求定积分∫（0，1）√1－x²dx
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令x=sint ,dx=costdt 当x=0时,t=0,当x=1时,t=π/2
∫（0,1）根号下[1-x^2]dx
= ∫(0,π/2)√(1-sin²t)costdt
= ∫(0,π/2)√(1-sin²t)costdt
=∫(0,π/2)cos²tdt
=∫(0,π/2) [(cos2t +1)/2]dt
=1/2∫(0,π/2)cos2t+1dt
=1/2[1/2∫(0,π/2)cos2td2t+∫(0,π/2)dt]
=1/2[1/2(sinπ-sin0)+(π/2-0)]
=1/2×π/2
=π/4
∫（0,1）根号下[1-x^2]dx
= ∫(0,π/2)√(1-sin²t)costdt
= ∫(0,π/2)√(1-sin²t)costdt
=∫(0,π/2)cos²tdt
=∫(0,π/2) [(cos2t +1)/2]dt
=1/2∫(0,π/2)cos2t+1dt
=1/2[1/2∫(0,π/2)cos2td2t+∫(0,π/2)dt]
=1/2[1/2(sinπ-sin0)+(π/2-0)]
=1/2×π/2
=π/4
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