求定积分=∫sinx/(1+(tanx)^2)dx(-π/4

问题描述:

求定积分=∫sinx/(1+(tanx)^2)dx(-π/4

数学人气:366 ℃时间:2020-05-20 20:01:47
优质解答
=∫sinx/(1+(tanx)^2)dx(-π/4=∫sinx(cosx)^2dx(-π/4=-∫(cosx)^2)dsinx(-π/4=-1/3*(cosx)^3(-π/4=0
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=∫sinx/(1+(tanx)^2)dx(-π/4=∫sinx(cosx)^2dx(-π/4=-∫(cosx)^2)dsinx(-π/4=-1/3*(cosx)^3(-π/4=0