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. $$\int \frac{\sinh (\ln x)}{x} d x$$

Short answer

In conclusion, we found that the integral $$\int \frac{\sinh (\ln x)}{x} dx$$ can be computed using both methods: Method 1: Simplify the integrand first and then integrate, resulting in the expression $$\frac{1}{2}x + \frac{1}{2x} + C_1$$. Method 2: Change variables by letting \(u = \ln x\), integrate with respect to \(u\), and then simplify the answer, resulting in the expression $$\frac{1}{2}x + \frac{1}{2x} + C_2$$. Both methods give the same result, confirming the equivalence of these approaches in solving the given integral.

Step-by-step solution

Simplify the integrand

Let's simplify the integrand using the definition of \(\sinh\): $$\sinh(\ln x) = \frac{e^{\ln x} - e^{-\ln x}}{2}$$ Now we can substitute this into the integral expression: $$\int \frac{\sinh (\ln x)}{x}dx = \int \frac{\frac{e^{\ln x} - e^{-\ln x}}{2}}{x}dx$$ Simplify further: $$\int \frac{e^{\ln x} - e^{-\ln x}}{2x}dx$$

Integrate

Now we can integrate with respect to \(x\): $$\int \frac{e^{\ln x} - e^{-\ln x}}{2x}dx = \frac{1}{2}\int \frac{e^{\ln x}}{x}dx - \frac{1}{2}\int \frac{e^{-\ln x}}{x}dx$$ Notice that \(e^{\ln x} = x\) and \(e^{-\ln x} = \frac{1}{x}\). So we have: $$\frac{1}{2}\int \frac{x}{x}dx - \frac{1}{2}\int \frac{\frac{1}{x}}{x}dx = \frac{1}{2}\int dx - \frac{1}{2}\int \frac{1}{x^2}dx$$ Now we can integrate: $$\frac{1}{2}x - \frac{1}{2}\int \frac{1}{x^2}dx = \frac{1}{2}x - \frac{1}{2}(-\frac{1}{x}) + C_1 = \frac{1}{2}x + \frac{1}{2x} + C_1$$ Method 2: Change variables (let \(u=\ln x\)), integrate, and then simplify.

Change of variables

Let \(u=\ln x\). Then we have \(x=e^u\) and \(\frac{dx}{du}=e^u\). So the integral becomes: $$\int \frac{\sinh (\ln x)}{x}dx = \int \frac{\sinh (u)}{e^u}e^u du = \int \sinh(u) du$$

Integrate with respect to \(u\)

Now integrate with respect to \(u\): $$\int \sinh(u) du = \cosh(u) + C_2$$

Substitute back \(x\)

Now we substitute \(u=\ln x\) back into the integral: $$\cosh(u) + C_2 = \cosh(\ln x) + C_2$$ Using the definition of \(\cosh\): $$\frac{e^{\ln x} + e^{-\ln x}}{2} + C_2 = \frac{x + \frac{1}{x}}{2}+ C_2 = \frac{1}{2}x + \frac{1}{2x} + C_2$$ Comparing the two methods, we see that the results are the same: $$\frac{1}{2}x + \frac{1}{2x} + C_1 = \frac{1}{2}x + \frac{1}{2x} + C_2$$ So, we have verified that both methods give the same answer for the given integral.

Key concepts

Integration by substitution

Integration by substitution is a widely used method in integral calculus that involves changing variables to simplify the integration process. It is especially useful when dealing with complex integrands or functions that are not easily integrable in their original form. The method works similarly to the chain rule in differentiation, aiming to transform the integral into a simpler or standard form.
This technique involves choosing a new variable, usually denoted by \( u \), which is a function of the original variable, often \( x \). Then, the differential \( dx \) is rewritten in terms of \( du \), changing the entire integrand into an expression involving \( u \) which is simpler to integrate.
Once integrated, it's important to substitute back the original variable using the inverse of the substitution we performed to complete the process. This technique was used in Method 2 of the solution, where \( u = \ln x \) was substituted to simplify the hyperbolic function integration.

Hyperbolic functions

Hyperbolic functions are analogs of ordinary trigonometric functions but are based on hyperbolas instead of circles. The most common ones are the hyperbolic sine, \( \sinh(x) \), and hyperbolic cosine, \( \cosh(x) \). These functions are valuable in many areas of mathematics, including calculus and complex analysis.
The hyperbolic sine function, \( \sinh(x) \), is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). This definition was crucial in simplifying the integrand in the initial step of the solution. Similarly, the hyperbolic cosine function, \( \cosh(x) \), is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). In the context of our problem, substituting back into this definition verified the equivalence of the two integration methods.

• These functions behave somewhat similar to their trigonometric counterparts but are fundamentally different and have unique properties and uses.

Change of variables

The change of variables is a fundamental strategy in calculus that involves replacing a variable with a new one to simplify the problem at hand. In integration, this technique often takes the form of substitution which can transform complex integrals into more manageable forms. It allows mathematicians and scientists to handle problems that would be otherwise cumbersome or impossible to solve directly.
For the original exercise, the change of variable \( u = \ln x \) was employed. This choice allowed the integral involving the hyperbolic function \( \sinh(\ln x) \) to be reinterpreted as a function of \( u \), specifically \( \sinh(u) \), which is straightforward to integrate. The integral of \( \sinh(u) \) is known to be \( \cosh(u) \), simplifying the integration significantly.
After integrating, we switch back to the original variable \( x \) by substituting \( u = \ln x \), ensuring that the answer remains consistent with the original form of the integral.

Integration techniques

Various integration techniques are utilized depending on the form and complexity of the function involved. They include standard methods like basic integration and substitution, as well as more advanced methods such as integration by parts and partial fraction decomposition.
In solving the integral \( \int \frac{\sinh(\ln x)}{x} dx \), two techniques were applied:

• **Simplifying the integrand:** This involved using definitions of hyperbolic functions to directly transform and reduce complexity.
• **Substitution:** In this problem, substitution facilitated moving the integral problem from one involving a tricky hyperbolic function to a standard integral \( \int \sinh(u) \, du \), which is well-known.

Both techniques ultimately provided the same solution, affirming their consistency and accuracy. The choice of method often depends on the individual's preference or the specific features of the integral being solved.