formula

$&&\int_0^t (X_s^{(\theta)})^{1/3}d\langle Z^{(\theta)}_s\\ &=&\int_0^t(X_s^{(\theta)})^{1/3}ds-2\int_0^t(X_s^{(\theta)})^{1/3}d\langle W,A^{(\theta)}\rangle _s+\int_0^t(X_s^{(\theta)})^{1/3}d\langle A^{(\theta)}\rangle_s$