Question

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

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}-1$
• C. $\frac{{\pi }^4}{32}$
• D. $\frac{{\pi }^4}{32}+\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{2}$

Let $I=\int {(x+\pi )}^{3}dx=\frac{1}{4}{(x+\pi )}^4$
And $J=\int co{s}^2(x+3\pi )dx=\int co{s}^{2}xdx=\int (\frac{1}{2}cos2x+\frac{1}{2})dx=\frac{1}{2}\int cos2xdx+\frac{1}{2}\int dx=\frac{1}{4}sin2x+\frac{x}{2}$
Therefore,
${\int }_{-\frac{3\pi }{2}}^{-\frac{\pi }{2}}[{(x+\pi )}^3+co{s}^2(x+3\pi )]dx={[\frac{1}{4}{(x+\pi )}^4+\frac{1}{4}sin2x+\frac{x}{2}]}_{-\frac{3\pi }{2}}^{-\frac{\pi }{2}}=\frac{\pi }{2}$