I wonder if there is a way (if any) to prove that the inner product of the derivative and indefinite integral of any periodic signal (with period $2\pi$) from $0$ to $2\pi$ is zero, i.e., they are always orthogonal. I tried a simple case and found that this is not always possible. For example, for a signal $v(t)=\cos t + \cos 2t$, we can derivate and integrate
$$\frac{d v(t)}{dt}=-\sin t - 2\sin 2t\\ \int{v(t)}=\sin t + \frac{1}{2}\sin 2t$$
so we can perform the inner product as
$$\int_0^{2\pi}{(-\sin t - 2\sin 2t)(\sin t + \frac{1}{2}\sin 2t)\neq0}$$
So they are not orthogonal!! Am I doing something wrong?