就是關(guān)于那個(gè)定積分“求定積分:∫ln(tanx)dx (o≤x≤π/2),積分是限是π/2,下限是0"的一些問(wèn)題
∫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
您的答案中,=∫[0,π/4]ln(tanx)dx+∫[π/4,0]lncotud(π/2-u)
=∫[0,π/4]ln(tanx)dx+∫[π/4,0]lntanudu
這兩步中,第二步是如何變化的來(lái)的,是不是又經(jīng)過(guò)一次換元,那么前面的積分上下限不用變換嗎?