15_3_6 - 6. is the same curve as in Exercise 5. We have 2 e...

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6. is the same curve as in Exercise 5. We have Z e z ds = Z 2 π 0 e t p 1 + 2 e 2 t dt Let 2 e t = tan θ 2 e t dt = sec 2 θ d θ = 1 2 Z t = 2 π t = 0 sec 3 θ d θ = 1 2 2 ± sec θ tan θ + ln | sec θ + tan θ | ² ³ ³ ³ ³ t = 2 π t = 0 = 2 e t 1 + 2 e 2 t + ln ( 2 e t + 1 + 2 e 2 t ) 2 2 ³ ³ ³ ³ 2 π 0 =
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