Calculus of a Single Variable, Chapter 10, 10.3, Section 10.3, Problem 48

Arc length of a curve C described by the parametric equations x=f(t) and y=g(t), a<=t<=b where f' and g' are continuous on [a,b] and C is traversed exactly once as t increases from a to b, then the length of the curve is given by,
L=int_a^bsqrt((dx/dt)^2+(dy/dt)^2)dt
We are given:x=arcsin(t),y=ln(sqrt(1-t^2)), 0<=t<=1/2
x=arcsin(t)
dx/dt=1/sqrt(1-t^2)
y=lnsqrt(1-t^2)
dy/dt=1/sqrt(1-t^2)d/dt(sqrt(1-t^2))
dy/dt=1/sqrt(1-t^2)(1/2)(1-t^2)^(1/2-1)(-2t)
dy/dt=-t/(1-t^2)
Now let's evaluate arc length by using the stated formula,
L=int_0^(1/2)sqrt((1/sqrt(1-t^2))^2+(-t/(1-t^2))^2)dt
L=int_0^(1/2)sqrt(1/(1-t^2)+t^2/(1-t^2)^2)dt
L=int_0^(1/2)sqrt((1-t^2+t^2)/(1-t^2)^2)dt
L=int_0^(1/2)sqrt(1/(1-t^2)^2)dt
L=int_0^(1/2)1/(1-t^2)dt
L=int_0^(1/2)1/((1+t)(1-t))dt
Using partial fractions integrand can be written as :
L=int_0^(1/2)1/2(1/(1+t)+1/(1-t))dt
Take the constant out and use the standard integral:int1/xdx=ln|x|+C
L=1/2int_0^(1/2)(1/(1+t)+1/(1-t))dt
L=1/2[ln|1+t|+ln|1-t|]_0^(1/2)
L=1/2{[ln|1+1/2|+ln|1-1/2|]-[ln1+ln1]}
L=1/2[ln|3/2|+ln|1/2|]
L=1/2[0.4054651081+0.69314718056]
L=1/2[1.09861228867]
L=0.54930614433
Arc length of the curve on the given interval is ~~0.549