Answer
$$\int^{\pi/2}_0\theta^2\sin2\theta d\theta=\frac{\pi^2}{8}-\frac{1}{2}$$
Work Step by Step
$$A=\int^{\pi/2}_0\theta^2\sin2\theta d\theta$$
Take $u=\theta^2$ and $dv=\sin2\theta d\theta$
We would have $du=2\theta d\theta$ and $v=-\frac{\cos2\theta}{2}$
Applying the formula $\int^b_a udv=uv]^b_a-\int^b_a vdu$, we have
$$A=-\frac{1}{2}\theta^2\cos2\theta\Big]^{\pi/2}_0-\int^{\pi/2}_0-\Big(\frac{\cos2\theta}{2}\Big)\times2\theta d\theta$$ $$A=-\frac{1}{2}\Big(\frac{\pi^2}{4}\cos\pi\Big)+\int^{\pi/2}_0\theta\cos2\theta d\theta$$ $$A=-\frac{1}{2}\Big(\frac{\pi^2}{4}(-1)\Big)+\int^{\pi/2}_0\theta\cos2\theta d\theta$$ $$A=\frac{\pi^2}{8}+\int^{\pi/2}_0\theta\cos2\theta d\theta$$
Take $u=\theta$ and $dv=\cos2\theta d\theta$
We would have $du=d\theta$ and $v=\frac{\sin2\theta}{2}$
Applying the formula $\int^b_a udv=uv]^b_a-\int^b_a vdu$, we have
$$A=\frac{\pi^2}{8}+\Big(\frac{1}{2}\theta\sin2\theta\Big)\Big]^{\pi/2}_0-\frac{1}{2}\int^{\pi/2}_0\sin2\theta de\theta$$ $$A=\frac{\pi^2}{8}+\frac{1}{2}\Big(\frac{\pi}{2}\sin\pi\Big)+\Big(\frac{1}{4}\cos2\theta\Big)\Big]^{\pi/2}_0$$ $$A=\frac{\pi^2}{8}+\frac{1}{2}\Big(\frac{\pi}{2}\times0\Big)+\frac{1}{4}(\cos\pi-\cos0)$$ $$A=\frac{\pi^2}{8}+\frac{1}{4}(-1-1)$$ $$A=\frac{\pi^2}{8}-\frac{1}{2}$$
