Problem 38 - Essential Calculus

Question

Find the integral. \(\int \frac{\cosh \sqrt{x}}{\sqrt{x}} d x\)

Step-by-Step Solution

Answer

The answer to the integral \( \int \frac{\cosh \sqrt{x}}{\sqrt{x}} d x \) is \( 2\sinh(\sqrt{x}) + C \).

Step 1: Choose a suitable substitution

Let \( u=\sqrt{x} \). This would transform our integral into a simpler form.

Step 2: Make substitution and differential replacements

Replace \( dx \) by \( 2u du \) and the given integral become \( \int \frac{\cosh(u)}{u} \cdot 2u du \).

Step 3: Simplify the integral

Simplify the equation to get : \( \int 2 \cosh(u) du \).

Step 4: Integrate the equation

Integration of the equation gives \( 2\sinh(u) + C \), where C is the constant of integration.

Step 5: Substitute back the original variable

Re-substitute original variable back into the equation: \( 2\sinh(\sqrt{x}) + C \)

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the ordinary trigonometric, or circular, functions. They are defined through the exponential function. For example, the hyperbolic cosine and sine functions are defined as:
\(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\), respectively.

Just as trigonometric functions arise naturally in the context of circles, hyperbolic functions have a natural relationship to hyperbolas. They possess similar properties to trigonometric functions, like their own identities for addition and double angles. Moreover, the derivatives of the hyperbolic functions are particularly elegant, and closely resemble the derivatives of their trigonometric counterparts. For instance, the derivative of \(\cosh x\) is \(\sinh x\), and the derivative of \(\sinh x\) is \(\cosh x\), which makes integration and differentiation involving these functions relatively straightforward.

Integration by Substitution

Integration by substitution, also known as u-substitution, is a method for finding integrals that is analogous to the chain rule in differentiation. When a function is composed of a function inside another function, substitution can simplify integration by reducing the compound function to a simpler one.

To effectively use integration by substitution:

• Identify a substitution that simplifies the integral.
• Replace all instances of the original variable with the new variable, including differentials.
• Perform the integration with respect to the new variable.
• Substitute back to express the antiderivative in terms of the original variable.

In the given exercise, a substitution \( u = \sqrt{x} \) transforms the original integral into an easier one, allowing us to integrate \( \cosh(u) \) without the complication of the square root.

Constant of Integration

In integral calculus, when we integrate a function, the result is termed an antiderivative or indefinite integral. This antiderivative is not unique; in fact, there are infinitely many functions that differentiate to the same original function because differentiation wipes out any constant term.

The constant of integration, often denoted as \(C\), represents this unknown constant that could be any real number. It is essential to include \(C\) in the solution to an indefinite integral to account for all possible antiderivatives. It is only when we are dealing with definite integrals, which have specified limits of integration, that this constant disappears because we are interested in the accumulated change over an interval, not the specific value of the antiderivative.