1. We need to evaluate $I=\int_{-3\pi/2}^{-\pi/2}\left[(x+\pi)^3+\cos^2(x+3\pi)\right]dx.$

2. Simplify the trigonometric part. Since cosine has period $2\pi$, $\cos(x+3\pi)=\cos(x+\pi)=-\cos x,$ so $\cos^2(x+3\pi)=\cos^2 x.$ Hence, $I=\int_{-3\pi/2}^{-\pi/2}(x+\pi)^3\,dx+\int_{-3\pi/2}^{-\pi/2}\cos^2 x\,dx.$

3. Evaluate the first integral. Let $u=x+\pi.$ Then when $x=-3\pi/2$, $u=-\pi/2$, and when $x=-\pi/2$, $u=\pi/2$. Therefore, $\int_{-3\pi/2}^{-\pi/2}(x+\pi)^3dx=\int_{-\pi/2}^{\pi/2}u^3\,du.$ Now $u^3$ is an odd function, and the interval is symmetric about $0$, so $\int_{-\pi/2}^{\pi/2}u^3\,du=0.$

4. Evaluate the second integral. $\int_{-3\pi/2}^{-\pi/2}\cos^2 x\,dx.$ Using the identity $\cos^2 x=\frac{1+\cos 2x}{2},$ we get $\int_{-3\pi/2}^{-\pi/2}\cos^2 x\,dx=\frac12\int_{-3\pi/2}^{-\pi/2}(1+\cos 2x)dx.$ So, $=\frac12\left[x+\frac{\sin 2x}{2}\right]_{-3\pi/2}^{-\pi/2}.$ Now, $\sin(-\pi)=0,\qquad \sin(-3\pi)=0.$ Thus, $\int_{-3\pi/2}^{-\pi/2}\cos^2 x\,dx=\frac12\left[\left(-\frac\pi2\right)-\left(-\frac{3\pi}{2}\right)\right]=\frac12(\pi)=\frac\pi2.$

5. Add both parts: $I=0+\frac\pi2=\frac\pi2.$

6. Compare with options:

• A: $\dfrac{\pi^4}{32}$
• B: $\dfrac{\pi^4}{32}+\dfrac\pi2$
• C: $\dfrac\pi2$
• D: $\dfrac\pi4-1$

Therefore, the correct option is $\boxed{\text{C }\frac\pi2}.$