costdcost不定积分
costdcost 不定积分 -
不定积分=x^4/8-(4x^3sin2x+6x^2cos2x-6xsin2x-3cos2x)/16 代入上下限x=2pi可得2pi^4-(24pi^2-3)/16 =2pi^4-(3/2)pi^2+3/16 x=0可得0-(0+0-0-3)/16 =3/16 相减可得积分=2pi^4-(3/2)pi^2 其中pi=3.1415.
令x=cost 则原式=∫√(1-cost)(1+cost)dcost =∫√(1-cos^2t)(1+cost)^2dcost ∫-sin^2t/(1+cost)dt =∫(cos^2t-1)(1+cost)dt =∫cost-1dt =-sint-t+c =-√(1-x^2) - arccost + c
不定积分=x^4/8-(4x^3sin2x+6x^2cos2x-6xsin2x-3cos2x)/16 代入上下限x=2pi可得2pi^4-(24pi^2-3)/16 =2pi^4-(3/2)pi^2+3/16 x=0可得0-(0+0-0-3)/16 =3/16 相减可得积分=2pi^4-(3/2)pi^2 其中pi=3.1415.
令x=cost 则原式=∫√(1-cost)(1+cost)dcost =∫√(1-cos^2t)(1+cost)^2dcost ∫-sin^2t/(1+cost)dt =∫(cos^2t-1)(1+cost)dt =∫cost-1dt =-sint-t+c =-√(1-x^2) - arccost + c
还有什么疑问吗?
原式=coste^cost-∫e^costdcost =coste^cost-e^cost+c
求一道不定积分题的解答 -
详情如图所示有任何疑惑,欢迎追问,
∫tantsin²tdt=-∫sin²t/cost d(cost)=-∫(1/cost - cost)d(cost) = - (ln|cost| - 1/2 * cos²t) + c = 1/2 * cos²t - ln|cost| + c, 望采纳,
不定积分求解谢谢 -
主要步骤如下:∫sinxdx/cosx√1+sin^2x =-∫dcosx/cosx√2-cos^2x(设√2cosx=cost),则:-∫(1/√2)dcost/[(cost/√2)√2-2cos^2t]=-∫(1/√2)dcost/[(cost/√2)√2sint]=∫√2dt/(cost)=∫√2dsint/(cost)^2 =∫√2dsint/[(1-sint)(1+sint)]=(√2/2)*ln(1+是什么。
= -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 + 后面会介绍。
高数不定积分 -
利用二倍角公式,
利用分部积分与凑微分,过程与结果如图所示,