is there a faster/easier way to compute $\int_a^bW(t)dW(t)$, $a<b$ where W(t) is the wiener process?
My method is the following:
Define $dX(t)=0dt+1dW(t)$ ($X(t)=W(t)$) and apply ito to $f(x)=x^2$.
$dW^2(t)=dt+2W(t)dW(t)$
$W(t)dW(t)=\frac12[dW^2(t)-dt]$
$\int_a^bW(t)dW(t)=\int_a^b\frac12[dW^2(t)-dt]=\frac12[W^2(b)-W^2(a)-(a-b)]$
This may be faster, it's likely about the same though. Stochastic integration by parts looks like $$\int_a^b f(t)g'(W_t)dW_t = f(t)g(W_t)|_a^b - \int_a^b f'(t)g(W_t)dt - \frac{1}{2}\int_a^b f(t)g''(W_t)dt,$$
where $f(t)$ is a deterministic function and $g(W_t)$ is a function of a Wiener process. For the integral you want we can let $f(t) = 1$ and we get
$$\int_a^b g'(W_t)dW_t = g(W_t)|_a^b - \frac{1}{2}\int_a^b g''(W_t)dt.$$
So for the integral $\int_a^b W_tdW_t$ we can let $g(x) = \frac{x^2}{2}$ and this gives
$$\int_a^b W_tdW_t = \frac{W_t^2}{2}|_a^b - \frac{1}{2}\int_a^b dt.$$