With cal 2 exercise involving integrals?
i have to find the arc length of y = lnx given that 1 ≤ x ≤ sqrt (3)
here is my procedure:

i found the derivative of y = lnx
and got, y' = 1 / (x)
1 + (1/(x))^2 = 1 + (1/(x^2)) = (x^2+1)/(x^2)
integral from 1 to sqrt 3 of sqrt((x^2+1/(x^2)) dx = sqrt ( x ^ 2 + 1 ) / ( x ) dx
x = tan b
dx = (sec^2)bdb
sqrt(x^2+1) = sqrt ( (tan^2)b + 1) = sqrt ( (sec^2)b) = secb

integral = ((sec)/(tanb)) times (sec^2)bdb

and here comes the part i don't understand, in my exercise it becomes

integral = ((sec^2)b)/((tan^2)b)) times secbtanbdb

why did this happen, me figure out how.