定积分,求答案和方法名,有过程最好

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设A=∫(-π/2→π/2)cosx/(e^x+1)dx
则A=∫(-π/2→π/2)cosx/(e^(-x)+1)dx
所以2A=∫(-π/2→π/2)[cosx/(e^x+1)+cosx/(e^(-x)+1)]dx
=∫(-π/2→π/2)cosx(e^x+e^(-x)+2)/(e^x+e^(-x)+2)dx
=∫(-π/2→π/2)cosxdx
=sinx|(-π/2→π/2)
=2所以原式=A=1