Stratonovich integral of $\sin(W^2)$ I have to solve the following Stratonovich integral:
$$\int_{0}^{t}\sin(W^2_s)\circ{dW_s}$$ 
First of all I use the conversion from Stratonovich to Ito, obtaining 
$$\int_{0}^{t}\sin(W^2_s)dW_s+\int_{0}^{t}W_s{\cos(W_s)}ds$$
Is it sufficient? Or can I rewrite it in a better or more creative way? 
I hope that someone could give me a hint because now I don't see anything in this expression. 
Thank you all for your attention and your help! 
 A: As @Did already pointed out, the Stratonovich integral obeys the usual rules from (deterministic) differential calculus, i.e.
$$\int_0^t f'(W_s) \, \circ dW_s = f(W_t)-f(W_0);$$
that's exactly how the Stratonovich integrla is defined. Therefore, it is usually more convenient to use this property instead of transforming the Stratonovich integral to an Itô integral. But here is the way how to do it by hand:
Ito's formula states that
$$f(W_t)-f(W_0) = \int_0^t f'(W_s) \, dW_s + \frac{1}{2} \int_0^t f''(W_s) \, ds.$$
If we want to solve
$$\int_0^t \sin(W_s^2) \, dW_s + \int_0^t W_s \cos(W_s) \,ds,$$
we have to choose a function $f$ such that
$$f'(x) = \sin(x^2) \qquad f''(x) = 2 x \cos(x).$$
Obviously,
$$f(x) := \int_0^x \sin(y^2) \, dy$$
is the only choice. Consequently, by Itô's formula,
$$\int_0^t \sin(W_s^2) \, dW_s + \int_0^t W_s \cos(W_s) \,ds = f(W_t)= \int_0^{W_t} \sin(y^2) \, dy.$$
Since there is no explicit formula for $f$, it is questionable whether we gained anything at all. But, in fact, these transformations are often helpful in order to calculate certain characteristics of the distribution (expectation, variance,...) and derive sample path properties.
A: For every regular function $u$,
$$
\int_0^tu'(W_s)\circ\mathrm dW_s=u(W_t)-u(W_0).
$$
Here one can use
$$
u(w)=\int_0^w\sin(x^2)\mathrm dx=S(w),
$$
which is the Fresnel integral.
