Given a Brownian motion $W(t)$ under $P$ and the stochastic process $\hat{W}(t) := W(t) - \int_{0}^{t}\theta(s)ds$ which is a Brownian motion under an equivalent measure $Q \sim P$.
The stochastic integral with respect to $dW(t)$ is $$\int_{0}^{t}dW(s) = W(t).$$ However, can one integrate with respect to $\hat{W}(t) + \int_{0}^{t}\theta(s)ds$ instead? I.e, $$ \int_{0}^{t} dW(s) = \int_{0}^{t}(d\hat{W}(s) + \int_{0}^{t}\theta(s) ds)$$
Does the expression above make sense, and if so, should it be evaluated as
$$\int_{0}^{t}d\hat{W}(s) + \theta(s) ds ?$$
Concretely, I have the following dynamics under P:
$$ X(t) = X(0)\exp(t\int_{0}^{t}\frac{1}{2}\sigma^2(s)ds + \int_{0}^{t}\sigma(s)dW(s)). $$
I want to describe the dynamics under Q~P under which $\hat{W}(t) = W(t) -\int_{0}^{t}\theta(s)ds$ is a Brownian motion.
Making the substitution seemingly gives $$(Q): X(t) = X(0)\exp(t\int_{0}^{t} \frac{1}{2}\sigma^2(s) ds + \int_ {0} ^{t} \sigma(s)(d\hat{W}(s) +\theta(s)ds)): $$