Question

求不定积分x^2√a^2-x^2

Answer 1

∫x²*√(a²-x²)dx
令x=asint，那么dx=d(asint)=acostdt
∴原式=∫(asint)²*√[a²-(asint)²]*acostdt
=∫a²sin²t*acost*acostdt
=∫a^4*sin²t*cos²tdt
=∫a^4*(sintcost)²dt
=∫1/4*a^4*sin²2tdt
=∫1/8*a^4*(1-cos4t)dt
=1/8*a^4*t-1/32*a^4*sin4t+C
而∵x=asint，∴sint=x/a，∴cost=[√(a²-x²)]/a
∴sin2t=2sintcost=[2x√(a²-x²)]/a²
cos2t=cos²t-sin²t=(a²-2x²)/a²
∴sin4t=2sin2tcos2t=[4x(a²-2x²)√(a²-x²)]/a^4
∴原式=1/8*a^4*arcsin(x/a)-1/8*x(a²-2x²)√(a²-x²)+C

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Answer 2

∫x^2dx/√（a^2-x^2)
=-∫(a^2-x^2)dx/√（a^2-x^2)+∫a^2dx/√（a^2-x^2）
=-∫√（a^2-x^2)dx+a^2∫dx/√（a^2-x^2）
对∫√（a^2-x^2)，设x=asint,dx=acostdt,t=arcsin(x/a),
sin2t=2sintcost=2(x/a)*√（1-x^2/a^2)=2(x/a^2)√（a^2-x^2)
∫√（a^2-x^2)=∫acost*acostdt=(a^2/2)∫(1+cos2t)dt=a^2t/2+a^2sin2t/4+C1
=a^2arcsin(x/a)/2+x√（a^2-x^2)/2+C1
∴原式=-a^2arcsin(x/a)/2-x√（a^2-x^2)/2+a^2arcsinx/a+C.
=a^2arcsin(x/a)/2-x√（a^2-x^2)/2+C.
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