Answer
$$ -\frac{1}{2}-\frac{\pi}{4}$$
Work Step by Step
Let $x=\theta^{2} \ \Rightarrow dx=2 \theta d \theta $. Thus,
\begin{align*}
\int_{\sqrt{\pi / 2}}^{\sqrt{\pi}} \theta^{3} \cos \left(\theta^{2}\right) d \theta
&=\int_{\sqrt{\pi / 2}}^{\sqrt{\pi}} \theta^{2} \cos \left(\theta^{2}\right) \cdot \frac{1}{2}(2 \theta d \theta)\\
&=\frac{1}{2} \int_{\pi / 2}^{\pi} x \cos x d x
\end{align*}
Let
\begin{align*}
u&=x\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\cos x dx\\
u&= dx\ \ \ \ \ \ \ \ \ \ \ \ \ v=\sin x
\end{align*}
Then
\begin{align*}
\frac{1}{2} \int_{\pi / 2}^{\pi} x \cos x d x &=\frac{1}{2}\left([x \sin x]_{\pi / 2}^{\pi}-\int_{\pi / 2}^{\pi} \sin x d x\right)\\
&=\frac{1}{2}[x \sin x+\cos x]\bigg|_{\pi / 2}^{\pi} \\
&=\frac{1}{2}(\pi \sin \pi+\cos \pi)-\frac{1}{2}\left(\frac{\pi}{2} \sin \frac{\pi}{2}+\cos \frac{\pi}{2}\right)\\
&= -\frac{1}{2}-\frac{\pi}{4}
\end{align*}
