求∫（0-1）x^2根号下1-x^2dx
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∫(0->1)x^2√(1-x^2)dx
let
x= siny
dx = cosy dy
∫(0->1)x^2√(1-x^2)dx
=∫(0->π/2)(siny)^2(cosy)^2 dy
=(1/4)∫(0->π/2)(sin2y)^2 dy
=(1/8)∫(0->π/2) (1-cos4y) dy
=(1/8) [ y - sin(4y)/4 ](0->π/2)
=π/16
let
x= siny
dx = cosy dy
∫(0->1)x^2√(1-x^2)dx
=∫(0->π/2)(siny)^2(cosy)^2 dy
=(1/4)∫(0->π/2)(sin2y)^2 dy
=(1/8)∫(0->π/2) (1-cos4y) dy
=(1/8) [ y - sin(4y)/4 ](0->π/2)
=π/16
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