Calculate autocorrelation of a sinus I am a beginer in signal processing.
I am currently studying a signal $s(t) = \cos(h(t))$, and I have to calculate its autocorellation.
Since $h(t)$ is periodic of period $T_s$, I calculated the value of 

$$R_s(\tau) = \frac{1}{T_s} \int_0^{T_s} \cos(h(t)) \cos(h(t-\tau)) dt$$

Honestly, this calculation is not easy (since $h(t)$ is also a cosinus), and in order to simplify my final document, I would like to use complex representation by writing $s(t) = Re\left\{ e^{ih(t)} \right\}$. However, This cannot work for the autocorrelation since $$Re\left\{ e^{ih(t)} e^{-ih(t-\tau)}  \right\} = s(t)s(t-\tau) + Re\left\{ \sin(h(t)) \sin(h(t-\tau))  \right\}$$

So, do you know a way to calculate $R_s$ by calculating $\frac{1}{T_s} \int_0^{T_s} e^{ih(t)} e^{-ih(t-\tau)} dt$ ? 

I was thinking about a relation like $R_s(\tau) = \frac{1}{2} Re \left\{\frac{1}{T_s} \int_0^{T_s} e^{ih(t)} e^{-ih(t-\tau)} dt \right\}$, but I cannot prove it :(
It's way easier to use the complex representation of cosine not involving the real part operator. In your case this translates into expressing

$$ \cos(h(t))=\frac{e^{jh(t)} + e^{-jh(t)}}{2} $$

Note that this yields an autocorrealtion of
$$ R_s(\tau)=\int_{0}^{T_s} \frac{e^{jh(t)} + e^{-jh(t)}}{2}\frac{e^{jh(t-\tau)} + e^{-jh(t-\tau)}}{2} dt $$