Question

求不定积分:∫x^2dx/根号(a^2-x^2)=

求不定积分:∫x^2dx/根号(a^2-x^2)=

Answer 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

追问

答案是a^2/2(arcsin(x/a)-x√(a^2-x^2)/a^2)+c.....刚忘记发了。。。我就是用你的方法来做的然后做到a²∫(1-cos2x)/2 dt 再往后做就做不对了= =！

追答

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/2-1/4·sin2x]+C
=a²[arcsin(x/a)/2-1/2·x/a·√(1-x²/a²)]+C
=a²/2·arcsinx/a-1/2·x√(a²-x²)+C