19/08/29 16:30:25.44 V/0HLVAJ.net
>>144 >>174
θ(t) = arccos{t/√(3/4 -t)},
θ'(t) = - (3/2 -t)/{2(3/4 -t)√((3/2 +t)(1/2 -t))} < 0,
部分積分で
∫(3/4 -t)[π-θ(t)] dt = - (1/2)(3/4 -t)^2･[π-θ(t)] - (1/2)∫(3/4 -t)^2 θ'(t)dt
= - (1/2)(3/4 -t)^2 [π-θ(t)] + (3/4)arcsin(1/2 +t) + ((6-t)/8)√{(3/2 +t)(1/2 -t)},
t: -3/2→1/2 のとき θ(t): π→0 だから
∫[-3/2,1/2] (3/4 -t)[π-θ(t)]dt = (23/32)π,
(3/4 -t)cosθsinθ = t√{(3/2 +t)(1/2 -t)} ゆえ
∫(3/4 -t)cosθsinθ dt = ∫t√{(3/2 +t)(1/2 -t)}dt
= (1/3)(tt +t/4 -9/8)√{(3/2 +t)(1/2 -t)} - (1/4)[π- arccos(1/2 +t)],
∫[-3/2,1/2] (3/4 -t)cosθsinθ dt = - (1/4)π,
∴ S = (23/32)π - (1/4)π = (15/32)π.