anonymous
  • anonymous
i'm unable to find the integral of sqrt(1+y'(x)^2) when y=ln(cosx)...please help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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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,?)

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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

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