Which of the following integrals represents the area of the surface obtained by rotating the curve $y=\sec(x),\;\;0 \le x \le \pi/4,$ about the $y$-axis?

A. $\displaystyle 2\pi \int_{0}^{\,\pi/4} {x\sqrt{1+(\tan^2(x))^2}}\,dx$
B. $\displaystyle 2\pi \int_{0}^{\,\pi/4} {\sec(x)\sqrt{1+(\sec(x)\tan(x))^2}}\,dx$
C. $\displaystyle 2\pi \int_{0}^{\,\pi/4} {x\sqrt{1+(\sec(x)\tan(x))^2}}\,dx$
D. $\displaystyle 2\pi \int_{0}^{\,\pi/4} {\sec(x)\sqrt{1+(2\sec(x)\tan(x))^2}}\,dx$
E. $\displaystyle 2\pi \int_{0}^{\,\pi/4} {\sec(x)\sqrt{1+(\tan^2(x))^2}}\,dx$
F. $\displaystyle 2\pi \int_{0}^{\,\pi/4} {x\sqrt{1+(2\sec(x)\tan(x))^2}}\,dx$