9. - / 2.08 points first make a substitution and then use integration by parts to evaluate the integral…

Explanation:

Step1: Make a substitution

Let $u = \theta^{2}$, then $du=2\theta d\theta$ and $\theta^{3}d\theta=\frac{1}{2}u du$. When $\theta=\sqrt{\frac{\pi}{2}}$, $u = \frac{\pi}{2}$; when $\theta=\sqrt{\pi}$, $u=\pi$. The integral $\int_{\sqrt{\pi/2}}^{\sqrt{\pi}}3\theta^{3}\cos(\theta^{2})d\theta$ becomes $\frac{3}{2}\int_{\pi/2}^{\pi}u\cos(u)du$.

Step2: Apply integration - by - parts

The integration - by - parts formula is $\int v dw=vw-\int w dv$. Let $v = u$ and $dw=\cos(u)du$. Then $dv = du$ and $w=\sin(u)$. So $\frac{3}{2}\int_{\pi/2}^{\pi}u\cos(u)du=\frac{3}{2}\left[u\sin(u)\big|{\pi/2}^{\pi}-\int{\pi/2}^{\pi}\sin(u)du\right]$.

Step3: Evaluate definite integrals

First, evaluate $u\sin(u)\big|{\pi/2}^{\pi}=\pi\sin(\pi)-\frac{\pi}{2}\sin(\frac{\pi}{2})=0 - \frac{\pi}{2}=-\frac{\pi}{2}$. Second, evaluate $\int{\pi/2}^{\pi}\sin(u)du=-\cos(u)\big|_{\pi/2}^{\pi}=-\cos(\pi)+\cos(\frac{\pi}{2})=1 + 0 = 1$.

Step4: Calculate the final result

$\frac{3}{2}\left[-\frac{\pi}{2}-1\right]=\frac{-3\pi - 6}{4}$.

Answer:

$\frac{-3\pi - 6}{4}$