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For $k\geq0$ a continuous function on $\mathbb{R}_+$ and $\{W_t\}$ a standard Brownian motion, could you help me find the Taylor's expansion of the following exponential: $e^{-\int_0^t k(s) dW_s}.$

For the case where $e^{-k W_t}$ where $k>0$ is a constant, I was able to recompute its Taylor's expansion as

$$

e^{-k W_t} = 1 - k\int_0^t e^{-kW_s}dW_s + \frac{1}{2}k^2\int_0^t e^{-kW_s}ds.

$$

But in the first exponential above, we have an integral exponent with a nonconstant $k,$ but a function. Please help me on this.

Thanks a lot in advance.