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#
