Turn $\int_{0}^{\infty}e^{-x^4}dx$ into a proper integral

Question

I'm asked to transform the improper integral $$\int_{0}^{\infty}e^{-x^4}dx$$ into a proper integral.

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What I tried:

Substituting $x = \tan(x)$ and $\frac{dx}{ds} = \sec^2(s)$

The integral becomes $$\int_{0}^{+ \frac{\pi}{2}} e^{- \tan^4(s)} \sec^2{(s)} ds$$

Is that correct ? Is there a way to go further ?