Why is there only one integration in the solution if there is two integral in the formula?
You have a single random variable $\Theta$ in this example, so taking the expectation with respect to that random variable results in a single integral. The formula for $R_X(\tau)$ in your question is rather messy and full of mistakes; it should be as follows:

$$\begin{align}R_X(\tau)&amp;=E[X(t+\tau)X(t)]\\&amp;=E[a\sin(\omega_0t+\omega_0\tau+\Theta)\cdot a\sin(\omega_0t+\Theta)]\\&amp;=\frac{a^2}{2}E[\cos(\omega_0\tau)-\cos(2\omega_0t+\omega_0\tau+2\Theta)]\\&amp;=\frac{a^2}{2}\frac{1}{2\pi}\int_{0}^{2\pi}[\cos(\omega_0\tau)-\cos(2\omega_0t+\omega_0\tau+2\Theta)]d\Theta\\&amp;=\frac{a^2}{2}\cos(\omega_0\tau)\end{align}$$