I don't understand when I should be using Ito's formula.

Question

The question give was "consider the Ito integral: $\int_0^T sin(B_t)dB_t $ ."

I thought I had to use Ito's formula which is: $df(B_t)= f'(B_t)dB_t + \frac{1}{2} f''(B_t)dt $ hence I got: $sin(B_t) = \int_0^T cos(B_t)dB_t - \frac 12 \int_0^T sin(B_t)dt $

But the answer given was $1 - cos(t)$ which was done by integrating Sin. So I don't understand why wasn't Ito's formal used?

Answer 1

Your answer $1-\cos(t)$ can not be correct as the integral to be calculated is a stochastic integral. Using Ito's lemma with $f(x) = \cos(x)$ you got $$\mathrm{d}\cos(B_t) = -\sin(B_t)\,\mathrm{d} B_t - \frac 12\cos(B_t)\,\mathrm{d}t.$$ Now integrating both sides from $0$ to $T$, can you conclude?

Answer 2

The basic recipe for using Ito's formula to solve $\int_0^T f(B_t) dB_t$ when $f$ is a $C^1$ function is to look at $F(B_t)$ where $F$ is an antiderivative of $f$ and then compute the differential of this process by Ito's formula. What you get is

$$dF(B_t)=f(B_t) dB_t + \frac{1}{2} f'(B_t) dt.$$

Therefore by moving a term over:

$$\int_0^T f(B_t) dB_t = F(B_T)-F(0)-\frac{1}{2} \int_0^T f'(B_t) dt.$$