Question

高数求详解

求积分

Answer 1

令x=tant，则dx=sec^2tdt

原式=∫(0,π/4) ln(1+tant)/sec^2t*sec^2tdt
=∫(0,π/4) ln(1+tant)dt
令u=π/4-t，则t=π/4-u，dt=-du

∫(0,π/4) ln(1+tant)dt=∫(π/4,0) ln[1+tan(π/4-u)](-du)
=∫(0,π/4) ln[1+(1-tanu)/(1+tanu)]du
=∫(0,π/4) ln[2/(1+tanu)]du
=∫(0,π/4) ln2du-∫(0,π/4) ln(1+tanu)du
=ln2*(π/4)-∫(0,π/4) ln(1+tant)dt
所以2*∫(0,π/4) ln(1+tant)dt=ln2*(π/4)
∫(0,π/4) ln(1+tant)dt=ln2*(π/8)
所以原式=ln2*(π/8)