Consider the Black-Scholes Model where we have the following risky asset
$dS_t = \mu S_t dt + \sigma S_t dW_t, t\in[0,T] t≥0 ,S_0 = s >0 $ where $\mu,\sigma$ are positive constants
and a risk-free asset $dB_t = rB_tdt , B_0 = 1 $
$(W_t)_{t≥0}$ denotes a standard brownian motion with its natural filtration $(F_t)_{t≥0}$.
My attempt : Consider $\gamma_{t} = \exp(-\frac 12 \int_{0}^{t} (\frac{r-\mu} {\sigma})^2 ds + \int_{0}^{t} (\frac{r-\mu} {\sigma})dW_{s}) .$
We need to show that $E[\gamma_{t}] = 1$; then Girsanov's theorem states that $W^{*}_t$ is a standard brownian motion on $(\Omega,f,(f_t)_{t≥0},P^{*})$ where the new probability measure $P^{*}$ is defined by $dP^{*} = \gamma_{t}dP$.
$$E\left[\gamma_{t}\right] = E\left[\exp\left(-\frac 12 \int_{0}^{t} \left(\frac{r-\mu} {\sigma}\right)^2 ds + \int_{0}^{t} \left(\frac{r-\mu} {\sigma}\right)dW_{s}\right)\right]$$
$$=E\left[\exp\left(-\frac 12\left(\frac{r-\mu} {\sigma}\right)^2 t + \frac{r-\mu}{\sigma}W_t\right)\right]$$
$$=\exp\left(-\frac 12\left(\frac{r-\mu} {\sigma}\right)^2 t\right) E\left[\exp\left(\frac{r-\mu}{\sigma}W_t\right)\right]$$
$$=\exp\left(-\frac 12\left(\frac{r-\mu} {\sigma}\right)^2 t\right) \cdot \exp\left(\frac 12\left(\frac{r-\mu} {\sigma}\right)^2 t\right) = 1$$
Hence $\gamma_{t}$ is the Radon-Nikodym density .
Is this following method correct and also how can we find what $\gamma_{t}$ is when it is not given from the following problem.
Thank you in advance!