How to determine the leading-order asymptotic behaviour of this integral? $\int^{1}_{0} \cos(xt^{3})\tan(t)\, dt$
as x → ∞
I am stuck on how to apply the stationary phase method when $\tan(t)$ vanishes at the stationary point. Should I expand tan in its Taylor series? Is the stationary point $0\,$? Thanks in advance

Just to show something different from @joriki's answer to [this question], working first the antiderivative
$$I=\int\cos(xt^{3})\tan(t)\, dt$$ let $x t^3=u^3$ to make
$$I=\frac 1{x^{1/3}}\int\cos \left(u^3\right) \tan \left(\frac{u}{x^{1/3}}\right)\,du$$ Since $x\to \infty$, using $\tan(\epsilon)\sim \epsilon$,we have
$$I\sim \frac 1{x^{2/3}}\int\ u\, cos \left(u^3\right)\,du$$ The integral can be computed using the gamma function or the exponential integral function to make
$$\int\ u\, cos \left(u^3\right)\,du=-\frac{1}{6} u^2 \left(E_{\frac{1}{3}}\left(-i
   u^3\right)+E_{\frac{1}{3}}\left(i u^3\right)\right)$$ Back to $t$
$$I \sim -\frac{1}{6} t^2 \left(E_{\frac{1}{3}}\left(-i t^3
   x\right)+E_{\frac{1}{3}}\left(i t^3 x\right)\right)$$ Using the bounds
$$J=\int_0^1\cos(xt^{3})\tan(t)\, dt\sim\frac{1}{6} \left(\frac{\Gamma
   \left(\frac{2}{3}\right)}{x^{2/3}}-E_{\frac{1}{3}}(-i x)-E_{\frac{1}{3}}(i x)\right)$$
Limited to the first order
$$E_{\frac{1}{3}}(-i x)+E_{\frac{1}{3}}(i x)\sim -\frac {e^{-i x }} x$$
So the leading order of $J$ is
$$\frac{\Gamma \left(\frac{2}{3}\right)}{6\, x^{2/3}}$$
