I am not sure whether I have solved it correctly. I need guidance. Thanks
To solve the stochastic integral
$$ \int_0^T (W_t)^5 dW_t, $$
we use Itô's formula. Let $( F(W_t) = \frac{1}{6} (W_t)^6 ).$ Then, Itô's formula gives
$$ dF(W_t) = \frac{\partial F}{\partial t} dt + \frac{\partial F}{\partial W_t} dW_t + \frac{1}{2} \frac{\partial^2 F}{\partial W_t^2} (dW_t)^2. $$
Since $( F(W_t) = \frac{1}{6} (W_t)^6 )$, we have
$$ \frac{\partial F}{\partial W_t} = (W_t)^5, \quad \frac{\partial^2 F}{\partial W_t^2} = 5(W_t)^4. $$
Also, $( \frac{\partial F}{\partial t} = 0 )$ because $( F )$ does not explicitly depend on $( t )$. Recall that $( (dW_t)^2 = dt )$. Thus,
$$ dF(W_t) = (W_t)^5 dW_t + \frac{1}{2} \cdot 5(W_t)^4 dt. $$
Integrating both sides from 0 to ( T ),
$$ \int_0^T dF(W_t) = \int_0^T (W_t)^5 dW_t + \frac{5}{2} \int_0^T (W_t)^4 dt. $$
The left-hand side becomes
$$ F(W_T) - F(W_0) = \frac{1}{6} (W_T)^6 - \frac{1}{6} (W_0)^6 = \frac{1}{6} (W_T)^6, $$
since $( W_0 = 0 )$. Therefore,
$$ \int_0^T (W_t)^5 dW_t = \frac{1}{6} (W_T)^6 - \frac{5}{2} \int_0^T (W_t)^4 dt. $$