Show that $\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$ for $n\ge 5.$ Question is to show that 
$$\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$$
when $n\geq5$, $x \in \mathbb R $ and $i$ is an imaginary unit.  
My effort:
$$|\int_{-n}^{n}e^{iy^2}dy|\leq \int_{-n}^{n}|e^{iy^2}|dy=\int_{-n}^{n}|\cos(y^2)+i\sin(y^2)|dy$$
$$ \leq \int_{-n}^{n}|\cos(y^2)|dy+\int_{-n}^{n}|i||\sin(y^2)|dy$$
$$\leq \int_{-n}^{n}|\cos(y^2)|dy+\int_{-n}^{n}|\sin(y^2)|dy$$  
It is also known that $|\cos(x)|,|\sin(x)|\leq1$
but its leading nowhere since integral then evalutes to $0$..  
Any tips?

A Contour Integral Estimate
This estimate is valid for all $n\gt0$, and shows that the bound is $2$ for $n\ge4.5$ (since $\sqrt\pi\doteq1.77245385$). The contours in the complex plane are straight lines, and are parametrized linearly.
$$
\begin{align}
\left|\,\int_{-n}^ne^{iy^2}\,\mathrm{d}y\,\right|
&\le\left|\,\int_{-n(1+i)}^{n(1+i)}e^{iy^2}\,\mathrm{d}y\,\right|
+\left|\,\int_n^{n(1+i)}e^{iy^2}\,\mathrm{d}y\,\right|
+\left|\,\int_{-n(1+i)}^{-n}e^{iy^2}\,\mathrm{d}y\,\right|\\
&\le\sqrt2\int_{-n}^ne^{-2y^2}\,\mathrm{d}y
+2\int_0^1e^{-2n^2t}n\,\mathrm{d}t\\[3pt]
&\le\sqrt\pi+\frac1n
\end{align}
$$

This is pretty good
Here is the plot of the actual value of the integral vs the estimate above vs $2$:
 
Note that the first maximum after $n=4.5$ is the first one that is less than $2$.
