Question

(4/π到-4/π)∫(cosx)^2/(1+e^x)dx的定积分是多少谢谢各位啦

Answer 1

是这个积分吗？∫[-π/4-->π/4] (cosx)^2/(1+e^x)dx，如果上下限颠倒了，就是差个负号。
令x=-u，则dx=-du，u：π/4-->-π/4
∫[-π/4-->π/4] (cosx)^2/(1+e^x)dx
=-∫[π/4-->-π/4] (cosu)^2/(1+e^(-u))du
分子分母同乘以e^u，然后上下限交换，前面负号消去
=∫[-π/4-->π/4] e^u(cosu)^2/(e^u+1)du
定积分可以随便换积分变量
=∫[-π/4-->π/4] e^x(cosx)^2/(e^x+1)du

这样证明了∫[-π/4-->π/4] (cosx)^2/(1+e^x)dx=∫[-π/4-->π/4] e^x(cosx)^2/(e^x+1)du
将两个积分相加得：
∫[-π/4-->π/4] (cosx)^2/(1+e^x)dx+∫[-π/4-->π/4] e^x(cosx)^2/(e^x+1)du
=∫[-π/4-->π/4] (cosx)^2dx
=2∫[0-->π/4] (cosx)^2dx
=∫[0-->π/4] (1+cos2x)dx
=x+1/2sin2x [0-->π/4]
=π/4+1/2
由于上面两个积分相等，因此每个都等于上面这个结果的一半。
∫[-π/4-->π/4] (cosx)^2/(1+e^x)dx=π/8+1/4

Answer 2

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