第1个回答 2020-02-15
练习4. 令 x = 4sint， 则 dx = 4costdt
I = ∫4cost 4costdt/(4sint) = 4∫[(cost)^2/sint]dt = 4∫[sint(cost)^2/(sint)^2]dt
= 4∫{-(cost)^2/[1-(cost)^2]}dcost = 4∫{-1-1/[1-(cost)^2]}dcost
= -4cost - 2∫[1/(1-cost)+1/(1+cost)]dcost
= -4cost - 2ln[(1+cost)/(1-cost)] + C
cost = (1/4)√(16-x^2) 代入即得。
练习5. 令 x = atant， 则 dx = a(sect)^2dt
I = ∫ a(sect)^2dt/[a^3(sect)^3] = (1/a^2)∫costdt = (1/a^2)sint + C
= (1/a^2)x/√(a^2+x^2) + C本回答被网友采纳
I = ∫4cost 4costdt/(4sint) = 4∫[(cost)^2/sint]dt = 4∫[sint(cost)^2/(sint)^2]dt
= 4∫{-(cost)^2/[1-(cost)^2]}dcost = 4∫{-1-1/[1-(cost)^2]}dcost
= -4cost - 2∫[1/(1-cost)+1/(1+cost)]dcost
= -4cost - 2ln[(1+cost)/(1-cost)] + C
cost = (1/4)√(16-x^2) 代入即得。
练习5. 令 x = atant， 则 dx = a(sect)^2dt
I = ∫ a(sect)^2dt/[a^3(sect)^3] = (1/a^2)∫costdt = (1/a^2)sint + C
= (1/a^2)x/√(a^2+x^2) + C本回答被网友采纳
第2个回答 2020-02-15
练习4，再采用降幂法，将
16cos^2X化为8(1+cos2X)
16cos^2X化为8(1+cos2X)
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