Actually, this is the case for any periodic function. Consider $f(x)$ being periodic with period $T$. That is, $f(x+T)=f(x)$. In particular, $f(T)=f(0)$. Then, integrating by parts,
$$\int\limits_0^Tf(x)f'(x)dx = f(x)f(x)\big|_0^T-\int\limits_0^Tf'(x)f(x)dx$$
Thus,
$$2\int\limits_0^Tf(x)f'(x)dx = f(x)f(x)\big|_0^T = f(T)^2-f(0)^2=0$$
Hence
$$\int\limits_0^Tf(x)f'(x)dx = 0$$