## Example 7.2.4(e): Applying the Substitution Rule

Now it gets interesting again. In a previous example we had the square root of*r*, and we used the substitution

^{2}- x^{2}*x = sin(t)*because of the equality

*sin*.

^{2}(t) + cos^{2}(t) = 1For the first integral we use the substitution

because of the hyperbolic trig equationx = sinh(t) [ = 1/2 (eso that^{t}- e^{-t}) ]

dx/dt = 1/2 (eor^{t}+ e^{-t}) = cosh(t)dx = cosh(t) dt

(which you should verify). With that substitution we have that:cosh^{2}(t) - sinh^{2}(t) = 1

where

*arc sinh*is the inverse function of

*sinh*. Note that it is not hard to show that

Evaluating the second integral is left as an exercise (try the substitution

*x = cosh(t)*).