求不定积分:∫x^2dx⼀根号(a^2-x^2)=

求不定积分:∫x^2dx/根号(a^2-x^2)=
2024-12-27 13:24:16
推荐回答(1个)
回答1:

解:
令x=asint,则dx=acost dt
∫x²/√(a²-x²) dx
=∫a²sin²t/(acost)·acost dt
=a²∫sin²t dt
=a²∫(1-cos2x)/2 dt
=a²[t-1/4·sin2x]+C
=a²[arcsin(x/a)-1/2·x/a·√(1-x²/x²)]+C