos^4t dt=\int_{\pi}^{x+ \pi}\cos^4t dt$$

Putting $u=t-\pi, dt=du$ and $\cos t=\cos(u+\pi)=-\cos u$

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

Answer 2

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 $x=0$.

Answer 3

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.

Answer 4

Hint:compute integral $$f(x)=\int_{0}^x \cos^4 t\, dt=\int_{0}^x\left(\dfrac{\cos2x+1}{2}\right)^2=\frac14 \int_0^{x}\left(\cos^22x+2\cos2x+1 \right)dx=\\=\dfrac14 \int_0^{x}\left(\frac{1+\cos4x}{2}+2\cos2x+1 \right)dx=\left(\frac38x+\frac{\sin4x}{32}+\frac{\sin2x}{4}\right)$$ nw its easy to show $$f(x+\pi)+f(\pi)=f(x+\pi)$$