令 x = sin u,dx = cos u du
原积分 = ∫cosudu/sinu×cosu
= ∫du/sinu
= ∫sinudu/sin²u
=-∫dcosu/(1+cosu)(1-cosu)
=-½∫[1/(1+cosu) + 1/(1-cosu)]dcosu
=-½[ln|1+cosu| - ln|1-cosu|] + C
=-½[ln|(1+cosu)/(1-cosu)| + C
= ½[ln|(1-cosu)/(1+cosu)| + C
= ½[ln|[1-√(1-x²)]/[1+√(1-x²)]| + C
= ½[ln|[1-√(1-x²)]²/x²| + C
= [ln|[1-√(1-x²)]/x| + C