Method of stationary phase when the stationary point is neither minimum nor maximum. I am trying to evaluate the leading order behaviour of $$I(x) = \int_{0}^{1} e^{ix(t-\sin(t))} dt,$$ using the method of stationary phase. The way we have been taught to solve these types of integrals is find all stationary points in the range, Taylor expand $t-\sin(t)$ about the maxima and minima, and evaluate the integral at small intervals around these points.
I'm running into trouble though because the only stationary point of $$f(t) = t - \sin(t)$$ in the interval $[0,1]$ is $t=0$, which isn't a minimum or a maximum but an inflection point. If anyone could help me out on how to solve this integral I would be very grateful.

Define an entire function
$$f(z)~:=~z - \sin z , \qquad z~\in~\mathbb{C}.\tag{1}$$
The integration contour of OP's integral $$I ~:=~ \int_0^1 \! \mathrm{d}z ~\exp\left(ix f(z)\right) ~=~J(1)-J(0) ,\tag{2}$$ 
is deformed along the negative imaginary axis of the complex $z$-plane, where the integrand becomes exponentially small. Here
$$J(a)~:=~\int_{-i\infty}^0   \mathrm{d}z ~\exp\left\{ix f(a\!+\!z)\right\}.
\tag{3} $$
OP is interested in the asymptotic expansion for $x\to \infty$.
The idea is now to choose a steepest descent contour for the integral (3) out of the endpoint $z\!=\!0$. If the function $$z~~\mapsto~~ f(a\!+\!z)\tag{4}$$ has a zero of order $n$ at $z\!=\!0$, the leading contribution will be of order $x^{-\frac{1}{n}}$.
It turns out that the leading term of $I$ in an asymptotic expansion for $x\to \infty$ comes from the integral (3) associated with the lower endpoint $a\!=\!0$, since the function (4) has a zero of order $n=3$ at $z\!=\!0$, as already noted by OP. The three angular steepest decent directions are $-\frac{\pi}{2}$, $\frac{\pi}{6}$, and $\frac{5\pi}{6}$. The first angle is the relevant one, since we have deformed the integration contour along the negative imaginary axis.
$$ J(0)~\stackrel{(3)}{=}~ \int_{-i\infty}^0   \mathrm{d}z ~\exp\left\{ix f(z)\right\}$$
$$~=~
i\int_{-\infty}^0   \mathrm{d}y ~\exp\left\{ -x\left(y-\sinh y\right) \right\} 
~=~i\int_0^{\infty}   \mathrm{d}y ~\exp\left\{ -x\left(\sinh y-y\right) \right\} $$
$$~\sim~i\int_0^{\infty}   \mathrm{d}y ~\exp\left( -\frac{xy^3}{6} \right)
~\stackrel{u=\frac{xy^3}{6} }{=}~
\frac{i}{3}\sqrt[3]{\frac{6}{x}}\int_0^{\infty}   \mathrm{d}u ~u^{-\frac{2}{3}}\exp\left( -u \right)~=~ i \sqrt[3]{\frac{2}{9x}}\Gamma\left(\frac{1}{3}\right)$$ $$ \qquad\text{for}\qquad x\to \infty. \tag{5}$$
This leads to OP's sought-for expansion

$$ I~\sim~  -i \sqrt[3]{\frac{2}{9x}}\Gamma\left(\frac{1}{3}\right) +O(x^{-1}) \qquad\text{for}\qquad x\to \infty. \tag{6}$$

