Question

i'm unable to find the integral of sqrt(1+y'(x)^2) when
y=ln(cosx)...please help

Answer 1

Start with the derivative of y, sin/cos, then use trig identity to get the two term into one. You will be left taking the integral of one trig function

Answer 2

it goes to sqrt(1-tanx).....and this is where i have a problem.....

Answer 3

do u have any idea ???find the next term in the sequence (6,7,9,8,6,8,?)

Answer 4

You forgot to square it after you ploged it in
One minus tan squared is sec squared I think, take the root and then the integral of sec

Answer 5

That sequence is letters in the days of the week

Answer 6

y' = -sin/cos
y' ^2 = sin^2/cos^2 = tan^2
trig identity
sec^2 = 1+tan^2
sqrt(1+y'^2) = sqrt(sec^2) = sec
integral sec = ln(sec+tan) +c