we can see that $\left\langle \int_0^t \! W_s \, \mathrm{d} s ,W_t \right\rangle_t = 0$
However if I am to use the expression $$\int_0^t \! W_s \, \mathrm{d} s= t W_t - \int_0^t \! s\, \mathrm{d} W_s$$ then $$\left\langle \int_0^t \! W_s \, \mathrm{d} s ,W_{t}\right\rangle_t =\left\langle t W_t - \int_0^t \! s\, \mathrm{d} W_s ,W_t \right\rangle_t$$ thus $$\left\langle \int_0^t \! W_s \, \mathrm{d} s ,W_t\right\rangle_t=t\left\langle W_t ,W_t\right\rangle - \left\langle \int_0^t \! s\, \mathrm{d} W_s ,W_t\right\rangle_t$$ which means that $$\left\langle \int_0^t \! W_s \, \mathrm{d} s ,W_t\right\rangle_t=t^2-\int_0^t \! s \, \mathrm{d} s=\frac {t^2} {2}$$
What is the flaw with my argument?
Many thanks in advance