Calculate $E\left(\tilde Y_{k}^{2}\right)$ where $ \tilde Y_{k}=\int_{(2(k-1)+\frac{1}{2}) \pi}^{\left(2 k+\frac{1}{2}\right) \pi} \cos s d \Theta(s)$

Question

Let $B(t)$ be a 1-dimensional Brownian motion. Define a sequence of random variables $\left\{Y_{k}\right\}_{k \geq 0}$ by $$ Y_{k}= \int_{\left(2(k-1)+\frac{1}{2}\right) \pi}^{\left(2 k+\frac{1}{2}\right) \pi} \cos s d B(s) . $$ $\left\{Y_{k}\right\}_{k \geq 1}$ is a sequence of independent random variables, because of the independence of the increments of the Brownian motion on disjoint intervals. Moreover, each $Y_{k}$ is normally distributed with mean zero and variance $$ E\left(Y_{k}^{2}\right)=\int_{\left(2(k-1)+\frac{1}{2}\right) \pi}^{\left(2 k+\frac{1}{2}\right) \pi} \cos ^{2} s d s=\pi $$ I would now define another sequence of random variables $\left\{\tilde Y_{k}\right\}_{k \geq 0}$ by $$ \tilde Y_{k}= \int_{\left(2(k-1)+\frac{1}{2}\right) \pi}^{\left(2 k+\frac{1}{2}\right) \pi} \cos s d \Theta(s) . $$ where $\Theta$ is the Heaviside step function with the distributional derivative $\dot \Theta=\delta$. Is $\left\{\tilde Y_{k}\right\}_{k \geq 0}$ still a sequence of independent random variables? What is $E\left(\tilde Y_{k}^{2}\right)$?