anonymous
  • anonymous
i'm unable to find the integral of sqrt(1+y'(x)^2) when y=ln(cosx)...please help
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
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
anonymous
  • anonymous
it goes to sqrt(1-tanx).....and this is where i have a problem.....
anonymous
  • anonymous
do u have any idea ???find the next term in the sequence (6,7,9,8,6,8,?)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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
anonymous
  • anonymous
That sequence is letters in the days of the week
dumbcow
  • dumbcow
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.