∫cosx/(sinx^3+sinx)dx在π/6到π/2上的积分
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令 sinx = u， 则
∫<π/6, π/2>[cosx/(sinx^3+sinx)]dx
= ∫<π/6, π/2>dsinx/(sinx^3+sinx)
= ∫<1/2, 1>du/(u^3+u) = ∫<1/2, 1>[1/u - u/(u^2+1)du
= [lnu - (1/2)ln(u^2+1)]<1/2, 1>
= -(1/2)ln2 + ln2 + (1/2)ln(5/4)
= -(1/2)ln2 + (1/2)ln5
∫<π/6, π/2>[cosx/(sinx^3+sinx)]dx
= ∫<π/6, π/2>dsinx/(sinx^3+sinx)
= ∫<1/2, 1>du/(u^3+u) = ∫<1/2, 1>[1/u - u/(u^2+1)du
= [lnu - (1/2)ln(u^2+1)]<1/2, 1>
= -(1/2)ln2 + ln2 + (1/2)ln(5/4)
= -(1/2)ln2 + (1/2)ln5
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