Find arc length given $x=t\sint$ , $y=t\cost$ and $0\let\le1$ ?

Question

I got up to $\int_0^1\sqrt(1+t^2)dt$ using the parametric arc length formula $\color(indianred)(\int_a^b\sqrt((\dotx)^2+(\doty)^2)dt$ , but a solution I found online changes the bounds from $\int_0^1$ to $\color(palevioletred)(\int_0^(\tan^-1(1))$ ...

I assume the 0 is because $\tan^-1(0)=0$ , but why arctangent? Is it because $t=\tan\theta$ from trigonometric substitution?

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Answer 1 · 2018-05-28T11:58:12

$L=1/2(sqrt2+ln(1+sqrt2))$ units.

$x=tsint$
$x'=sint+tcost$

$y=tcost$
$y'=cost-tsint$

Arc length is given by:

$L=int_0^1sqrt((sint+tcost)^2+(cost-tsint)^2)dt$

Expand and simplify:

$L=int_0^1sqrt(1+t^2)dt$

Apply the substitution $t=tantheta$ :

$L=int_0^(tan^(-1)(1))sec^3thetad theta$

This is a known integral. If you do not have it memorized look it up in a table of integrals or apply integration by parts:

$L=1/2[secthetatantheta+ln|sectheta+tantheta|]_0^(tan^(-1)(1))$

Insert the limits of integration:

$L=1/2(sqrt2+ln(1+sqrt2))$