SDE
$dX_t=-a^2\sin X_t\cos^3X_tdt+a\cos^2X_tdW_t$ with $X_0=x_0$
I think this is inverse type of SDE, refer to Solve the SDE $dX_t = \frac{1}{2}\sigma(X_t)\sigma'(X_t)dt+\sigma(X_t)dW_t$.
However, I can't find the inverse funcition.
My try is $(\cos^2X_t)'=-\sin X_t\cos^3X_t$ ,but I can't go forward anymore.
Could you tell me how to solve this?
If we set
$$\sigma(x) = a \cdot \cos(x)^2$$
then $\sigma'(x) = -a \cdot \cos(x) \cdot \sin(x)$, i.e. we can rewrite the SDE in the following way:
$$dX_t = \sigma(X_t) \, dW_t + \sigma(X_t) \sigma'(X_t) \, dt.$$
This means that this SDE is in fact an inverse type SDE. Using the approach discussed in this answer (see Solution 2), we find that for
$$g(x) := \int_0^x \frac{1}{a \cos^2(y)} \, dy = \frac{1}{a} \tan(x)$$
we have
$$\frac{1}{a} \tan(X_t) - \frac{1}{a} \tan(x_0)= g(X_t) - g(x_0) = W_t.$$
Consequently, we obtain
$$X_t = \arctan \bigg( a W_t + \tan(x_0) \bigg).$$
It remains to check that $(X_t)_{t \geq 0}$ is in fact a solution to the SDE. This follows by applying Itô's formula.