Question

$\underset{-3\pi /2}{\overset{-\pi /2}{\int }}$$\left[{\left(x+\pi \right)}^3+co{s}^2\left(x+3\pi \right)\right]dx$ is equal to :

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

Answer 1

Answer: (B)

Let $\underset{-3\pi /2}{\overset{-\pi /2}{\int }}$$\left[{\left(x+\pi \right)}^3+co{s}^2\left(x+3\pi \right)\right]dx...(i)$
and $I=\underset{-3\pi /2}{\overset{-\pi /2}{\int }}$$\left[{\left(-\frac{\pi }{2}-\frac{3\pi }{2}-x+\pi \right)}^3+co{s}^2\left(-\frac{\pi }{2}-\frac{3\pi }{2}-x+3\pi \right)\right]dx$
$\Rightarrow$ $I=\underset{-3\pi /2}{\overset{-\pi /2}{\int }}$$\left[-{\left(x+\pi \right)}^3+co{s}^2\left(\pi -x\right)\right]dx...\left(ii\right)$
On adding Eqs. (i) and (ii), we get
$2I=\underset{-3\pi /2}{\overset{-\pi /2}{\int }}$$2co{s}^{2}xdx$
$=\underset{-3\pi /2}{\overset{-\pi /2}{\int }}$$\left(1+cos2x\right)dx$
$={\left[x+\frac{sin2x}{2}\right]}_{{}_{{}_{-3\pi /2}}}^{-\pi /2}$
$=-\frac{\pi }{2}+\frac{3\pi }{2}=\pi$
$\Rightarrow I=\frac{\pi }{2}$