麻烦下列式子详细过程

Question

如题所述

Answer 1

令u=π/2-v
则可以得到∫(0,π/2)cosudu/(sinu+cosu)=∫(0,π/2)sinudu/(sinu+cosu)
则设为a
所以2a=∫(0,π/2)cosudu/(sinu+cosu)+∫(0,π/2)sinudu/(sinu+cosu)
=∫(0,π/2)du
=π/2
所以原式=π/4来自：求助得到的回答

Answer 2

原式+sinu/(sinu+cosu)=pi/2，
原式=(cosu-sinu+sinu)/(sinu+cosu)du=ln(sinu+cosu)+pi/2-原式=ln1-ln1+pi/2-原式，
得原式=pi/4