Question

求解∫√x²-a²/x dx 谢谢

Answer 1

∫x^2.√(a^2-x^2) dx
let
x= asiny
dx=acosy dy

∫x^2.√(a^2-x^2) dx
=a^4∫(siny)^2.(cosy)^2 dy
=(a^4/4)∫(sin2y)^2 dy
=(a^4/8)∫ (1- cos4y) dy
=(a^4/8) [ y- (1/4)sin4y ] + C
=(a^4/8) { arcsin(x/a) - (x/a) .(√(a^2-x^2)/a) [ 1- 2(x/a)^2 ] } + C

where
(1/4)sin4y
=(1/2) sin2y. cos2y
=(1/2) [ 2siny. cosy ( 1- 2(siny)^2 ) ]
=siny. cosy ( 1- 2(siny)^2 ) ]
=(x/a) .(√(a^2-x^2)/a) [ 1- 2(x/a)^2 ]