Set up, but do not evaluate, an integral for the length of the curve.
$$y = 6 e ^ { x } \sin x , \quad 0 \leq x \leq \frac { 9 \pi } { 2 }$$
A)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 - 36 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$
B)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$
C)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 - 36 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
D)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
E)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 - \sin x ) } d x$

Question

Quizplus · Accepted Answer

The Answer of Set up, but do not evaluate, an integral for the length of the curve.
$$y = 6 e ^ { x } \sin x , \quad 0 \leq x \leq \frac { 9 \pi } { 2 }$$
A)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 - 36 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$
B)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$
C)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 - 36 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
D)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
E)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 - \sin x ) } d x$

Set Up, but Do Not Evaluate, an Integral for the Length