Consider the integral 1-1 But now we try the hyperbolic trigonometric substitution dx Since = du 2*cosh(U) I= cosh (u) V/1+sinh2(u) The reason this substitution works so well is because the identity -S u a = 2sinh(u). we can rewrite this integral as: 1 /2+x -dz. Then as a function in terms of the original variable z I= cosh? (u) sinh?(u)=1 can be rearranged to simplify the square root. The integral can be evaluated first as a function of u I= +C. Note: the Maple syntax for cosh(u) is arccosh (u) du. +C