两道丧心病狂的高数题.....

Answer 1

1、原式=∫[√(x^2+1)+√(x^2-1)]/(x^2+1-x^2+1)dx
=(1/2)*∫[√(x^2+1)+√(x^2-1)]dx
=(1/2)*[(x/2)*√(x^2+1)+(1/2)*ln|x+√(x^2+1)|+(x/2)*√(x^2-1)-(1/2)*ln|x+√(x^2-1)|]+C，其中C是任意常数
2、原式=(1/2)*∫2sinxcosx/(sinx+cosx)dx
=(1/2)*∫[sin^2x+2sinxcosx+cos^2x-1]/(sinx+cosx)dx
=(1/2)*∫(sinx+cosx)^2/(sinx+cosx)dx-(1/2)*∫dx/(sinx+cosx)
=(1/2)*∫(sinx+cosx)dx-(√2/4)*∫dx/sin(x+π/4)
=(1/2)*(sinx-cosx)-(√2/4)*∫csc(x+π/4)d(x+π/4)
=(1/2)*(sinx-cosx)-(√2/4)*ln|csc(x+π/4)-cot(x+π/4)|+C，其中C是任意常数

Answer 2

(1)
∫dx/[√(x^2+1)-√(x^2-1)]
=(1/2)∫[√(x^2+1)+√(x^2-1)] dx

let
x=tany
dx = (secy)^2 dy

∫√(x^2+1) dx
=∫(secy)^3 dy
=∫secy dtany
=secy.tany -∫ secy (tany)^2 dy
=secy.tany -∫ secy [(secy)^2 -1] dy
2∫(secy)^3 dy =secy.tany +∫ secydy
∫(secy)^3 dy =(1/2) [secy.tany +ln|secy+tany| ] + C'
=(1/2) [x.√(1+x^2) +ln|√(1+x^2)+x| ] + C'

let
x= secz
dx= secz.tanz dz

∫√(x^2-1) dx
=∫ secz.(tanz)^2 dz
=∫ secz.[(secz)^2-1] dz
=(1/2) [secz.tanz +ln|secz+tanz| ] -(1/2) [ln|secz+tanz| ] +C''
=(1/2)secz.tanz +C''
=(1/2)x.√(x^2-1) + C''

∫dx/[√(x^2+1)-√(x^2-1)]
=(1/2)∫[√(x^2+1)+√(x^2-1)] dx
=(1/4)[x.√(1+x^2) +ln|√(1+x^2)+x| +x.√(x^2-1) ] + C

(2)
(2)
∫ sinx.cosx/(sinx+cosx) dx
=(1/2)∫ (sinx+ cosx)^2/(sinx+cosx) dx - (1/2)∫ dx/(sinx+cosx)
=(1/2)(-cosx + sinx) -∫ dx/(sinx+cosx)
=(1/2)(-cosx + sinx) -(1/√2)∫ dx/sin(x+π/4)
=(1/2)(-cosx + sinx) -(1/√2)∫ csc(x+π/4) dx
=(1/2)(-cosx + sinx) -(1/√2)ln|csc(x+π/4) -cot(x+π/4)| +C

追问

求∫√(x^2-1) dx
=∫ secz.(tanz)^2 dz
=∫ secz.[(secz)^2-1] dz
=(1/2) [secz.tanz +ln|secz+tanz| ] -(1/2) [ln|secz+tanz| ] +C'' 这里的后半部分为什么有个系数1/2呢。。

追答

你对
∫√(x^2-1) dx
=∫ secz.(tanz)^2 dz
=∫ secz.[(secz)^2-1] dz
=∫ (secz)^3 dz -∫(secz) dz
=(1/2) [secz.tanz -ln|secz+tanz| ] + C''

consider
∫ (secz)^3 dz =(1/2) [secz.tanz +ln|secz+tanz| ]
∫(secz) dz = ln|secz+tanz|

追问

这样算来就不和你的第一次过程相同了，应该这个第二次才是正确的吧。还有第二道题，
∫ sinx.cosx/(sinx+cosx) dx
=(1/2)∫ (sinx+ cosx)^2/(sinx+cosx) dx - (1/2)∫ dx/(sinx+cosx)
=(1/2)(-cosx + sinx) -∫ dx/(sinx+cosx)
第三步后半步是不是把系数1/2弄掉了。

追答

不用, 不好意思运算太差
(1/2)(sinx+cosx)^2
=(1/2) (1 + 2sinxcosx)
= sinx.cosx + 1/2
sinx.cosx =(1/2)(sinx+cosx)^2 - 1/2

∫ sinx.cosx/(sinx+cosx) dx
=(1/2)∫ (sinx+ cosx)^2/(sinx+cosx) dx - (1/2)∫ dx/(sinx+cosx)
=(1/2)(-cosx + sinx) -(1/2)∫ dx/(sinx+cosx)
=(1/2)(-cosx + sinx) -(√2/4)∫ dx/sin(x+π/4)
=(1/2)(-cosx + sinx) -(√2/4)∫ csc(x+π/4) dx
=(1/2)(-cosx + sinx) -(√2/4)ln|csc(x+π/4) -cot(x+π/4)| +C