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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?

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  • $\begingroup$ @ Chung-Han Hsieh : no it is absolutely wrong that is not the way it works even for standard integral. Regards $\endgroup$ – TheBridge Mar 17 '14 at 22:18
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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.

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