Question

${\int }_{-\pi /2}^{\pi /2}\sqrt{cosx-co{s}^{3}x}d{x}_{=}$

• A. $1$
• B. $\frac{4}{3}$
• C. $\frac{-1}{3}$
• D. $0$

Answer 1

Answer: (B)

$\frac{4}{3}$

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\sqrt{cosx-co{s}^{3}xdx}$
${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\sqrt{cosx}|sinx|dx$
$-{\int }_{\frac{-\pi }{2}}^0\sqrt{cosx}sinxdx+{\int }_0^{\frac{\pi }{2}}\sqrt{cosx}sinxdx$
${\int }_{\frac{-\pi }{2}}^0\sqrt{cosx}d\left(cosx\right)-{\int }_0^{\frac{-\pi }{2}}\sqrt{cosx}d\left(cosx\right)$
$\frac{2}{3}co{s}^{\frac{3}{2}}x{\int }_{\frac{-\pi }{2}}^0-\frac{2}{3}co{s}^{\frac{3}{2}}x{\int }_0^{\frac{\pi }{2}}$
$\frac{2}{3}\left[\right(1-0)-(0-1\left)\right]$
$=\frac{4}{3}$