Question

∫（1-tanx）/（1+tanx）dx求導?

Answer 1

原式=∫(tanπ/4-tanx)/(1+tanπ/4tanx）dx

=∫tan(π/4-x)dx

=∫sin(π/4-x)/cos(π/4-x) dx

=-∫sin(π/4-x)/cos(π/4-x) d(π/4-x)

=∫dcos(π/4-x)/cos(π/4-x)

=ln|cos(π/4-x)| +C