求不定积分∫x^2/√(a^2-x^2)dx=?
求不定积分∫x^2/√(a^2-x^2)dx=?
用分部积分法
∫x^2/√(a^2-x^2)dx=x^2*arcsin(x/a)-∫2x/√(a^2-x^2)dx=x^2*arcsin(x/a)-2xarcsin(x/a)+2arcsin(x/a)
=(x^2-2x+2)arcsin(x/a)+C
这样做有什么不对？
用分部积分法
∫x^2/√(a^2-x^2)dx=x^2*arcsin(x/a)-∫2x/√(a^2-x^2)dx=x^2*arcsin(x/a)-2xarcsin(x/a)+2arcsin(x/a)
=(x^2-2x+2)arcsin(x/a)+C
这样做有什么不对？
数学人气：457 ℃时间：2020-03-27 09:02:03
优质解答
令x=asint,则dx=acost dt ∫x²/√(a²-x²) dx=∫a²sin²t/(acost)·acostdt=a²∫sin²t dt=a²∫(1-cos2t)/2 dt=a²∫1/2dt-a²∫cos2tdt=a²t/2-1/2·a²sin2t+C=1/2·a²arcsin(x/a)-x·√(a²-x²)+C
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