Question

Evaluate each integral. $$\int_{0}^{\ln 2} \tanh x d x$$

Short answer

Question: Evaluate the integral \(\int_{0}^{\ln 2} \tanh x d x\). Answer: \(\frac{3}{5}\)

Step-by-step solution

Write down the integral and definition of \(\tanh x\)

The integral we want to evaluate is: $$\int_{0}^{\ln 2} \tanh x d x$$ Where the function \(\tanh x\) is defined as the ratio of the hyperbolic sine and hyperbolic cosine functions, which is: $$\tanh x = \frac{\sinh x}{\cosh x}$$

Find the derivative of \(\tanh x\) with respect to x

To find the derivative of \(\tanh x\), we can use the quotient rule, which states that the derivative of \(\frac{u}{v}\) with respect to x is \(\frac{u'v - uv'}{v^2}\). In this case, let \(u = \sinh x\) and \(v = \cosh x\). Then, we have $$\frac{d}{dx } (\tanh x) = \frac{d}{dx}\left(\frac{\sinh x}{\cosh x}\right) = \frac{\cosh x\cdot\frac{d}{dx}(\sinh x) - \sinh x \cdot\frac{d}{dx}(\cosh x)}{(\cosh x)^2}$$ We know that \(\frac{d}{dx}(\sinh x) = \cosh x\) and \(\frac{d}{dx}(\cosh x) = \sinh x\), so we can substitute these in the equation above, which gives us: $$\frac{d}{dx}(\tanh x) = \frac{\cosh x(\cosh x) - \sinh x(\sinh x)}{(\cosh x)^2}$$ Now, use the identity \(\cosh^2 x - \sinh^2 x = 1\) to simplify the equation, and we get: $$\frac{d}{dx}(\tanh x) = \frac{1}{\cosh^2 x}$$

Integrate \(\tanh x\) with respect to x

Now that we've found the derivative of \(\tanh x\), we can use it to integrate the function. Since we know that \(\frac{d}{dx}(\tanh x) = \frac{1}{\cosh^2 x}\), we can say that the antiderivative of \(\frac{1}{\cosh^2 x}\) is \(\tanh x + C\). Therefore, we have: $$\int \tanh x dx = \int \frac{1}{\cosh^2 x} dx = \tanh x + C$$

Evaluate the definite integral

Now that we have the antiderivative of \(\tanh x\), we can evaluate the definite integral over the specified limits. Using the Fundamental Theorem of Calculus, we have: $$\int_{0}^{\ln 2} \tanh x d x = [\tanh x]_0^{\ln 2} = \tanh(\ln 2) - \tanh(0)$$ We know that \(\tanh(0) = \frac{\sinh(0)}{\cosh(0)} = 0\). Meanwhile, to evaluate \(\tanh(\ln 2)\), we can use the property \(\tanh(\ln a) = \frac{a - \frac{1}{a}}{a + \frac{1}{a}}\). In our case, \(a = 2\), so we have: $$\tanh(\ln 2) = \frac{2 - \frac{1}{2}}{2 + \frac{1}{2}} = \frac{3}{5}$$

Calculate the final result

Now, we can substitute the values of \(\tanh(\ln 2)\) and \(\tanh(0)\) into the definite integral to find the final result: $$\int_{0}^{\ln 2} \tanh x d x = \tanh(\ln 2) - \tanh(0) = \frac{3}{5} - 0 = \frac{3}{5}$$ Therefore, the value of the integral is \(\frac{3}{5}\).

Key concepts

Definite Integral

A definite integral is a powerful tool in calculus, used to calculate the accumulation of quantities. When evaluating a definite integral, such as \( \int_{0}^{\ln 2} \tanh x \, dx \), we essentially measure the total area under the curve of the function \( \tanh x \) from \( x = 0 \) to \( x = \ln 2 \).

In general, a definite integral has the form \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. These limits specify the interval over which the function is being integrated. Unlike an indefinite integral, which represents a family of functions, a definite integral gives a single numerical value.

We approach definite integrals using antiderivatives and sometimes need to apply specific properties of the functions involved to find exact values. One key point when solving a definite integral is the importance of recognizing properties of the specific functions included in the integral, which could simplify the problem significantly.

Hyperbolic Functions

Hyperbolic functions resemble trigonometric functions but are based on hyperbolas instead of circles. They are used frequently in mathematical analyses involving Fourier transformations, complex numbers, or non-Euclidean geometries.

The hyperbolic tangent, \( \tanh x \), specifically is defined by the relation \( \tanh x = \frac{\sinh x}{\cosh x} \). Here, \( \sinh x \) is the hyperbolic sine and \( \cosh x \) is the hyperbolic cosine, where:

• \( \sinh x = \frac{e^x - e^{-x}}{2} \)
• \( \cosh x = \frac{e^x + e^{-x}}{2} \)

Hyperbolic functions have properties similar to those of trigonometric functions. For example, \( \cosh^2 x - \sinh^2 x = 1 \), which mirrors the Pythagorean identity from trigonometry. These similarities make them useful in various fields of science and engineering, especially when dealing with growth processes or wave-like phenomena.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges integral and differential calculus, linking antiderivatives to definite integrals. It comprises two parts: one that allows us to find antiderivatives and another enabling the evaluation of definite integrals.

When evaluating \( \int_{0}^{\ln 2} \tanh x \, dx \), the theorem simplifies the process. Once we find the antiderivative of \( \tanh x \), represented as \( F(x) \), the theorem states:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

This means that we can compute the definite integral by evaluating the antiderivative at the upper and lower bounds and subtracting them. This principle greatly simplifies the evaluation of definite integrals since it turns the problem into one of basic subtraction, once an antiderivative is known. Understanding this theorem is critical for mastering calculus, making it easier to solve complex real-world problems using integration.