※ 引述《plover (>//////<)》之銘言：
: π/2 1
: Prove that π/2 ≦ ∫ -------------------------- dx < π/√2.
: 0 √( 1 - (1/2)(sin^2 x) )
證明 : -1 ≦ sinx ≦ 1
2
=> 0 ≦ sin x ≦ 1
2
sin x 1
=> 0 ≦ ------- ≦ ---
2 2
2
1 sin x
=> - --- ≦ - ------- ≦ 0
2 2
2
1 sin x
=> --- ≦ 1 - ------- ≦ 1
2 2
2
√2 sin x 1
=> ----- ≦ (1 - -------)^(---) ≦ 1
2 2 2
1
=> 1 ≦ ----------------------- ≦ √2
√(1 - (1/2)((sinx)^2))
π/2 π/2 1 π/2
=> ∫ 1 dx ≦ ∫ ----------------------- dx ≦ ∫ꄠ √2 dx 0 0 √(1 - (1/2)((sinx)^2)) 0
π/2 1
=> π/2 ≦ ∫ ------------------------ dx ≦ π/√2
0 √(1 - (1/2)((sinx)^2))
π/2 1
=> π/2 ≦ ∫ ------------------------ dx < π/√2
0 √(1 - (1/2)((sinx)^2))
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 61.66.173.21