Question

不定积分∫√(x²-a²)dx可以求吗？？怎么求？？

Answer 1

令x=asect,dx=sect tant dt
原式=∫atant*d(asect)
=a^2∫tant sect tant dt (1)
=a^2∫tant d sec t
=a^2tant sect -a^2∫sect sec^2 t dt
=a^2tant sect -a^2∫ sect(tan^2 t+1) dt
=a^2tant sect -a^2∫sect dt -a^2∫tant sect tant dt (2)
联立(1),(2)
所以
a^2∫tant sect tant dt=a^2/2[tant sect -∫sect dt]
=a^2/2[tant sect- ln|sect+tan t| ]+C
=a^2/2*x√(x²-a²)-a^2/2ln|x+√(x²-a²)|+c

Answer 2

可以，取x=asect,代换整理求导即可，很简单。

Answer 3

x=asect,t=arccosa/x,dx=a(sect)^2dt
原积分=Satant*a(sect)^2dt=a^2*Stantdtant=1/2*a^2*(tant)^2+c=1/2*x^2-1/2*a^2+c