Question

高等数学，求定积分

Answer 1

大致就这样

追答

将（-π，0）区间作x=-u代换就行了

Answer 2

令t=-x，则dx=-dt
∫(-π,π)sin^4x/(1+e^x)dx
=∫(π,-π)sin^4t/[1+e^(-t)](-dt)
=∫(-π,π)sin^4t*e^t/(e^t+1)dt
所以2*∫(-π,π)sin^4x/(1+e^x)dx
=∫(-π,π)sin^4x/(1+e^x)dx+∫(-π,π)sin^4t*e^t/(e^t+1)dt
=∫(-π,π)sin^4xdx
=2∫(0,π)sin^4xdx
=(1/2)*∫(0,π)(1-cos2x)^2dx
=(1/2)*∫(0,π)[1-2cos2x+cos^2(2x)]dx
=(1/2)*∫(0,π)[1-2cos2x+(1/2)*(1+cos4x)]dx
=(1/2)*[3x/2-sin2x+(1/8)*sin4x]|(0,π)
=3π/4
所以原式=3π/8