Problem 68 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(67-74\) in terms of a. inverse hyperbolic functions. b. natural logarithms. $$\int_{0}^{1 / 3} \frac{6 d x}{\sqrt{1+9 x^{2}}}$$

Step-by-Step Solution

Answer

\( \int_{0}^{1/3} \frac{6 \, dx}{\sqrt{1+9x^2}} = 2 \ln(1 + \sqrt{2}) \).

Step 1: Recognizing the Integral Form

The integral \( \int \frac{dx}{\sqrt{1+x^2}} \) is a standard form that resembles the derivative of the inverse hyperbolic sine function, \( \sinh^{-1}(x) \). Our integral is similar to this form with a constant factor and a squared term, \( 9x^2 \), in the denominator.

Step 2: Rewriting the Integral

Rewrite the integral as \( \int \frac{6}{\sqrt{1+(3x)^2}} \, dx \), which matches the form \( \int \frac{du}{\sqrt{1+u^2}} \) if we perform a substitution.

Step 3: Substitution for Simplicity

Let \( u = 3x \), then \( du = 3 \, dx \) or \( dx = \frac{du}{3} \). Substitute these into the integral, resulting in \( 6 \int \frac{1}{\sqrt{1+u^2}} \cdot \frac{du}{3} = 2 \int \frac{du}{\sqrt{1+u^2}} \).

Step 4: Integration using Inverse Hyperbolic Function

The integral \( \int \frac{du}{\sqrt{1+u^2}} \) is known to be \( \sinh^{-1}(u) + C \). Apply this to our current expression: \( 2 \sinh^{-1}(u) + C \).

Step 5: Back-Substitution

Replace \( u \) back with \( 3x \), resulting in \( 2 \sinh^{-1}(3x) + C \). Now, substitute back the limits of integration. When \( x = 0 \), \( 3x = 0 \); when \( x = \frac{1}{3} \), \( 3x = 1 \).

Step 6: Evaluate Definite Integral

Evaluate the antiderivative at the upper and lower bounds: \[ 2 \sinh^{-1}(1) - 2 \sinh^{-1}(0) = 2 \ln(1 + \sqrt{2}) \](since \( \sinh^{-1}(1) = \ln(1 + \sqrt{2}) \) and \( \sinh^{-1}(0) = 0 \)).

Step 7: Change Integral Form to Natural Logarithms

Using the known transformations between inverse hyperbolic functions and logarithms, we conclude:\[ \int \frac{6 \, dx}{\sqrt{1+9x^2}} = 2 \ln(1 + \sqrt{2}) \].

Key Concepts

Integrals

Integrals are a fundamental concept in calculus that involve finding the area under a curve on a graph, the accumulated quantity, or the total change over an interval. In the exercise we are looking at, an integral is expressed in the form \( \int \frac{6 \ dx}{\sqrt{1+9x^2}} \). This particular integral is a definite integral, meaning it has specified upper and lower limits of integration, in this case from 0 to \( \frac{1}{3} \). Definite integrals calculate the exact area under the curve between these two points. They can be seen as the summation of infinitesimally small areas. To solve the integral, several steps are typically followed:

• Recognize the standard form that relates to known functions, like the inverse hyperbolic functions.
• Rewrite and simplify the expression to facilitate easier calculation.
• Carry out substitution to simplify the integral into a more recognizable form.
• Apply specific integral formulas or properties, such as those for inverse hyperbolic functions.
• Evaluate the result over the given interval to find the exact value.

Understanding integrals in this way is crucial for solving many mathematical problems that involve cumulative or changing quantities.

Inverse Hyperbolic Sine Function

The inverse hyperbolic sine function, denoted as \( \sinh^{-1}(x) \), is a function that plays a key role in solving certain integrals, like the one in our exercise. The inverse hyperbolic sine is related to the hyperbolic sine function, \( \sinh(x) \), such that\( \sinh(y) = x \) implies \( y = \sinh^{-1}(x) \). The derivative of the inverse hyperbolic sine function is known to be \( \frac{d}{dx}[\sinh^{-1}(x)] = \frac{1}{\sqrt{1+x^2}} \), which matches the form of many integrals involving \( \sqrt{1+x^2} \) in the denominator.Utilizing \( \sinh^{-1} \) is very useful in integration, because:

• It allows for simple expressions of integrals that would otherwise seem complex.
• The known derivative is used directly in definite integrals to find antiderivatives.
• Inverse hyperbolic functions can often be converted into more familiar forms such as logarithms.

The inverse hyperbolic sine function is particularly helpful because of its straightforward relationship to natural logarithms, which often simplifies results as you evaluate expressions for real-world applications.

Natural Logarithms

Natural logarithms come into play when we simplify expressions, especially after integration involving inverse hyperbolic functions. A natural logarithm is denoted as \( \ln(x) \) and is the inverse operation of exponentiation with base \( e \), where \( e \approx 2.718 \). In the context of the exercise, after simplifying and evaluating the integral with the inverse hyperbolic sine function, we convert it to \( \ln \) form: \( \sinh^{-1}(x) = \ln(x + \sqrt{1+x^2}) \).Here's why they are important:

• Natural logarithms are highly used in calculus and mathematical applications because they solve exponential equations with base \( e \).
• They simplify expressions that result from integration of inverse hyperbolic functions.
• Understanding their relation to inverse hyperbolic functions aids in transitioning between different forms and expressions.

In solving the integral \( \int \frac{6 \ dx}{\sqrt{1+9x^2}} \), using natural logarithms helped us express the solution in a form \( 2 \ln(1 + \sqrt{2}) \), which is often easier to interpret and apply within further mathematical contexts.