int_{frac{-3pi}{2}}^{frac{-pi}{2}} left[ (x+p | vedclass

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(C) Let $I = \int_{-\frac{3\pi}{2}}^{-\frac{\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx$. Substitute $t = x + \pi$,then $dt = dx$. When $x =

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$\frac{\pi}{2}$

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(C) Let $I = \int_{-\frac{3\pi}{2}}^{-\frac{\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx$. Substitute $t = x + \pi$,then $dt = dx$. When $x = -\frac{3\pi}{2}$,$t = -\frac{\pi}{2}$. When $x = -\frac{\pi}{2}$,$t = \frac{\pi}{2}$. Since $\cos(x+3\pi) = \cos(x+\pi) = -\cos x$,we have $\cos^2(x+3\pi) = \cos^2 x$. Thus,$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (t^3 + \cos^2 t) dt$. Since $t^3$ is an odd function and $\cos^2 t$ is an even function,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t^3 dt = 0$. Therefore,$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 t dt = 2 \int_{0}^{\frac{\pi}{2}} \cos^2 t dt$. Using the identity $\cos^2 t = \frac{1 + \cos 2t}{2}$,we get $I = 2 \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos 2t}{2} dt = \int_{0}^{\frac{\pi}{2}} (1 + \cos 2t) dt$. Evaluating the integral: $[t + \frac{\sin 2t}{2}]_{0}^{\frac{\pi}{2}} = (\frac{\pi}{2} + 0) - (0 + 0) = \frac{\pi}{2}$.

$\int_{\frac{-3\pi}{2}}^{\frac{-\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx = $

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