${\int }_{-3\mathrm{\pi }/2}^{-\mathrm{\pi }/2} \left[(\mathrm{x}+\mathrm{\pi }{)}^3+{\cos }^2(\mathrm{x}+3\mathrm{\pi })\right]\mathrm{dx}$ is equal to

$\mathrm{I}={\int }_{-3\mathrm{\pi }/2}^{-\mathrm{\pi }/2} \left[(\mathrm{x}+\mathrm{\pi }{)}^3+{\cos }^2(\mathrm{x}+3\mathrm{\pi })\right]\mathrm{dx}$
Put x + π = t

diff w.r.to x on both sides

$$\begin{gathered} \Rightarrow dx=dt \\ U.L,x=-\frac{\mathrm{\pi }}{2}\Rightarrow t=\frac{\mathrm{\pi }}{2} \\ L.L,x=-\frac{3\mathrm{\pi }}{2}\Rightarrow t=-\frac{\mathrm{\pi }}{2} \end{gathered}$$
$$\begin{gathered} \begin{array}{l}\mathrm{Given} \mathrm{integral} \mathrm{becomesI}={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \left[{\mathrm{t}}^3+{\cos }^2\mathrm{t}\right]\mathrm{dt} \\ ={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}{t}^3 dt +{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} {\cos }^2\mathrm{tdt}\\ =0+2{\int }_0^{\mathrm{\pi }/2} {\cos }^2\mathrm{tdt}\\ ={\int }_0^{\mathrm{\pi }/2} (1+\cos 2\mathrm{t})\mathrm{dt}=\frac{\mathrm{\pi }}{2}+0.\end{array} \end{gathered}$$