# What is the arclength of f(t) = (sin^2t/sin(2t),tant-sec^2t) on t in [pi/12,pi/4]?

Jun 11, 2018

$\approx 0.5007429256$

#### Explanation:

We have

$x \left(t\right) = {\sin}^{2} \frac{t}{\sin} \left(2 t\right)$
then
$x ' \left(t\right) = \frac{1}{2 \cos \left(t\right)} ^ 2$

$y \left(t\right) = \tan \left(t\right) . \sec {\left(t\right)}^{2}$
then
$y ' \left(t\right) = - \frac{\cos \left(t\right) + 2 \sin \left(t\right)}{\cos} ^ 3 \left(t\right)$
so we get the integral

int_(pi/12)^(pi/4)sqrt((1/(2cos^2(t))^2+(-(cos(t)+2sin(t))/cos(t)^3)dt
which gives $\approx 0.5007429256$