Answer
$$\int^\sqrt[3]{\pi^2}_0\sqrt\theta\cos^2(\theta^{3/2})d\theta=\frac{\pi}{3}$$
Work Step by Step
$$A=\int^\sqrt[3]{\pi^2}_0\sqrt\theta\cos^2(\theta^{3/2})d\theta$$
We set $u=\theta^{3/2}$, which means $$du=\frac{3}{2}\theta^{1/2}d\theta=\frac{3}{2}\sqrt\theta d\theta$$ $$\sqrt\theta d\theta=\frac{2}{3}du$$
For $\theta=\sqrt[3]{\pi^2}=\pi^{2/3}$, we have $$u=(\pi^{2/3})^{3/2}=\pi$$
For $\theta=0$, we have $$u=0^{3/2}=0$$
Therefore, $$A=\frac{2}{3}\int^\pi_0\cos^2udu$$
Recall the identity: $$\cos^2x=\frac{1+\cos2x}{2}$$
So, $$A=\frac{2}{3}\int^\pi_0\frac{1+\cos2u}{2}du$$ $$A=\frac{2}{3}\Big(\int^\pi_0\frac{1}{2}du+\frac{1}{2}\int^\pi_0\cos2udu\Big)$$ $$A=\frac{2}{3}\Big(\frac{u}{2}\Big]^\pi_0+\frac{1}{2}\times\frac{1}{2}\sin2u\Big]^\pi_0\Big)$$ $$A=\frac{2}{3}\Big(\frac{\pi}{2}+\frac{1}{4}(\sin2\pi-\sin0)\Big)$$ $$A=\frac{2}{3}\Big(\frac{\pi}{2}+\frac{1}{4}(0-0)\Big)$$ $$A=\frac{2}{3}\Big(\frac{\pi}{2}\Big)=\frac{\pi}{3}$$
