Question

Evaluate the following integrals. $$\int_{\ln 2}^{\ln 3} \frac{e^{x}+e^{-x}}{e^{2 x}-2+e^{-2 x}} d x$$

Short answer

The integral equals $\ln(2) - (\arctan(3) - \arctan(2))$.

Step-by-step solution

Change of Variables

Let's substitute \(t = e^x\). This means \(x = \ln t\) and \(d x = \frac{1}{t} dt\). We also need to change the limits of integration. When \(x = \ln 2\), we have \(t = e^{\ln 2}=2\). And when \(x = \ln 3\), we have \(t = e^{\ln 3}=3\). Now the integral becomes: $$\int_{2}^{3} \frac{t + \frac{1}{t}}{t^2 - 2 + \frac{1}{t^2}} \frac{1}{t} d t$$

Simplify the integral

To simplify the integral, let's first combine the terms and multiply \(\frac{1}{t}\) with the function within the integral. $$\int_{2}^{3} \frac{t^2 + 1}{t^3 - 2t + 1} d t$$

Integrate using Partial Fractions

Now, let's try to integrate this function. We can use partial fractions to break down the fraction and obtain an easier integrand. Let the fraction be equal to the sum of two other fractions: $$\frac{t^2 + 1}{t^3 - 2t + 1} = \frac{A}{t - 1} + \frac{Bt + C}{t^2 + 1}$$ Multiplying both sides by the denominator \((t^3 - 2t + 1)\), we get: $$t^2 + 1 = A(t^2 + 1) + (Bt + C)(t - 1)$$

Solve for A, B, and C

To find the values of A, B, and C, we can plug in specific values of t. For \(t=1\), we get \(A(1^2 + 1) = 1 + 1 \Rightarrow A = 1\). Now we let \(t=0\) and we have \(1 = -C \Rightarrow C = -1\). Finally, from the original equation, we know that the coefficient of \(t^2\) on both sides should be equal, thus: $$1 = A + B \Rightarrow B = 0$$. Now, we rewrite the integrand: $$\frac{t^2 + 1}{t^3 - 2t + 1} = \frac{1}{t - 1} - \frac{1}{t^2 + 1}$$.

Integrate terms separately

Now we can integrate the two terms separately: $$\int_{2}^{3} \frac{t^2 + 1}{t^3 - 2t + 1} d t = \int_{2}^{3} \frac{1}{t - 1} d t -\int_{2}^{3} \frac{1}{t^2 + 1} d t$$ The first integral is the natural logarithm function. The second integral can be solved using the inverse tangent integration formula: $$\int_{2}^{3} \frac{1}{t - 1} d t -\int_{2}^{3} \frac{1}{t^2 + 1} d t = [\ln(|t - 1|)]_{2}^{3} - [\arctan(t)]_{2}^{3}$$

Evaluate the integral at the limits

Now, we substitute the limits of integration and find the final result: $$[\ln(|t - 1|)]_{2}^{3} - [\arctan(t)]_{2}^{3} = \ln(2) - \ln(1) - [\arctan(3) - \arctan(2)]$$ Combine the terms to get the final result: $$\int_{\ln 2}^{\ln 3} \frac{e^{x}+e^{-x}}{e^{2 x}-2+e^{-2 x}} d x = \ln(2) - (\arctan(3) - \arctan(2))$$

Key concepts

Change of Variables

Change of variables is a powerful tool in integral calculus. It allows us to transform a given integral into a form that's easier to evaluate. In this problem, we start with an integral that involves exponential functions. We substitute a new variable, which simplifies the integrand. The substitution here is \( t = e^x \), resulting in \( x = \ln t \). The differential also changes accordingly: \( dx = \frac{1}{t}dt \).

Beyond changing the variable, we also need to adjust the limits of integration. For our original limits, \( x = \ln 2 \) becomes \( t = 2 \), and \( x = \ln 3 \) becomes \( t = 3 \). This substitution makes the integrand more manageable and easier to solve. The technique of changing variables is commonly applied when integrals include exponential or trigonometric functions, as it simplifies expressions by transforming them into familiar forms.

Partial Fractions

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions. Once simplified, these fractions are easier to integrate. For our specific problem, we're dealing with the integrand \( \frac{t^2 + 1}{t^3 - 2t + 1} \).

The goal is to express this fraction as a sum of simpler fractions. Here, it's broken down into two terms: \( \frac{A}{t - 1} + \frac{Bt + C}{t^2 + 1} \).

• We solve for \( A \), \( B \), and \( C \) by equating coefficients or substituting specific values that simplify the equations.
• In our exercise, substituting values like \( t = 1 \) and \( t = 0 \) gave us solutions for \( A \) and \( C \)> \( B \) was then determined through algebraic manipulation.

This decomposition allows us to integrate each term separately, simplifying what could seem a daunting task. It's a fundamental technique in calculus that finds application in various fields, such as control theory and electrical engineering.

Definite Integration

Definite integration provides us with the area under the curve for a specific interval. This concept is central in the exercise, where we evaluate the integral from one boundary to another. After simplifying the integrand using a substitution and partial fractions, we integrate over an interval from \( t = 2 \) to \( t = 3 \).

The definite integral of \( f(t) \) over \([a, b] \) is given by evaluating the antiderivative, \( F(t) \), at these bounds: \[ F(b) - F(a) \].

• In the exercise, once our function was broken down, we took antiderivatives of each component separately.
• After obtaining these antiderivatives, the limits \( t = 2 \) and \( t = 3 \) were substituted to find the areas under these transformed curves.

In summary, definite integration transforms an integrand into a solvable form over a given interval. It's essential in calculating exact quantities, such as work or probability, in applied mathematics.

Trigonometric Substitution

While trigonometric substitution wasn't specifically used in this exercise, it is a valuable method worth understanding. It's particularly useful when dealing with integrals involving algebraic expressions that contain trigonometric identities.

This method uses trigonometric identities to simplify integrands containing square roots or other complex expressions. Here's a simple breakdown:

• Choose a trigonometric identity that matches the form of the integrand. For example, \( \sqrt{a^2 - x^2} \) can be simplified using the identity \( x = a \sin(\theta) \).
• Substitute and convert the original variable into a trigonometric function.
• Change the limits of integration if it's a definite integral, based on the substitution.

Using these identities makes it easier to evaluate certain integrals by transforming them into a more straightforward form that often can be solved using basic calculus techniques. Understanding trigonometric substitution adds another tool to handle complex integration challenges.