Find $D \int_{0}^{t}W^{2}_{s}dW_{s}$ My solution is next: variance = $E(\int_{0}^{t}W^{2}_{s}dW_{s})^{2} - (E\int_{0}^{t}W^{2}_{s}dW_{s})^{2}$ Which is equal to (using Ito's isometry principle) = $\int_{0}^{t}EW_{s}^{4}dW_{s} - (\int_{0}^{t}EW_{s}^{2}dW_{s})^{2}$
And that is easily calculable. But i wonder if all steps are correct
$$\int_0^t X_s \, dW_s \sim \mathcal{N}(\mu,\sigma^2)$$ for any stochastic process in $\mathcal{L}^2$ (this comes comes from the fact that the previous r.v. is a countable sum of normal distributions).
$$\mu=\mathbb{E}[\int_0^t W_s^2 \, dW_s]=0 \text{, by Itô's first isometry}$$
$$\sigma^2=\mathbb{E}[(\int_0^t W_s^2 \, dW_s)^2]-\mu^2=\int_0^t \mathbb{E}[W_s^4] \, ds-0=\int_0^t 3s^2 ds = t^3 $$ by the second isometry
Thus $$\int_0^t W_s^2 \, dW_s \sim \mathcal{N}(0,t^3)$$