Question

${\int }_0^{\pi /2}\sqrt{\cos x}{\sin }^{5}xdx=$

• A. $\frac{64}{231}$
• B. $\frac{30}{321}$
• C. $\frac{34}{231}$
• D. $\frac{128}{231}$

Answer 1

Answer: (A)

${\int }_0^{\frac{\pi }{2}}\sqrt{cosx}si{n}^{5}xdx$
${\int }_0^{\frac{\pi }{2}}\sqrt{cosx}\left(si{n}^{4}x\right)sinxdx$
$-{\int }_0^{\frac{\pi }{2}}\sqrt{cosx}(1-co{s}^{2}x{)}^{2}dcosx$
$-[{\int }_0^{\frac{\pi }{2}}\sqrt{cosx}dcosx+{\int }_0^{\frac{\pi }{2}}\sqrt{cosx}co{s}^{4}xdx-2{\int }_0^{\frac{\pi }{2}}\sqrt{cosx}co{s}^{2}xdx]$
$-[\frac{2}{3}co{s}^{\frac{3}{2}}x{\int }_0^{\frac{\pi }{2}}+\frac{2}{11}co{s}^{\frac{11}{2}}x{\int }_0^{\frac{\pi }{2}}-2(\frac{2}{7}\left)co{s}^{\frac{7}{2}}x{\int }_0^{\frac{\pi }{2}}\right]$
$=-\left[\frac{2}{3}\right(-1)+\frac{2}{11}(-1)-2(\frac{2}{7}\left)\right(-1\left)\right]$
$=\frac{64}{231}$