高等数学,求如图中积分的详细过程
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I = (1/8) sintcost ∫<0 r^4 √(1+4r^2) d(1+4r^2)
= (1/12) sintcost ∫ r^4 d(1+4r^2)^(3/2) (u = r^2)
= (1/12) sintcost ∫ u^2 d(1+4u)^(3/2)
I1 = ∫ u^2 d(1+4u)^(3/2)
= u^2(1+4u)^(3/2) - 2∫ u(1+4u)^(3/2)du
= u^2(1+4u)^(3/2) - (1/5)∫ ud(1+4u)^(5/2)
= u^2(1+4u)^(3/2) - (1/5)u(1+4u)^(5/2) + (1/5)∫ (1+4u)^(5/2)du
= u^2(1+4u)^(3/2) - (1/5)u(1+4u)^(5/2) + (2/35)(1+4u)^(7/2)
I = (1/12) sintcost [u^2(1+4u)^(3/2) - (1/5)u(1+4u)^(5/2) + (2/35)(1+4u)^(7/2)]<0, 1>
= (1/12) sintcost = [5^(5/2) / 42] sintcost
= (1/12) sintcost ∫ r^4 d(1+4r^2)^(3/2) (u = r^2)
= (1/12) sintcost ∫ u^2 d(1+4u)^(3/2)
I1 = ∫ u^2 d(1+4u)^(3/2)
= u^2(1+4u)^(3/2) - 2∫ u(1+4u)^(3/2)du
= u^2(1+4u)^(3/2) - (1/5)∫ ud(1+4u)^(5/2)
= u^2(1+4u)^(3/2) - (1/5)u(1+4u)^(5/2) + (1/5)∫ (1+4u)^(5/2)du
= u^2(1+4u)^(3/2) - (1/5)u(1+4u)^(5/2) + (2/35)(1+4u)^(7/2)
I = (1/12) sintcost [u^2(1+4u)^(3/2) - (1/5)u(1+4u)^(5/2) + (2/35)(1+4u)^(7/2)]<0, 1>
= (1/12) sintcost = [5^(5/2) / 42] sintcost
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