Integral of Brownian Motion w.r.t Time: what is wrong with this solution?

Question

My question is about a stochastic integral of brownian motion w.r.t time.

Let $W(t)$ the Wiener process (or brownian motion). I want to calculate this: \begin{eqnarray} X(t)=\int_{0}^t dt' W(t'). \end{eqnarray} My strategy:

1) Itô's Fórmula: \begin{eqnarray} d(tW(t))=tdW(t)+W(t)dt \implies W(t)dt=d(tW(t))-tdW(t). \end{eqnarray} 2) Integrate: \begin{eqnarray} X(t)=\int_{0}^t dt'W(t')=\int_0^t d(t'W(t'))+\int_{0}^tdW(t')t'=tW(t)-\frac{t}{\sqrt{3}}W(t)=\left(1-\frac{1}{\sqrt{3}}\right)tW(t). \end{eqnarray} I used: \begin{eqnarray} \int_{0}^t dW(t')f(t')=\left(\frac{1}{t}\int_{0}^t dt'|f(t')|^2\right)^{1/2}W(t)\implies \int_0^t dW(t')t'=\frac{t}{\sqrt{3}}W(t), \end{eqnarray} because... \begin{eqnarray} \int_{0}^t dW(t')f(t')\sim \mathcal{N}\left(0,\int_0^t dt'|f(t')|^2\right), \hspace{0.5cm} W(t)\sim \mathcal{N}(0,t). \end{eqnarray}

The "problem" is: \begin{eqnarray} \sigma^2_X=\left(1-\frac{1}{\sqrt{3}}\right)^2 t^3 \end{eqnarray} But the correct is: \begin{eqnarray} \sigma^2_X=\frac{t^3}{3} \end{eqnarray} Can some illuminated mind tell me where the error is?

Answer 1

Let me know if any of the steps is not clear.

Answer 2

Re-comment and your second question as to why your transformation leads to a different result, it is because the transformation does not take into account the covariance between the two terms. As you can see below, the covaraince is distorted;