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求定积分(sinx)^3/[1+sinx]dx 上限4/派 下限0

发布时间:2019-07-07 11:09

∫[0:2][x³/(4+x²)]dx
=∫[0:2][(x³+4x-4x)/(4+x²)]dx
=∫[0:2][x -4x/(4+x²)]dx
=∫[0:2]xdx -2∫[0:2][2x/(4+x²)]dx
=½x²|[0:2]-2∫[0:2][1/(4+x²)]d(4+x²)
=½(2²-0²)-2ln|4+x²||[0:2]
=2-2(ln|4+2²|-ln|4+0²|)
=2-2(ln8-ln4)
=2-2(3ln2-2ln2)
=2-2ln2

回复:

∫[1→√3] 1/[x²√(1+x²)] dx
令x=tanu,则√(1+x²)=secu,dx=sec²udu,u:π/4→π/3
=∫[π/4→π/3] [1/(tan²usecu)](sec²u) du
=∫[π/4→π/3] secu/tan²u du
=∫[π/4→π/3] cosu/sin²u du
=∫[π/4→π/3] 1/sin²u dsinu
=-1/sinu ||[π/4→π/3]
=√2 - 2/√3

回复:

令t=√x换元即可

回复:

∫[0:2][x³/(4+x²)]dx =∫[0:2][(x³+4x-4x)/(4+x²)]dx =∫[0:2][x -4x/(4+x²)]dx =∫[0:2]xdx -2∫[0:2][2x/(4+x²)]dx =½x²|[0:2]-2∫[0:2][1/(4+x²)]d(4+x²) =½(2²-0²)-2ln...

回复:

∫√(16-x²)dx =16∫√(1-sin²t)dsint =16∫cos²tdt =8∫1+cos2tdt =8t+4sin2t+C =8arcsin(x/4)+8x√(16-x²)+C

回复:

计算过程如下: ∫[0~π/2]sinφcos∧3φdφ =-∫[0~π/2]cos∧3φdcosφ =-(1/4)cos^4φ(0~π/2) =1/4

回复:

∫(π/4,π/3) 1/(sin²xcos²x) dx = ∫(π/4,π/3) 1/[(1/2)sin2x]² dx = ∫(π/4,π/3) 4csc²2x dx 令u = 2x,du = 2 dx = 4(1/2)∫(π/2,2π/3) csc²u du = 2[- cotu] |(π/2,2π/3) = - 2[cot(2π/3) - cot(π/2)] = 2/√3

回复:

∫[1→√3] 1/[x²√(1+x²)] dx 令x=tanu,则√(1+x²)=secu,dx=sec²udu,u:π/4→π/3 =∫[π/4→π/3] [1/(tan²usecu)](sec²u) du =∫[π/4→π/3] secu/tan²u du =∫[π/4→π/3] cosu/sin²u du =∫[π/4→π/3] 1/sin²u d...

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