In this question assume we have a suitable probability space and everything is well defined...
Let $X$ be the solution to some SDE, and $\langle , \rangle$ the Euclidean inner product.
Consider the integral for $f,W \in \mathbb{R}^d$
$$ I:=\int_0^t \langle f(X_s)dW_s\rangle, $$
formally (if $W_s$ was differentiable with derivative $W'$) :
\begin{align*} I=&\int_0^t \langle f(X_s),W'_s\rangle ds \\ \leq& \int_0^t|f(X_s)||W'_s| ds\\ \leq& 1/2 \left( \int_0^t|f(X_s))|^2ds+\int_0^t|W'_s|^2 ds \right). \end{align*}
Is there a way to obtain a bound like this rigorously? Is it essential to have an expected value $\mathbb{E}$ somewhere to use BDG inequality?
Obviously we cant make $W'$ appear, so by a 'bound like this' I mean :
$$I\leq 1/2 \left( \int_0^t|f(X_s))|^2ds+|W_t|^2\right).$$