Let $H$ be the class of general integrands, so $H := \{(h_t)_{t\ge0} : h_t $ is adapted, $ E \int^\infty_0h^2_tdt < \infty \}$.
I need to show that $\sin(2W_t)\mathbb{1}_{[0,T]}(t) \in H$, I have proven $\mathbb{1}_{[0,T]}(t)$ is in $H$, I just don't know how to go about proving $\sin(2W_t)$ is in $H$, where $W_t$ for $t \geq 0$ is a standard Brownian motion. Any help is much appreciated, thanks.
One has that $\sin(2W_t)$ is not in $H$ because $E \int_0^\infty \sin(2W_t)^2\,dt = \infty$. However, $\sin(2W_t)1_{[0,T]}(t)$ is in $H$ because the product of adapted things is adapted and since $|\sin| \leq 1$ one has $$E \int_0^\infty (\sin(2W_t)1_{[0,T]})^2\,dt =E \int_0^T \sin(2W_t)^2\,dt \leq E \int_0^T 1\,dt = T.$$