∫±π/2 (sinx/1+cos平方x)dx

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∫±π/2 (sinx/1+cos平方x)dx
∫±π/2 (sinx/1+cos平方x)dx

∫±π/2 (sinx/1+cos平方x)dx
解法一:∫[sinx/(1+cos²x)]dx=∫d(cosx)/(1+cos²x)
=[arctan(cosx)]│
=arctan(cos(π/2))-arctan(cos(-π/2))
=arctan(0)-arctan(0)
=0;
解法二:∫[sinx/(1+cos²x)]dx=∫[sinx/(1+cos²x)]dx+∫[sinx/(1+cos²x)]dx
=∫[sin(-x)/(1+cos²(-x))]d(-x)+∫[sinx/(1+cos²x)]dx
(第一个积分用-x代换x)
=-∫[sinx/(1+cos²x)]dx+∫[sinx/(1+cos²x)]dx
=0.