Question

求不定积分，换元法，如图

Answer 1

令x=sint
1-x²=1-sin²t=cos²t
dx=costdt
原式=∫costdt/cos^4 t
=∫dt/cos³t
=∫sec³dx
=∫sectdtant
=secttant-∫tantdsect
=secttant-∫(sec²t-1)sectdt
=secttant+∫sectdt-∫sec³tdt
2∫sec³tdt=secttant+∫sectdt=secttant+ln|sect+tant|
∫sec³tdt=(1/2)secttant+(1/2)ln|sect+tant|+C
x=sint
cost=√(1-x²)
sect=1/cost=1/√(1-x²)
tant=sint/cost=x/√(1-x²)
把这些再代入上式即可。