Question

求不定积分dx/(a^2-x^2)^(3/2),其中a>0

求不定积分dx/(a^2-x^2)^(3/2),其中a>0
哪位高手给我动笔算一下,我要答案和过程
我想了两天了还是没做出来

Answer 1

[(a^2-x^2)^(-1/2)]'
=-1/2*(a^2-x^2)^(-3/2)*(-2x)
所以(a^2-x^2)^(-3/2)=[(a^2-x^2)^(-1/2)]'*1/x
用分部积分法
∫(a^2-x^2)^(-3/2)dx
=(a^2-x^2)^(-1/2)*1/x-∫(-1/x^2)*(a^2-x^2)^(-1/2)dx
这时可令x=asint
后一项=∫-1/[a^2*(sint)^2*cost]dsint
=-1/a^2∫(sint)^(-2)dt
再往后就好算了，自己算吧

Answer 2

设t=(a^2-x^2)^(1/2),式子=1/t^2dt,可以得出y=-1/t 把t代了 可得解

Answer 3

设x=atant 则dx=a(sect)^2dt

I=a(sect)^2dt/(a^2-(atant)^2)^(3/2)
=a(sect)^2dt/(asect)^3
=dt/a^2*sect
=costdt/a^2
=sint/a^2

Answer 4

是[(a^2-x^2)^(3/2)]^(-1)*dx吧
其实这题我猜是要用x=asint来带的
给个思路你试试

Answer 5

设x=asint 则dx=acostdt

I=acostdt/(a^2-(acost)^2)^(3/2)
=acostdt/(asin)^3
=d(sint)/a^2*(sint)^3
=(-1/2)*(sint)^(-2)/a^2
=-1/(2a^2*(sint)^2)