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

∫[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

令t=√x换元即可

∫[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...

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

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

∫(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

∫[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...