Question

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

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

Answer 1

Answer: (A)

$\left({\pi }^4/32\right)+\left(\pi /2\right)$

Let $I={\int }_{\frac{-3\pi }{2}}^{\frac{-\pi }{2}}[{(x+\pi )}^3+{\cos }^2(x+3\pi )]dx$

Put $x+\pi =z$

Also,$x=\frac{-\pi }{2}\Rightarrow x=\frac{\pi }{2}$ and $x=\frac{-3\pi }{2}\Rightarrow z=\frac{-\pi }{2}$

$\therefore dx=dz$ and $x+3\pi =z+2\pi$

$\therefore I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}[z^3+{\cos }^2(2\pi +z)]dz$

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{z}^{3}dz+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\cos }^2(2\pi +z)dz$

$=0\left(anoddfunction\right)+2{\int }_0^{\frac{\pi }{2}}{\cos }^{2}zdz$

$=0+2\times \frac{1}{2}\times \frac{\pi }{2}$ using fact ${\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx=\{\begin{array}{c}\frac{n-1}{n}.\frac{n-3}{n-2}...\frac{1}{2}.\frac{\pi }{2}ifn=2m\\ \frac{n-1}{n}.\frac{n-3}{n-2}...\frac{2}{2}ifn=2m+1\end{array}$

$=\frac{\pi }{2}$