Yahoo Answers: Answers and Comments for ∫(1→9) 1 / (1+√t )dt [數學]
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From 阿阿 黃
zhHantTW
Mon, 19 May 2014 00:38:48 +0000
3
Yahoo Answers: Answers and Comments for ∫(1→9) 1 / (1+√t )dt [數學]
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https://tw.answers.yahoo.com/question/index?qid=20140519000010KK00387
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From 少年時:
令 t = tan^4 θ，所以 dt = 4 tan^3 θ sec^2 θ dθ
當 ...
https://tw.answers.yahoo.com/question/index?qid=20140519000010KK00387
https://tw.answers.yahoo.com/question/index?qid=20140519000010KK00387
Mon, 19 May 2014 11:22:07 +0000
令 t = tan^4 θ，所以 dt = 4 tan^3 θ sec^2 θ dθ
當 t = 1 時，θ = π/4；當 t = 9 時，θ = π/3；所以
f(1)
= ∫(1→9) 1 / (1 + √t) dt
= ∫(π/4→π/3) 4 tan^3 θ sec^2 θ / (1 + tan^2 θ) dθ
= ∫(π/4→π/3) 4 tan^3 θ dθ
= ∫(π/4→π/3) 4 tan θ sec^2 θ dθ  ∫(π/4→π/3) 4 tan θ dθ
= ∫(π/4→π/3) 4 tan θ d(tan θ)  4 ∫(π/4→π/3) tan θ dθ
= (π/4→π/3) [2 tan^2 θ  4 ln(tan θ + sec θ)]
= [6  4 ln(2 + √3)]  [2  4 ln(1 + √2)]
= 4 + 4 ln[(1 + √2)/(2 + √3)]