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$\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{\pi}{2}$
$\frac{\pi}{2}$
$$
I=\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] d x
$$
Put $\mathrm{x}+\pi=\mathrm{t}$
$$
\begin{aligned}
& I=\int_{-\pi / 2}^{\pi / 2}\left[t^3+\cos ^2 t\right] d t=2 \int_0^{\pi / 2} \cos ^2 t d t \\
& =\int_0^{\pi / 2}(1+\cos 2 t) d t=\frac{\pi}{2}+0 .
\end{aligned}
$$
I=\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] d x
$$
Put $\mathrm{x}+\pi=\mathrm{t}$
$$
\begin{aligned}
& I=\int_{-\pi / 2}^{\pi / 2}\left[t^3+\cos ^2 t\right] d t=2 \int_0^{\pi / 2} \cos ^2 t d t \\
& =\int_0^{\pi / 2}(1+\cos 2 t) d t=\frac{\pi}{2}+0 .
\end{aligned}
$$
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