Answer
The length of the curve for $0 \le t \le \pi $ is
$s \simeq 6.0972$
Work Step by Step
We have
$x\left( t \right) = \sin 2t$, ${\ \ \ }$ $x'\left( t \right) = 2\cos 2t$,
$y\left( t \right) = 2\cos t$, ${\ \ \ }$ $y'\left( t \right) = - 2\sin t$.
By Theorem 1 of Section 12.2, the length of the curve for $0 \le t \le \pi $ is
$s = \mathop \smallint \limits_0^\pi \sqrt {{{\left( {2\cos 2t} \right)}^2} + {{\left( { - 2\sin t} \right)}^2}} {\rm{d}}t$
$s = \mathop \smallint \limits_0^\pi \sqrt {4{{\cos }^2}2t + 4{{\sin }^2}t} {\rm{d}}t$
Approximate it using a computer algebra system gives
$s \simeq 6.0972$
