Given f(x) = y = log(s∵ ec(x)) on [0,4π]
The arc length is given by L = ∫ab(f′(x))2+1dx
we have f(x) = log(sec(x))
⟹ f'(x) = sec(x)1⋅ (sec(x) ⋅ tan(x)) = tan(x)
∴ L = ∫04π(tan(x))2+1dx
= ∫04πtan2(x)+1dx
= ∫04πsec2(x)dx ( ∵ sec2(x) - tan2(x) = 1)
= ∫04πsec(x)dx
= [ log( sec(x) +tan(x) )]∣04π
= [log(tan(4π) + sec(4π)] - [log(tan(0) + sec(0)]
= [log(1+ 2 )] - [log(1)]
∴ L = log(1+ 2 )
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