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?
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![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?](/uploads/image/z/5244634-10-4.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%ABx%5E2%2F%E2%88%9A%28a%5E2-x%5E2%29dx%3D%3F%E7%94%A8%E5%88%86%E9%83%A8%E7%A7%AF%E5%88%86%E6%B3%95%E2%88%ABx%5E2%2F%E2%88%9A%28a%5E2-x%5E2%29dx%3Dx%5E2%2Aarcsin%28x%2Fa%29-%E2%88%AB2x%2F%E2%88%9A%28a%5E2-x%5E2%29dx%3Dx%5E2%2Aarcsin%28x%2Fa%29-2xarcsin%28x%2Fa%29%2B2arcsin%28x%2Fa%29%3D%28x%5E2-2x%2B2%29arcsin%28x%2Fa%29%2BC%E8%BF%99%E6%A0%B7%E5%81%9A%E6%9C%89%E4%BB%80%E4%B9%88%E4%B8%8D%E5%AF%B9%EF%BC%9F)
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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?
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?
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?
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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