In my book there's the following equality which is said to be easily derived from Ito's formula.
$$\int_0^1 \exp(B(t)^2) dB(t) = \int_0^{B(1)} e^{t^2} dt - \int_0^1 B(t) \exp(B(t)^2 dt$$
This may be trivial but after some trials I haven't been able to prove it. Could you please give me some help with this?
Thanks!!
Set $$F(x) := \int_0^x e^{t^2} \, dt.$$ Since $F'(x) = e^{x^2}$ and $F''(x) =2x e^{x^2}$, it follows from Itô's formula that
$$F(B_1)-F(B_0) = \int_0^1 e^{B_t^2} \, dB_t + \frac{1}{2} \int_0^1 2B_t e^{B_t^2} \, dt.\tag{1}$$
By definition, the left-hand side is $$F(B_1)-F(B_0) = \int_0^{B_1} e^{t^2} \, dt.$$
Plugging this into $(1)$ and rearranging the terms proves the assertion.