Which of the following integrals represents the area of the surface obtained by rotating the curve $y=\sin^2(x),\;\;0 \le x \le \pi/2,$ about the $x$-axis?

A. $\displaystyle 2\pi \int_{0}^{\,\pi/2} {\sin^2(x)\sqrt{1+(2\sin(x))^2}}\,dx$
B. $\displaystyle 2\pi \int_{0}^{\,\pi/2} {x\sqrt{1+(\sin(x)\cos(x))^2}}\,dx$
C. $\displaystyle 2\pi \int_{0}^{\,\pi/2} {\sin^2(x)\sqrt{1+(2\sin(x)\cos(x))^2}}\,dx$
D. $\displaystyle 2\pi \int_{0}^{\,\pi/2} {x\sqrt{1+(2\sin(x)\cos(x))^2}}\,dx$
E. $\displaystyle 2\pi \int_{0}^{\,\pi/2} {x\sqrt{1+2\sin(x)\cos(x)}}\,dx$
F. $\displaystyle 2\pi \int_{0}^{\,\pi/2} {\sin^2(x)\sqrt{1+(\sin(x)\cos(x))^2}}\,dx$