Answer
$L$ = $\frac{\sqrt 2}{2}+\frac{1}{2}{\ln}(1+\sqrt 2)$
Work Step by Step
$x$ = $t\sin t$
$y$ = $t\cos t$
$\frac{dx}{dt}$ = $t\cos t+\sin t$
$\frac{dy}{dt}$ = $-t\sin t+\cos t$
$\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2$ = $(t\cos t+\sin t)^2+(-t\sin t+\cos t)^2$ = $t^2\cos^2t+2t\cos t\sin t+\sin^2t+t^2\sin^2t-2t\sin t\cos t+\cos^2t$ = $t^2+1$
$L$ = $\int_0^1{\sqrt {t^2+1}}dt$
$L$ = $\left[\frac{1}{2}t\sqrt {t^2+1}+\frac{1}{2}{\ln}({t+\sqrt {t^2+1}})\right]_0^1$
$L$ = $\frac{\sqrt 2}{2}+\frac{1}{2}{\ln}(1+\sqrt 2)$
