Re: [積分] 一題三角函數的定積分
: π
: ∫ |sin(x/2) - cos(x/2)| dx
: 0
: π/2 π
: = ∫ |sin(x/2) - cos(x/2)| dx + ∫ |sin(x/2) - cos(x/2)| dx
: 0 π/2
: π/2 π
: = ∫ (cos(x/2) - sin(x/2)) dx + ∫ sin(x/2) - cos(x/2) dx
: 0 π/2
: (0≦x≦π/2 , 0≦ x/2 ≦ π/4 => sin(x/2) ≦ cos(x/2))
: (π/2 ≦ x ≦ π , 0 ≦ x/2 ≦ π/2 => sin(x/2) ≧ cos(x/2))
: x x |π/2 x x |π
: = (2)(sin(---) + cos(---)) | + (2)(-cos(---) - sin(---)) |
: 2 2 |0 2 2 |π/2
: 1 1 4
: = (2)(--- - 1) + (2)(1 + ---) = ---- = (2)(√2)
: √2 √2 √2
這裡我算好像是
√2 √2 √2 √2
= (2)(--- + --- - 1) + (2)(-1-(- --- - ---))=4√2-4
2 2 2 2
請問我哪裡有算錯嗎??
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 220.134.245.25
推
07/01 15:08, , 1F
07/01 15:08, 1F
討論串 (同標題文章)