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I want to calculate $\int ^t _0 W_tdW_t$

I know that the reasoning is the following:

Let $x(t)=W(t)$ with $a=0$ and $b=1$ in the definition of an Ito Process, and $f(t,x)=x^2$.

Then, applying Ito's formula:

$df=\frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial x}dW+\frac{1}{2}\frac{\partial ^2f}{\partial x^2}dt$

$dW^2=dt+2WdW$

$\int dW^2 = \int dt + 2\int WdW$

$\int WdW=\frac{1}{2}W^2-\frac{t}{2}$

My problem is that I don't understand why $df$ is equal to $dW^2$ since for me it would be $df=d(x^2)=d(W^2)$ and considering that I don't know over which variable I am deriving it, I don't know how the results above are explained.

It may be a really easy question but I am starting to study these things and I want to make sure that I understand the basics of it.

Thanks!

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    $\begingroup$ Yes, it is $d(W^2)$ $\endgroup$ – noob2 Nov 17 '16 at 20:47
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You are "deriving" with respect to $t$ (the time index in your stochastic process).

$f(t,x) = x^2$ so $f(t,W_t) = W_t^2$. And Ito's lemma tells you

$W_b^2 - W_a^2 = \int_{t=a}^b d(W_t^2) = \int_{t=a}^b df(t,W_t) = \int_{t=a}^b 2W_t dW_t + \int_{t=a}^b dt$ for all $0 \le a <= b$.

PS: Actually you are not deriving. The differential notation is just a notation to avoid having to write integrals. Ito's lemma is an identity between stochastic integrals.

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