$(2014-$Intl$-$CalAB$-$MCQ$-8)$

Using the substitution $u=sin(2x),{\displaystyle \underset{\frac{}{6}}{\overset{\frac{}{2}}{}}}si{n}^5(2x)cos(2x)dx$ is equivalent to $($A$)-2{\displaystyle \underset{\frac{1}{2}}{\overset{1}{}}}{u}^{5}du$

$($B$)\frac{1}{2}{\displaystyle \underset{\frac{1}{2}}{\overset{1}{}}}{u}^{5}du$

$($C$)\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\sqrt[\phantom{2}]{3}}{2}}{}}}{u}^{5}du$

$($D$)\frac{1}{2}{\displaystyle \underset{\frac{\sqrt[\phantom{2}]{3}}{2}}{\overset{0}{}}}{u}^{5}du$

$($E$)2{\displaystyle \underset{\frac{\sqrt[\phantom{2}]{3}}{2}}{\overset{0}{}}}{u}^{5}du$

$\frac{dn}{dx}=cos(2x)$

$=>dx=\frac{dv}{cos(2x)}$