求不定积分,第二换元法根号(1-x^2)dx/x^2

Question

求不定积分,第二换元法
根号(1-x^2)dx/x^2

Answer 1

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.