Question

$\int\limits^{-\pi/2}_{{-3\pi/2}}$$\left[\left(x+\pi\right)^{3}+cos^{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 $\int\limits^{-\pi/2}_{{-3\pi/2}}$$\left[\left(x+\pi\right)^{3}+cos^{2} \left(x+3\pi\right)\right]dx ...(i)$
and $I=\int\limits^{-\pi/2}_{{-3\pi/2}}$$\left[\left(-\frac{\pi}{2}-\frac{3\pi}{2}-x+\pi\right)^{3}+cos^{2}\left(-\frac{\pi}{2}-\frac{3\pi}{2}-x+3\pi\right)\right]dx$
$\Rightarrow $ $I=\int\limits^{-\pi/2}_{{-3\pi/2}}$$\left[-\left(x+\pi\right)^{3}+cos^{2}\left(\pi-x\right)\right]dx ...\left(ii\right)$
On adding Eqs. (i) and (ii), we get
$2I=\int\limits^{-\pi/2}_{{-3\pi/2}}$$2\,cos^{2}\,x\,dx$
$=\int\limits^{-\pi/2}_{{-3\pi/2}}$$\left(1+cos\,2x\right)dx$
$=\left[x+\frac{sin\,2x}{2}\right]^{-\pi/2}_{_{_{-3\pi/2}}}$
$=-\frac{\pi}{2}+\frac{3\pi}{2}=\pi$
$\Rightarrow I=\frac{\pi}{2}$