Let $R > 0$ and consider a curve $$\beta(t) = R\exp(it), \ \ \ \ \ \ \ \ \ 0 \leq t \leq \pi/4.$$ I need to show that $$\left|\int_{\beta}\exp(iz^2)\ dz \right| \leq \frac{\pi(1-\exp(-R^2))}{4R}.$$
Attempt: Well, I thought I will just use the ML-estimate, but I got stuck. I have calculated the length of the arc of the curve $\beta$: $$l(\beta) = \int_0^{\pi/4}|\beta^{\prime}(t)|dt = \int_0^{\pi/4}|iR\exp(it)|\ dt = \int_0^{\pi/4}R^2\ dt = \frac{\pi R^2}{4}.$$ Now I would like to find an $M > 0$ such that $|\exp(iz^2)|$ for any $z \in \ \text{Image}\ \beta $.
We have $|\exp(iz^2)| = \exp(-R^2\sin(2t))$. Using $\sin(2t) \geq \frac{4}{\pi}t$ (which I dont understand how that is true), for any $t \in [0, \pi/4]$, it yields that $$|\exp(iz^2)| \leq \exp\left(-R^2\frac{4}{\pi}t\right).$$ From here, I don't know how to proceed.
Note that we have
$$\begin{align} \left|\int_\beta e^{iz^2}\,dz\right|&=\left|\int_0^{\pi/4}e^{i(Re^{i\phi})^2}\,iRe^{i\phi}\,d\phi\right|\\\\ &=\left|\int_0^{\pi/4}e^{iR^2e^{i2\phi}}\,iRe^{i\phi}\,d\phi\right|\\\\ &=\left|\int_0^{\pi/4}e^{iR^2\cos(2\phi)}e^{-R^2\sin(2\phi)}\,iRe^{i\phi}\,d\phi\right|\\\\ &\le R\int_0^{\pi/4}e^{-(4R^2/\pi)\phi}\,d\phi \end{align}$$
where we used the triangle inequality along with the estimate $ \sin(2\phi)\ge \frac{4\phi}\pi$ for $0\le \phi\le \pi/4$.
And now you can finish.