推 hothero:OK~瞭了，3Q 220.134.245.25 07/01 11:22
※ 引述《hothero (hothero)》之銘言：
: π
: ∫ √(1-sinx)dx
: 0
: π
: =∫ √(sin(x/2)^2+cos(x/2)^2-2sin(x/2)cos(x/2))dx
: 0
: π π
: =∫ √(sin(x/2)-cos(x/2))^2dx=∫ (sin(x/2)-cos(x/2))dx
: 0 0
^^^^^^^^^^^^^^^^^^^^^^^^
這一步就錯了
去根號不確定正或負 , 所以要加絕對值
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
: |π
: =(-2cos(x/2)-2sin(x/2))| =(-2)-(-2)=0
: |0
: LuisSantos大大，這是我的算式..謝謝
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