定积分∫(上限π/3,下限π/4)x/(sin^2x)dx 答案（1/4-√3/9)π+1/2ln2/3
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∫(π/4→π/3) x/sin²x dx
= ∫(π/4→π/3) xcsc²x dx
= ∫(π/4→π/3) x d(- cotx)
= - xcotx + ∫(π/4→π/3) cotx dx
= - xcotx + ∫(π/4→π/3) 1/sinx d(sinx)
= - xcotx + ln(sinx) |(π/4→π/3)
= (- π/3)cot(π/3) + ln(sin(π/3)) - [(- π/4)cot(π/4) + ln(sin(π/4))]
= (1/36)(9 - 4√3)π + (1/2)ln(3/2) ≈ 0.383531
我的答案是对的，验算如下。
http://www.wolframalpha.com/input/?i=int_%28pi%2F4%29%5E%28pi%2F3%29+x%2Fsin%5E2x+dx
= ∫(π/4→π/3) xcsc²x dx
= ∫(π/4→π/3) x d(- cotx)
= - xcotx + ∫(π/4→π/3) cotx dx
= - xcotx + ∫(π/4→π/3) 1/sinx d(sinx)
= - xcotx + ln(sinx) |(π/4→π/3)
= (- π/3)cot(π/3) + ln(sin(π/3)) - [(- π/4)cot(π/4) + ln(sin(π/4))]
= (1/36)(9 - 4√3)π + (1/2)ln(3/2) ≈ 0.383531
我的答案是对的，验算如下。
http://www.wolframalpha.com/input/?i=int_%28pi%2F4%29%5E%28pi%2F3%29+x%2Fsin%5E2x+dx
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