r=5^theta theta goes from 0 to 2Pi
Length= integral between a and b of sqrt(r^2+(dr/dtheta)^2)dtheta
The Attempt at a Solution
r^2=25^theta or 5^(2theta) dr/dtheta=5^theta (ln 5) (dr/dtheta)^2=25^theta+10^theta (ln 5)+ (ln 5)^2
so the integral would be= sqrt(25^theta+25^theta+10^theta (ln 5)+ (ln 5)^2) dtheta
ive been working on this problem all day only to find no clue how to solve it,wolfram alpha keeps timing out and can't solve it,maple can't solve it
i also tried the length= integral(sqrt(dx/dtheta)^2+(dy/dtheta)^2) but both of these turn out to be ridiculously hard integrals to solve...can someone help please?