求定积分∫1/(1-sinx)dx

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求定积分∫1/(1-sinx)dx

∫ 1/(1-sinx) dx
= ∫ (1+sinx)/(1-sin²x) dx
= ∫ (1+sinx)/cos²x dx
= ∫sec²x dx + ∫secxtanx dx
= tanx + secx + C
咁简单嘅不定积分,不容许错误!

sinx=2sin(x/2)cos(x/2)
1-sinx=sin²(x/2)-2sin(x/2)cos(x/2)+cos²(x/2)=(sin(x/2)-cos(x/2))²=[√2sin(x/2-π/4)]²=2sin²(x/2-π/4)
原积分=(1/2)∫1/sin²(x/2-π/4) dx
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sinx=2sin(x/2)cos(x/2)
1-sinx=sin²(x/2)-2sin(x/2)cos(x/2)+cos²(x/2)=(sin(x/2)-cos(x/2))²=[√2sin(x/2-π/4)]²=2sin²(x/2-π/4)
原积分=(1/2)∫1/sin²(x/2-π/4) dx
=-(1/2)cot(x/2-π/4)+C

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