Question

定积分计算题求解答

Answer 1

令t=-x，则dx=-dt
∫(-π/4,π/4) sin^4x/(1+e^x)dx
=∫(-π/4,π/4) e^t*sin^4t/(1+e^t)dt
=∫(-π/4,π/4) e^x*sin^4x/(1+e^x)dx
2∫(-π/4,π/4) sin^4x/(1+e^x)dx
=∫(-π/4,π/4) sin^4x/(1+e^x)dx+∫(-π/4,π/4) e^x*sin^4x/(1+e^x)dx
=∫(-π/4,π/4) sin^4xdx
=(1/4)*∫(-π/4,π/4) (2sin^2x)^2dx
=(1/4)*∫(-π/4,π/4) (1-cos2x)^2dx
=(1/4)*∫(-π/4,π/4) [1-2cos2x+(cos2x)^2]dx
=(1/2)*∫(0,π/4) [1-2cos2x+(1/2)*(1+cos4x)]dx
=(1/2)*∫(0,π/4) [3/2-2cos2x+(1/2)*cos4x]dx
=(1/2)*[3x/2-sin2x+(1/8)*sin4x]|(0,π/4)
=(1/2)*(3π/8-1)
所以原式=(1/4)*(3π/8-1)=3π/32-1/4

Answer 2