Consider $$\int_0^t \sin(B_s) ds$$
where $B_s$ is standard Brownian motion,
I was wondering can I write
$$\int_0^t \sin(B_s) ds = – ( \cos(B_t) – \cos(B_0)) = – \cos(B_t) ? $$
by using the Calculus fact:
$$\int \sin(x) dx = -\cos(x) $$
or I should only remain $\int_0^t \sin(B_s) ds$ as the final result?
Best Answer
As @TheBridge already pointed out, this does not work; not even for determinstic "standard" integrals. Please note that the integral
$$\int_0^t \sin(B_s) \,ds \tag{1}$$
is not of the form
$$\int_0^x \sin(y) \, dy.$$
Using Itô's formula you may rewrite $(1)$ as
$$\int_0^t \sin(B_s) \, ds = 2\int_0^t \cos(B_s) \, dB_s - 2 \sin B_t.$$
If we consider a stochastic (Itô-)integral with respect to Brownian motion, i.e.
$$\int_0^t f(B_s) \, dB_s$$
for some function $f$, then Itô's formula tells us that we can not simply apply the classical rules of integration (e.g. the fundamental theorem of calculus). One of the most-well known examples for this fact is the equality
$$\int_0^t B_s \, dB_s = \frac{1}{2} (B_t^2 - t);$$
from the classical rules we would expect
$$\int_0^t B_s \, dB_s = \frac{B_t^2}{2}.$$
This leads to the so-called Stratonovich integral.