Question

${\int }_0^{\pi /2}{\cos }^5\left(\frac{x}{2}\right)\sin xdx$ is equal to

• A. $\frac{2}{7}\left(1-\frac{1}{8\sqrt{2}}\right)$
• B. $-\frac{4}{7}\left(1-\frac{1}{8\sqrt{2}}\right)$
• C. $\frac{4}{7}\left(1-\frac{1}{8\sqrt{2}}\right)$
• D. $-\frac{2}{7}\left(1-\frac{1}{8\sqrt{2}}\right)$

Answer 1

Answer: (C)

${\int }_0^{\pi /2}{\cos }^5\left(\frac{x}{2}\right)\sin xdx$
$={\int }_0^{\pi /2}{\cos }^5\left(\frac{x}{2}\right)\cdot 2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)dx$
$=2{\int }_0^{\pi /2}\sin \left(\frac{x}{2}\right){\cos }^6\left(\frac{x}{2}\right)dx$
By putting $\cos \frac{x}{2}=t$ , $-\frac{1}{2}\sin \frac{x}{2}dx=dt$
$\sin \frac{x}{2}dx=-2dt$
$\sin \frac{x}{2}dx=-2dt$
The given integral becomes
$2{\int }_0^{\pi /2}{t}^6(-2)dt$
$=-4{\int }_0^{\pi /2}{t}^{6}dt$
$=-4{\left[\frac{t^7}{7}\right]}_0^{\pi /2}$
$=-\frac{4}{7}{\left[{\cos }^7\left(\frac{x}{2}\right)\right]}_0^{\pi /2}$
$=\frac{-4}{7}\left[{\cos }^7\left(\frac{\pi }{4}\right)-{\cos }^7(0)\right]$
$=\frac{-4}{7}\left[{\left(\frac{1}{\sqrt{2}}\right)}^7-1\right]$
$=\frac{4}{7}\left[1-\frac{1}{8\sqrt{2}}\right]$