∫(sinx)^2⼀[1+e^(-x)] dx 积分上下限(π⼀4,π⼀4)怎么算？

注：此题的上下限有错，应该是积分上下限(-π/4,π/4)！
解：原式=∫(-π/4,π/4)(sinx)^2/[1+e^(-x)]dx (∫(-π/4,π/4)表示从-π/4到π/4积分)
=∫(-π/4,0)(sinx)^2/[1+e^(-x)]dx+∫(0,π/4)(sinx)^2/[1+e^(-x)]dx
=-∫(π/4,0)(sinx)^2/(1+e^x)dx+∫(0,π/4)(sinx)^2/[1+e^(-x)]dx (第一个积分用-x代换x得)
=∫(0,π/4)(sinx)^2/(1+e^x)dx+∫(0,π/4)e^x(sinx)^2/(1+e^x)dx (第二个积分分子分母同乘e^x得)
=∫(0,π/4)(1+e^x)(sinx)^2/(1+e^x)dx
=∫(0,π/4)(sinx)^2dx
=1/2∫(0,π/4)[1-cos(2x)]dx ()
=1/2[x-1/2sin(2x)]|(0,π/4)
=1/2(π/4-1/2)
=(π-2)/8

上下限相等？ 那不就是零么

如果是正负的话，那个，我也算不出来...用软件计算出来约等于 0.14...

所以猜测答案是 pi-3 ，哈哈