Let:
{(x=x(t)),(y=y(t)):}
with t_1 <= t <= t_2 be the equation of a curve, the length of the element of the curve is:
dl = sqrt(dx^2+dy^2) = sqrt(x'(t)^2+y'(t)) dt
and so the length is calculated with the integral:
L = int_(t_1)^(t_2) sqrt(x'(t)^2+y'(t)) dt
In this case (exercise 43):
{(x(t) = tsint),(y(t) = tcost):}
with 0<=t<=1
{(x'(t) = sint+tcost),(y'(t) = cost-tsint):}
L=int_0^1 sqrt( ( sint+tcost)^2 + (cost-tsint)^2)dt
L=int_0^1 sqrt( sin^2t+cancel(2tsintcost)+t^2cos^2t + cos^2 -cancel(2tsintcost) +t^2sin^2t)dt
L=int_0^1 sqrt( (sin^2t+cos^2t +t^2(sin^2t+cos^2t))dt
L=int_0^1 sqrt( 1+t^2)dt
Substitute:
t= tan theta
dt = sec^2 theta d theta
and the limits of integration become theta = 0 and theta = pi/4:
L=int_0^(pi/4) sec^2 theta sqrt( 1+tan^2 theta )d theta
Using the trigonometric identity:
1+tan^2 theta = sec^2 theta
and considering that for theta in [0,pi/4] the secant is positive:
sqrt( 1+tan^2 theta ) = sec theta
so:
L=int_0^(pi/4) sec^3 theta d theta
Integrate by parts:
L=int_0^(pi/4) sec theta * sec^2 theta d theta
L=int_0^(pi/4) sec theta d ( tantheta)
L=[sec theta tantheta ]_0^(pi/4) - int_0^(pi/4) tan theta d(sec theta)
L=sqrt2 - int_0^(pi/4) tan^2 theta sec theta d theta
L=sqrt2 - int_0^(pi/4) (sec^2 theta - 1)sec theta d theta
L=sqrt2 - int_0^(pi/4) sec^3 thetad theta + int_0^(pi/4) sec theta d theta
L=sqrt2 - L + [ln abs(sec theta +tan theta)]_0^(pi/4)
2L=sqrt2 + ln abs(sqrt2 +1)
L = (sqrt2+ln(sqrt2+1))/2
