An alternative way to evaluate 1 I = sech(u) du = du cosh (u) is to use another hyperbolic trig substitution t = sinh(u) where dt cosh(u) O du With this substitution the integral becomes I = g(t) dt where g( t) = 1/((sinh(t)^2+1))"cosh(t) X . Hence we can evaluate I using the table of standard integrals, first as a function of t: I = X P +C. Then as a function of the variable u: I = X P +C