Problem 71 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(67-74\) in terms of \begin{equation}\begin{array}{l}{\text { a. inverse hyperbolic functions. }} \\\ {\text { b. natural logarithms. }}\end{array}\end{equation} $$\int_{1 / 5}^{3 / 13} \frac{d x}{x \sqrt{1-16 x^{2}}}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \sinh^{-1}(\frac{12}{13}) - \sinh^{-1}(\frac{4}{5}) \) or \( \ln(\frac{12}{13} + \sqrt{\frac{144}{169} + 1}) - \ln(\frac{4}{5} + \sqrt{\frac{16}{25} + 1}) \).

Step 1: Identify the form of the integral

The given integral \( \int \frac{dx}{x \sqrt{1-16x^2}} \) suggests a form involving inverse hyperbolic functions because of the \( \sqrt{1-16x^2} \) term. This resembles the derivative form of inverse hyperbolic functions.

Step 2: Use the substitution for inverse hyperbolic functions

Consider the substitution \( x = \frac{1}{4} \sinh(t) \), giving us \( dx = \frac{1}{4} \cosh(t) \, dt \). This substitution turns \( \sqrt{1-16x^2} \) into \( \cosh(t) \). Thus, the integral becomes \( \int \frac{\frac{1}{4} \cosh(t)}{x \cosh(t)} = \int \frac{1}{4x}\ dt \).

Step 3: Substitute the boundary values

Since \( x = \frac{1}{4} \sinh(t) \), when \( x = \frac{1}{5} \), \( \sinh(t_1) = \frac{4}{5} \). When \( x = \frac{3}{13} \), \( \sinh(t_2) = \frac{12}{13} \). Solve these to find \( t_1 \) and \( t_2 \).

Step 4: Evaluate the integral in terms of inverse hyperbolic functions

The integral evaluates to \([ \sinh^{-1}(4x) ]_{1/5}^{3/13}\), which simplifies to \( \sinh^{-1}(\frac{12}{13}) - \sinh^{-1}(\frac{4}{5}) \).

Step 5: Convert results to natural logarithms

Recall the identity \( \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \). Apply this to both \( \sinh^{-1}(\frac{12}{13}) \) and \( \sinh^{-1}(\frac{4}{5}) \). The final expression in natural logarithms becomes: \( \ln(\frac{12}{13} + \sqrt{\frac{144}{169} + 1}) - \ln(\frac{4}{5} + \sqrt{\frac{16}{25} + 1}) \).

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions may sound complex, but they are essential when dealing with certain types of integrals in calculus, especially those resembling the form involving the expression \(\sqrt{1-x^2}\). Inverse hyperbolic functions are analogous to the typical trigonometric inverse functions; however, they apply to hyperbolic functions like sinh, cosh, and tanh rather than sin, cos, and tan.

• Hyperbolic Sine (sinh): The function \(\sinh(x)\) is defined as \(\frac{e^x - e^{-x}}{2}\).
• The Inverse Hyperbolic Sine, \(\sinh^{-1}(x)\), can be useful in integration because it helps transform complex radical expressions into something more manageable.

When you spot an integral with a form like \(\int \frac{dx}{x \sqrt{1-16x^2}}\), it suggests the potential use of an inverse hyperbolic function due to its resemblance to the derivative of \(\sinh^{-1}(x)\). A common strategy is to substitute with a hyperbolic identity, simplifying the integral into a form that's easier to manage. Once the integral is transformed and simplified, it can be expressed using inverse hyperbolic functions.

Natural Logarithms

Natural logarithms are fundamental in calculus. They appear frequently during the process of integration, especially when inverse hyperbolic functions are involved. The notation \(\ln(x)\) represents the natural logarithm of \(x\). Natural logarithms have the unique property of being the inverse operation of the exponential function, which provides a way to solve equations involving exponentials.When working with inverse hyperbolic functions, it's useful to recall that they can be written in terms of natural logarithms. For example, the identity for inverse hyperbolic sine, \(\sinh^{-1}(x)\), can be expressed as:\[\sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})\]This is particularly useful when converting the results of an integral into standard forms. By applying this identity, students can express complex trigonometric expressions as natural logarithms, allowing a more straightforward analysis or further simplification.

Definite Integrals

Definite integrals are an essential concept in calculus, used to calculate the area under a curve between two specified points on the x-axis. A definite integral has limits of integration, denoted as lower and upper bounds, which in this exercise are \(\frac{1}{5}\) and \(\frac{3}{13}\).

• A definite integral is written as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration.
• They yield a constant value, representing the accumulation of quantities described by the function \(f(x)\) between \(a\) and \(b\).

In the context of the exercise, the definite integral \(\int_{1 / 5}^{3 / 13} \frac{dx}{x \sqrt{1-16 x^{2}}}\) involves solving the integral between these boundaries.Evaluating definite integrals often involves substitution methods, especially when dealing with expressions that can be simplified by using known functions like inverse hyperbolic functions or converting results into natural logarithms.Once the integral is solved, you apply the boundaries to the expression to compute a specific value, which can represent anything from the physical area under a curve to solving real-world problems.