Question

Evaluate the integral
${\int }_0^{{\pi }_{/}2}{\cos }^{5}x.\sin 2xdx$

• A. $2/7$
• B. $1/7$
• C. $-1/7$
• D. $3/7$

Answer 1

The correct option is A $2/7$

${\int }_0^{\frac{\pi }{2}}co{s}^{5}x\cdot sin2xdx={\int }_0^{\frac{\pi }{2}}co{s}^{5}x\cdot 2sinxcosxdx$

$=2{\int }_0^{\frac{\pi }{2}}co{s}^{6}xsinxdx=-2{\int }_0^{\frac{\pi }{2}}co{s}^{6}x\cdot d\left(cosx\right)$

$=-2{\int }_0^{\frac{\pi }{2}}co{s}^{6}xdcosx=-2\left(\frac{co{s}^{7}x}{7}\right){|}_0^{\frac{\pi }{2}}$

$=-2\left[\frac{co{s}^{7}x}{7}{|}_0^{\frac{\pi }{2}}\right]$

$=-2\left(\frac{-1}{7}\right)=\frac{2}{7}$

Thus ${\int }_0^{\frac{\pi }{2}}co{s}^{5}x\times sin2xdx=\frac{2}{7}$