How to compute $E[W_t^4]$, with $W_t$ being a standard Wiener process I want to compute the fourth moment of a standard Wiener process: $E[W_t^4]$. My solution is not equal to the one in my textbook but I don't understand where I am wrong. I started by applying Ito's lemma to $g(W_t) = W_t^4$ and got 
$dg(W_t) = 6W_t^2dt  + 4W_t^3dW_t$,
implying that  
$W_t^4 = \int_0^t 6W_t^2ds + 4 \int_0^t W_s^3 dW_s$.
Then, I took expectations and got for the first term
$ E[\int_0^t 6W_t^2ds] = 6\int_0^t V[W_t]ds = 6\int_0^t sds = 3 t^2$.
In order to calculate the second term, I applied Ito's lemma to $f(W_t) = W_t^4/4$ and got 
$df(W_t) = \frac{3}{2}W_t^2dt  +  W_t^3dW_t$,
implying that  
$\int_0^t W_s^3 dW_s = W_t^4/4 - \int_0^t \frac{3}{2} W_s^2ds $.
So, I took again expectations and got 
$E[\int_0^t W_s^3 dW_s] = E[W_t^4]/4 - \frac{3}{2} \int_0^t  V[W_t]ds \\
= 3 t^4 /4 - 3 t^2 /4$.
Hence, my result is $E[W_t^4]=3 t^2  + 4(3 t^4/4  - 3 t^2/4)  = 3t^4 $.
However, in my textbook the second term was just dropped and the result was $3 t^2$. What's wrong with my way?
 A: First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write
$$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$
(I don't get what you did in this last step - you want to calculate $\mathbb{E}(W_t^4)$; so why replace it with $3t^4$?)

Note that applying Itô's lemma is overkill: Since $W(_t)_{t \geq 0}$ is a Wiener process, we know that $W_t \sim N(0,t)$ (i.e. $W_t$ is Gaussian with mean $0$ and variance $t$) and the moments of Gaussian random variables can be calculated explicitly. However, if you really want to invoke Itô's formula, then it goes like that: By Itô's formula, we have
$$W_t^4 = 4 \int_0^t W_s^3 \, dW_s + 6 \int_0^t W_s^2 \, ds. \tag{1}$$
Since $(W_s^3)_{s \geq 0}$ is properly integrable, we know that the stochastic integral
$$M_t := \int_0^t W_s^3 \, dW_s$$
is a martingale and therefore $\mathbb{E}M_t = \mathbb{E}M_0=0$. Taking expectation in $(1)$ yields
$$\mathbb{E}(W_t^4) = 6 \int_0^t \mathbb{E}(W_s^2) \, ds$$
by Fubini's theorem. Finally, since $\mathbb{E}(W_s^2)=s$, we get $\mathbb{E}(W_t^4) = 3t^2$.
