Question

数学微积分问题？

请问这道题（如图），如何用替换法x=coshu来解啊

Answer 1

用代换 x = coshu 解之：

Answer 2

let
x=secu
dx=secu.tanu du
∫x^5/√(x^2-1) dx
=∫ [(secu)^5/tanu] .[secu.tanu du]
=∫ (secu)^6 du
=(1/5)(secu)^4.tanu + (4/15)(secu)^2.tanu + (4/15)tanu + (4/15)u + C
=(1/5)x^4.√(x^2-1) + (4/15)x^2.√(x^2-1) + (4/15)√(x^2-1) + (4/15)arcsecx + C
//
I(2n)
=∫ (secu)^(2n) du
=∫ (secu)^(2n-2) dtanu
=(secu)^(2n-2) .tanu -(2n-2)∫ (secu)^(2n-2).(tanu)^2 du
=(secu)^(2n-2) .tanu -(2n-2)∫ (secu)^(2n-2).[(secu)^2 -1] du
(2n-1)I(2n)
=(secu)^(2n-2) .tanu +(2n-2)∫ (secu)^(2n-2) du
=(secu)^(2n-2) .tanu +(2n-2)I(2n-2)
I(2n) =[1/(2n-1)] +[(2n-2)/(2n-1)] I(2n-2)
ie
I6
=(1/5)(secu)^4.tanu + (4/5)I4
=(1/5)(secu)^4.tanu + (4/5)[ (1/3)(secu)^2.tanu + (2/3)I2]
=(1/5)(secu)^4.tanu + (4/15)(secu)^2.tanu + (8/15)I2
=(1/5)(secu)^4.tanu + (4/15)(secu)^2.tanu + (8/15)[ (1/2)tanu + (1/2)I0]
=(1/5)(secu)^4.tanu + (4/15)(secu)^2.tanu + (4/15)tanu + (4/15)I0
=(1/5)(secu)^4.tanu + (4/15)(secu)^2.tanu + (4/15)tanu + (4/15)u + C