Question

Perform the indicated integrations. $$ \int_{0}^{1} \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x $$

Short answer

The value of the integral is \( \frac{1}{2} \ln(\cosh(2)) \).

Step-by-step solution

Recognize the Function and Simplify

The given integral is: \[ \int_{0}^{1} \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \, dx \]Notice that the integral contains the expression \(\frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}\). This expression resembles the hyperbolic tangent function, \( \tanh(x) = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \). Therefore, redefine the function using \( \tanh(2x) \): \[ \int_{0}^{1} \tanh(2x) \, dx \]

Determine the Integral Formula for Hyperbolic Tangent

The antidifferentiation of \( \tanh(u) \) is known, which is \( \ln(\cosh(u)) \). We can use this formula to find: \[ \int \tanh(2x) \, dx = \frac{1}{2} \ln(\cosh(2x)) + C \] where \( C \) is the constant of integration. The factor \(\frac{1}{2}\) comes from the chain rule when integrating involving \(2x\) instead of \(x\).

Apply Bounds to Evaluate the Definite Integral

Evaluate the integral with the definite bounds from 0 to 1:\[ \left[ \frac{1}{2} \ln(\cosh(2x)) \right]_0^1 \]Calculate at the upper limit (1):\[ \frac{1}{2} \ln(\cosh(2 \cdot 1)) = \frac{1}{2} \ln(\cosh(2)) \]Calculate at the lower limit (0):\[ \frac{1}{2} \ln(\cosh(0)) = \frac{1}{2} \ln(1) = 0 \]

Compute the Result

Subtract the lower limit result from the upper limit result:\[ \frac{1}{2} \ln(\cosh(2)) - 0 = \frac{1}{2} \ln(\cosh(2)) \]Thus, the value of the integral from 0 to 1 is\[ \frac{1}{2} \ln(\cosh(2)) \].

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are based on the hyperbola, much like trigonometric functions are related to the circle. These functions often appear in calculations involving hyperbolic geometry, but they also find uses in other areas like calculus and complex analysis. Here, we are particularly interested in hyperbolic tangent, denoted as \( \tanh(x) \), which is defined as:\[ \tanh(x) = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \]This function is similar to the tangent function in trigonometry but relies on exponential expressions. When dealing with calculus problems involving hyperbolic functions, recognizing a pattern similar to \( \tanh(x) \) can greatly simplify the integration process. Instead of working through complex exponential expressions, you can switch to using hyperbolic function identities.

Definite Integral

A definite integral computes the accumulation of quantities, such as area under a curve, between two specific points on the x-axis. In mathematical notation, it looks like \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.Unlike indefinite integrals, which result in a function of \( x \) plus a constant integration (\( C \)), definite integrals yield a number, often representing area. The process involves finding the antiderivative and evaluating it at the given bounds.In our exercise, the definite integral is:\[ \int_{0}^{1} \tanh(2x) \, dx \]This integral evaluates the accumulated area of the \( \tanh(2x) \) function from \( x=0 \) to \( x=1 \), encapsulating the essence of definite integration.

Antidifferentiation

Antidifferentiation, or finding the indefinite integral, involves determining a function whose derivative is the given function. Essentially, it's the reverse process of differentiation. For hyperbolic functions like \( \tanh(u) \), this process can be simplified using known formulas.For example, the antidifferentiation of \( \tanh(u) \) is:\[ \int \tanh(u) \, du = \ln(\cosh(u)) + C \]Here, the constant \( C \) represents any constant since derivative differentiation of \( C \) is zero. When integrating \( \tanh(2x) \), the integral becomes:\[ \int \tanh(2x) \, dx = \frac{1}{2} \ln(\cosh(2x)) + C \]This result uses the substitution method implicitly because of the \( 2x \) inside the hyperbolic tangent, bringing us to the next essential point: the integration by substitution.

Integration by Substitution

Integration by substitution is a valuable technique used to simplify integration by changing variables. It is especially helpful when dealing with compositions of functions, like in our exercise. In our problem, substituting \( u = 2x \) simplifies \( \tanh(2x) \) to a form easier to integrate. This substitution changes the differential \( dx \) to \( \frac{1}{2} du \), aligning the integral for antidifferentiation:

• Let \( u = 2x \)
• Then \( du = 2 \, dx \), so \( dx = \frac{1}{2} \, du \)
• The integral becomes \( \frac{1}{2} \int \tanh(u) \, du \)

By using substitution, we reduce the complexity of the integral, making it easier to apply known antiderivatives. This leads to the final expression:\[ \frac{1}{2} \ln(\cosh(2x)) + C \]The definite integral benefits from substitution by computing the constant terms cleanly and directly.