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
|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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