证明∫(-a,a)f(x)d(x)=∫(-a,a)f(-x)d(x),并计算∫(-π/4,π/4)(cosx)^2/(1
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证明∫(-a,a)f(x)d(x)=∫(-a,a)f(-x)d(x),并计算∫(-π/4,π/4)(cosx)^2/(1+e^x)d(x)
做变量替换:x=-t,则:
∫(-a,a)f(-x)d(x)=∫(a,-a)f(t)d(-t)=-∫(a,-a)f(t)d(t)=∫(-a,a)f(t)d(t)
证毕
由上述结论知:
∫(-π/4,π/4)(cosx)^2/(1+e^x)d(x)=∫(-π/4,π/4)(cosx)^2/(1+e^(-x))d(x)
=∫(-π/4,π/4)e^x (cosx)^2/(1+e^x)d(x)(分子分母同乘以e^x)
所以
2∫(-π/4,π/4)(cosx)^2/(1+e^x)d(x)
=∫(-π/4,π/4)(cosx)^2/(1+e^x)d(x)+∫(-π/4,π/4)e^x (cosx)^2/(1+e^x)d(x)
=∫(-π/4,π/4) (cosx)^2d(x)
而∫(cosx)^2d(x)=∫(cos2x+1)d(x)/2=(sin2x)/4+x/2+C
故原积分=π/8+1/4.
∫(-a,a)f(-x)d(x)=∫(a,-a)f(t)d(-t)=-∫(a,-a)f(t)d(t)=∫(-a,a)f(t)d(t)
证毕
由上述结论知:
∫(-π/4,π/4)(cosx)^2/(1+e^x)d(x)=∫(-π/4,π/4)(cosx)^2/(1+e^(-x))d(x)
=∫(-π/4,π/4)e^x (cosx)^2/(1+e^x)d(x)(分子分母同乘以e^x)
所以
2∫(-π/4,π/4)(cosx)^2/(1+e^x)d(x)
=∫(-π/4,π/4)(cosx)^2/(1+e^x)d(x)+∫(-π/4,π/4)e^x (cosx)^2/(1+e^x)d(x)
=∫(-π/4,π/4) (cosx)^2d(x)
而∫(cosx)^2d(x)=∫(cos2x+1)d(x)/2=(sin2x)/4+x/2+C
故原积分=π/8+1/4.
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