Question

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

希望详细写清步骤，谢谢

Answer 1

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