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下面这个定积分问题,哪位大神能帮忙解答一下呀
2个回答
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∫ln(tanx)dx
=∫[0,π/2] ln(tanx)dx
=∫[0,π/4]ln(tanx)dx+∫[π/4,π/2]ln(tanx)dx
=∫[0,π/4]ln(tanx)dx+∫[π/4,π/2]lncot(π/2-x)dx
=∫[0,π/4]ln(tanx)dx+∫[π/4,0]lncotud(π/2-u)
=∫[0,π/4]ln(tanx)dx+∫[π/4,0]lntanudu
=∫[0,π/4]ln(tanx)dx-∫[0,π/4]ln(tanu)du
=0
=∫[0,π/2] ln(tanx)dx
=∫[0,π/4]ln(tanx)dx+∫[π/4,π/2]ln(tanx)dx
=∫[0,π/4]ln(tanx)dx+∫[π/4,π/2]lncot(π/2-x)dx
=∫[0,π/4]ln(tanx)dx+∫[π/4,0]lncotud(π/2-u)
=∫[0,π/4]ln(tanx)dx+∫[π/4,0]lntanudu
=∫[0,π/4]ln(tanx)dx-∫[0,π/4]ln(tanu)du
=0
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