Let $B_t$ be brownian motion at time $t$, and $x$ be some random variable.
For instance, I know that
$$\int_0^T 1 dB_t = 1(B_T-B_0)$$
And that
$$\int_0^T \cos(B_t) dB_t$$ cannot be directly integrated,
but what about
$$\int_0^T \sin(B_t) dx$$
Can this last one be integrated, treating $$\sin(B_t)$$ as if it were a constant, and thus yielding :
$$\int_0^T \sin(B_t) dx = T\sin(B_t)$$