L = ⌡ √ 1+[ f ' (x) ]² this breaks down to ⌡ √ 1 + y² dx I was wondering if I could use trig identities. I get it to sec(x)^3; the integral of that is ln(sec(x) + tan(x) but subing back in I'm not getting the correct answer.
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The integral of sec^3(x) is not ln|sec(x) + tan(x)|, check again... Use integration by parts making I = the integral of sec^3(x) and integrate twice so you can solve algebraically for I.
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Anytime.. No problem.
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