Question

求解释打问号的式子

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Answer 1

x-a/2=a/2sint
x=a/2(1+sint)
dx=a/2costdt
t=arcsin(2x/a-1)
t1=arcsin(0-1)
=-π/2
t2=arcsin(2a/a-1)
=arcsin1
=π/2
x^2=a^2/4(1+sint)^2
∫(0,a)x√[x/(a-x)]dx
=∫(0,a)x^2/√[(a/2)^2-(x-a/2)^2]dx
=∫(-π/2,π/2)[a^2/4(1+sint)^2]/[a/2√(1-sin^2t)]×(a/2cost)dt
=a^2/4∫(-π/2,π/2)(1+sint)^2/(a/2cost)×(a/2cost)dt
=a^2/4∫(-π/2,π/2)(1+sint)^2dt
注：^2——表示平方。