∫π/2 0 (cos2x/cosx+sinx)dx 的定积分
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∫π/2 0 (cos2x/cosx+sinx)dx
=∫π/2 0 (cos²x-sin²x)/(cosx+sinx)dx
=∫π/2 0 (cosx-sinx)dx
=sinx+cosx π/2 0
=(1+0)-(0+1)
=0
=∫π/2 0 (cos²x-sin²x)/(cosx+sinx)dx
=∫π/2 0 (cosx-sinx)dx
=sinx+cosx π/2 0
=(1+0)-(0+1)
=0
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∫[0,π/2] (cos2x/cosx+sinx)dx
=∫[0,π/2] (cos^2x-sin^2x)/(cosx+sinx)dx
=∫[0,π/2] (cosx-sinx)dx
=(sinx+cosx)[0,π/2]
=0
=∫[0,π/2] (cos^2x-sin^2x)/(cosx+sinx)dx
=∫[0,π/2] (cosx-sinx)dx
=(sinx+cosx)[0,π/2]
=0
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