HYPERBOLIC FUNCTIONS

DEFINING THE HYPERBOLIC FUNCTIONS

FACT: Every function f that is defined on an interval centered at the origin

can be written as the sum of one even function and one odd function.

Let us use this idea to define the exponential function ex.

We will let the even part be the hyperbolic cosine of x,

and the odd part be the hyperbolic sine of x,

The graph of the hyperbolic cosine is called a catenary curve. You can see this curve when you

look at two high line polls. The cable that strung between them forms the catenary curve. All

bridges must be designed with a catenary curve. It provides structural strength to the bridge.

Look at any bridge, and you will find a catenary curve.

DEFINITIONS

Hyperbolic cosine of x Hyperbolic sine of x

Hyperbolic tangent of x Hyperbolic cotangent of x

Hyperbolic secant of x Hyperbolic cosecant of x

BASIC IDENTITIES

DERIVATIVES OF THE HYPERBOLIC FUNCTIONS

EXAMPLE 1:

SOLUTION:

EXAMPLE 2:

SOLUTION:

EXAMPLE 3:

SOLUTION:

EXAMPLE 4:

SOLUTION:

EXAMPLE 5:

SOLUTION:

EXAMPLE 6:

SOLUTION:

EXAMPLE 7:

SOLUTION:

EXAMPLE 8:

SOLUTION:

INTEGRAL FORMULAS FOR HYPERBOLIC FUNCTIONS

EXAMPLE 9:

SOLUTION:

Let u = x 2, then du = 2x dx.

EXAMPLE 10:

SOLUTION:

Let u = 3x, then du = 3dx.

EXAMPLE 11:

SOLUTION:

EXAMPLE 12:

SOLUTION:

Let u = sinh 2x, then du = 2cosh 2x dx.

EXAMPLE 13:

SOLUTION:

Let u = cosh x, then du = sinh x dx.

EXAMPLE 14:

SOLUTION:

There is no direct antiderivative for the sech x, so we will substitute in the definition of the

hyperbolic secant in for sech x.

Now, we can simplify the above rational expression. Here is the work required to do

this.

We will now substitute this expression into the integral.

Let u = e x, then du = e

x dx.

EXAMPLE 15:

SOLUTION:

EXAMPLE 16:

SOLUTION:

Let u = e x - e

- x, then du = e

x + e

- x dx.

As you work through these examples, try to memorize the derivatives and the integrals of the

hyperbolic trig functions.