Question

Evaluate the following integrals two ways. a. Simplify the integrand first, and then integrate. b. Change variables (let \(u=\ln x\) ), integrate, and then simplify your answer. Verify that both methods give the same answer. \(\int_{1}^{\sqrt{3}} \frac{\operatorname{sech}(\ln x)}{x} d x\)

Short answer

Question: Evaluate the integral \(\int_{1}^{\sqrt{3}} \frac{\operatorname{sech}(\ln x)}{x} d x\) using two different methods and verify that the answers are equivalent. Answer: The results from both methods, simplifying the integrand first and changing variables (u-substitution), are equivalent and the final result is \(\frac{\pi}{2}\).

Step-by-step solution

Method 1: Simplify the integrand first

First, we are going to simplify the integrand, which contains the hyperbolic secant function. Remember that \(\operatorname{sech}(t) = \frac{2}{e^t + e^{-t}}\). We can rewrite the integrand with this definition: \(\frac{\operatorname{sech}(\ln x)}{x} = \frac{2}{x(e^{\ln x} + e^{-\ln x})}\) Since \(e^{\ln x} = x\) and \(e^{-\ln x} = \frac{1}{x}\), we can now rewrite again our integrand: \(\frac{\operatorname{sech}(\ln x)}{x} = \frac{2}{x(x + \frac{1}{x})}\) Simplify further: \(\frac{\operatorname{sech}(\ln x)}{x} = \frac{2}{x^2 + 1}\) Now we can integrate: \(\int_{1}^{\sqrt{3}} \frac{2}{x^2 + 1} d x\)

Integrate simplified expression

To integrate, we will use the following formula: \(\int \frac{A}{x^2 + B^2} dx = \frac{A}{B} \arctan(\frac{x}{B}) + C\) In our case, A = 2 and B = 1, so our integral evaluates to: \(2 \int_{1}^{\sqrt{3}} \frac{1}{x^2 + 1} d x = 2 (\arctan(\frac{x}{1}))\Big|_1^{\sqrt{3}}\) Substitute the upper and lower limits of the integral: \(2 (\arctan(\sqrt{3}) - \arctan(1))\) Now, since arctan(1) = \(\frac{\pi}{4}\) and arctan(√3) = \(\frac{\pi}{3}\), we can compute the result: \(2 (\frac{\pi}{3} - \frac{\pi}{4}) = \frac{\pi}{2}\) The result of method 1 is \(\frac{\pi}{2}\).

Method 2: Change of variables (u-substitution)

Now, let's solve the integral using the change of variables. We will substitute \(u = \ln x\). Then, we calculate the derivative: \(\frac{du}{dx} = \frac{1}{x} \Rightarrow du = \frac{1}{x}dx\) Next, we need to update the integral limits to match the new variable u. We substitute x with 1 and √3 and calculate the corresponding u values: \(u_1 = \ln(1) = 0\) and \(u_2 = \ln(\sqrt{3}) = \frac{\ln(3)}{2}\) Now, with the substitution, our integral looks like: \(\int_{0}^{\frac{\ln(3)}{2}} \frac{2}{e^u + e^{-u}} du\) Notice that we already have our integrand simplified, now we just need to integrate: \(\int_{0}^{\frac{\ln(3)}{2}} \frac{2}{e^u + e^{-u}} du = 2 (\arctan(e^u))\Big|_0^{\frac{\ln(3)}{2}}\) Substitute the limits: \(2 (\arctan(e^{\frac{\ln(3)}{2}}) - \arctan(e^0))\) Since \(e^{\frac{\ln(3)}{2}} = \sqrt{3}\) and \(e^0 = 1\), we can compute the result: \(2 (\arctan(\sqrt{3}) - \arctan(1)) = 2 (\frac{\pi}{3} - \frac{\pi}{4}) = \frac{\pi}{2}\) The result of method 2 is \(\frac{\pi}{2}\).

Comparing the results

Now, let's compare the results from both methods: Method 1: \(\frac{\pi}{2}\) Method 2: \(\frac{\pi}{2}\) Both methods give us the same answer, which verifies that the two methods are equivalent. The final result is \(\frac{\pi}{2}\).

Key concepts

u-substitution

Understanding u-substitution is crucial for evaluating integrals, especially when the integrand involves composite functions. The main idea is to simplify the integral by changing the variable of integration to make the function easier to integrate. This technique essentially involves choosing a new variable, say \( u \), to replace a part of the integrand. This is often done when the integrand contains a function whose derivative is present nearby within the integral.

For example, take the function \( \ln x \) in the exercise. By setting \( u = \ln x \), and recognizing that the derivative of \( \ln x \) is \( \frac{1}{x} \), we can effectively substitute \( \, du = \frac{1}{x} dx \). This substitution transforms an integral that was challenging to compute in terms of \( x \) into one in terms of \( u \), which often is more straightforward.

After changing variables, you will typically need to adjust the limits of integration to reflect the new variable \( u \). As shown in our example, if \( x = 1 \), then \( u = 0 \), and if \( x = \sqrt{3} \), then \( u = \frac{\ln(3)}{2} \). Once the substitution is complete, the integral can be completed within this simpler framework, eventually translating results back to the original variable's context (if needed). Remember that the primary goal of u-substitution is to simplify the integration process and solve the problem more efficiently.

Hyperbolic functions

Hyperbolic functions, such as hyperbolic secant (\(\operatorname{sech}\)), often appear in calculus and have similar properties to trigonometric functions but are defined in terms of exponential functions. The hyperbolic secant function is defined as \( \operatorname{sech}(t) = \frac{2}{e^t + e^{-t}} \). These functions find various applications in real-world scenarios like in the calculations of angles and distances in hyperbolic geometry and in the modeling of certain physical phenomena.

In solving integrals that involve hyperbolic functions, it's often helpful to first express these functions in terms of exponentials. This step can allow a simplification of the integrand, making the integral easier to solve. In our exercise, knowing the identity for \( \operatorname{sech}(\ln x) \) allows us to replace it with exponential terms, subsequently simplifying to a form that is easier to integrate, \( \frac{2}{x^2 + 1} \).

The simplification using the exponential form is vital because it transforms a potentially complex-looking expression into a more recognizable form that associates closely with inverse trigonometric functions. This process exemplifies the power of recognizing and converting function identities as necessary for integration.

Definite integrals

Definite integrals allow the computation of the area under the curve of a function, providing a numerical value as a result. The integral is evaluated over a specific interval, giving it limits. These limits are substituted after evaluating the indefinite integral, which usually involves a constant of integration.

In our exercise, by integrating \( \frac{2}{x^2 + 1} \) over the limits from \( x = 1 \) to \( x = \sqrt{3} \), we find that the outcome reflects an area bounded by these lines on a graph of the function. Calculating definite integrals fundamentally involves determining the antiderivative on this interval and substituting the upper and lower bounds to find the exact area.

For verification purposes like in our exercise, conducting the definite integration using two methods — simplifying first versus using substitutions — allows not only for solving the integral but also for verifying correctness of both approaches by comparing results. The methods showed, despite their differences in process, both arrive at the same correct result, illustrating that multiple valid strategies can be pursued in solving such problems.