Question

求不定积分，第二换元法

根号(1-x^2)dx/x^2

Answer 1

x=sint t=arcsinx dx=cost dt
S根号(1-x^2)dx/x^2
=Scost *cost *dt /(sint)^2
=S(1-(sint)^2)/(sint)^2 *dt
=S(cot t)^2 dt-Sdt
=cot t-t-t+c
=cot t-2t+c
最后代入化简即可
=-√（1-x^2)/x-2arcsinx+C.

Answer 2

∫√(1-x^2)dx/x^2)
x=cosu dx=-sinudu
sinu=√(1-x^2)
tanu=√(1-x^2)/x
=∫-(sinu)^2du/(cosu)^2
=∫((cosu)^2-1)du/(cosu)^2
=u-tanu+C
=arccosx -√(1-x^2)/x +C

Answer 3

设x=sint,dx=costdt,
cost=√（1-x^2),
cott=√（1-x^2)/x
原式=∫cost*costdt/(sint)^2
=∫(cott)^2dt
=∫[(csct)^2-1]dt
=-cott-t+C
=-√（1-x^2)/x-arcsinx+C.