Question

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

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

Answer 1

The correct option is C $\frac{\pi }{2}$

Let $I={\int }_{-3\pi /2}^{-\pi /2}[{(x+\pi )}^3+{\cos }^2(x+3\pi )]dx$
Substitute $x+\pi =t$ $\Rightarrow$ $dx=dt$
at $x=-\frac{\pi }{2}$, $t=-\frac{\pi }{2}+\pi =\frac{\pi }{2}$
at $x=-\frac{3\pi }{2}$, $t=-\frac{3\pi }{2}+\pi =-\frac{\pi }{2}$
$\therefore$ $I={\int }_{-\pi /2}^{\pi /2}[t^3+{\cos }^2(t+2\pi )]dt={\int }_{-\pi /2}^{\pi /2}[t^3+{\cos }^{2}t]dt$
$=2{\int }_0^{\pi /2}{\cos }^{2}tdt$
$={\int }_0^{\pi /2}(1+\cos 2t)dt=\frac{\pi }{2}+0$ $\therefore$ $I=\frac{\pi }{2}$
Ans: C