Question

求定积分？

Answer 1

令x=tanu dx=sec²udu
x=0,u=0,
x=1,u=π/4
∫<0,π/4>[ln(1+tanu)/(1+tan²u)]*sec²udu =∫<0,π/4>ln(1+sinu/cosu)du
=∫<0,π/4>ln[(sinu+cosu)/cosu]du
=∫<0,π/4>[ln(sinu+cosu)-ln(cosu)]du
=∫<0,π/4>ln(sinu+cosu)du-∫<0,π/4>ln(cosu)du
=∫<0,π/4>ln[√2sin(u+π/4)]du-∫<0,π/4>ln(cosu)du
=∫<0,π/4>ln(√2)du+∫<0,π/4>ln[sin(u+π/4)]du-∫<0,π/4>ln(cosu)du
=(π/4)ln(√2)+∫<π/4,0>ln[sin(π/2-y)]d(-y)-∫<0,π/4>ln(cosu)du
(在第二个积分中，令u=π/4-y)
=πln2/8+∫<0,π/4>ln(cosy)dy-∫<0,π/4>ln(cosu)du
=πln2/8+∫<0,π/4>ln(cosu)du-∫<0,π/4>ln(cosu)du
(在第一个积分中，令u=y)
=πln2/8

Answer 2

Answer 3