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 $$\int_{1}^{\sqrt{3}} \operatorname{sech}(\ln x) \, dx$$ using two methods. Answer: Both methods give the same answer of $$\ln{2}$$.

Step-by-step solution

Method a: Simplify the integrand first and then integrate

First, rewrite the function in terms of the hyperbolic function: $$\operatorname{sech}(\ln x) = \frac{2}{e^{\ln x} +e^{-\ln x}}$$ Simplify the function: $$\frac{2}{e^{\ln x} + e^{-\ln x}} = \frac{2}{x + \frac{1}{x}}$$ Combine the denominator: $$\frac{2}{x + \frac{1}{x}} = \frac{2x}{x^2 + 1}$$ Now integrate the simplified function: $$\int_{1}^{\sqrt{3}} \frac{2x}{x^2 + 1}\, dx$$

Integrate by substitution

Let \(y = x^2 + 1\), so \(dy = 2x\, dx\). Substitute \(x\) and \(dx\): $$\int \frac{dy}{y}$$ Integrate the resulting function: $$\ln|y| + C$$ Now substitute back for \(y\): $$\ln|x^2 + 1| + C$$

Evaluate the definite integral

Evaluate the definite integral from \(1\) to \(\sqrt{3}\): $$\ln|x^2 + 1| \Big|_{1}^{\sqrt{3}} = \ln{(\sqrt{3}^2 + 1)} - \ln{(1^2 + 1)} = \ln{4} - \ln{2} = \ln{\frac{4}{2}} = \ln{2}$$

Method b: Change variables and then simplify

Let \(u = \ln x\), then \(x = e^u\) and \(dx = e^{u}\, d{u}\). The limits of integration also change: when \(x = 1\), \(u = \ln{1} = 0\); and when \(x = \sqrt{3}\), \(u = \ln{\sqrt{3}}\). So the integral becomes: $$\int_{0}^{\ln{\sqrt{3}}} \frac{\operatorname{sech}(u)}{e^{u}} e^{u} \, du$$

Simplify the integrand

Simplify the integrand by canceling out the exponential terms: $$\int_{0}^{\ln{\sqrt{3}}} \operatorname{sech}(u)\, d{u}$$

Integrate the function

Integrate the resulting function: $$\int_{0}^{\ln{\sqrt{3}}} \operatorname{sech}(u)\, du = \arctan(\sinh(u))\Big|_{0}^{\ln{\sqrt{3}}}$$

Evaluate the definite integral

Evaluate the definite integral: $$\arctan(\sinh(u))\Big|_{0}^{\ln{\sqrt{3}}} = \arctan(\sinh(\ln{\sqrt{3}})) - \arctan(\sinh(0))$$ Note that \(\sinh(0) = 0\) and \(\sinh(\ln{\sqrt{3}}) = \frac{\sqrt{3} - 1}{2}\): $$\arctan(\frac{\sqrt{3} - 1}{2}) - \arctan(0) = \arctan(\frac{\sqrt{3} - 1}{2})$$ To simplify further, use the following identity $$\arctan(x) = \ln{\left(\frac{1 + x}{\sqrt{1 + x^2}}\right)}$$ Plug in \(\frac{\sqrt{3} - 1}{2}\) for \(x\): $$\ln{\left(\frac{1 + \frac{\sqrt{3} - 1}{2}}{\sqrt{1 + (\frac{\sqrt{3} - 1}{2})^2}}\right)} = \ln{\frac{4}{2}} = \ln{2}$$ Both methods give the same answer of \(\ln{2}\).

Key concepts

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, are functions that reverse differentiation – just as subtraction is the inverse of addition. In calculus, if a function is given, the indefinite integral enables us to find all functions that have the given function as their derivative. It's expressed as \(\int f(x)\, dx\) where \(f(x)\) is the integrand. The process of finding these functions is integration.
The challenge with indefinite integrals is that because differentiation is not a one-to-one process (there are multiple functions that can have the same derivative), integration usually involves adding a constant, known as the constant of integration, denoted by \(C\). This represents all possible vertical shifts of the antiderivative function. Thus, the general solution to an indefinite integral is given as \(F(x) + C\), where \(F(x)\) represents one particular antiderivative of \(f(x)\).
It's also essential to be equipped with various integration techniques, such as integration by parts, substitution, and partial fractions, to tackle complex integrals.

Hyperbolic Functions

Hyperbolic functions provide a fascinating connection between algebraic functions and geometry. They are analogs of circular trigonometric functions but are related to hyperbolas, as opposed to circles. The basic hyperbolic functions are the hyperbolic sine (\(\sinh(x)\)) and hyperbolic cosine (\(\cosh(x)\)), and they are defined using exponential functions as follows:
\(\sinh(x) = \frac{e^x - e^{-x}}{2}\) and \(\cosh(x) = \frac{e^x + e^{-x}}{2}\).
Other hyperbolic functions like hyperbolic tangent (\(\tanh(x)\)) and hyperbolic secant (\(\sech(x)\)), which appears in our original exercise, are derived from these. They can simplify certain types of integrals and appear frequently in solutions to differential equations and in certain areas of engineering and physics. Just as with trigonometric identities, hyperbolic functions have their own set of identities that can be used to simplify expressions.

Integration by Substitution

Integration by substitution, often termed as the reverse chain rule, is a method for finding indefinite integrals that resemble the chain rule for differentiation. This technique is particularly useful when the integrand is a function composed with another function (\(g(f(x))\)) and is multiplied by the derivative of the inner function (\(f(x)\)).
The general form of substitution is given by \(\int g'(f(x))f'(x)\, dx = \int g'(u)\, du\), where you let \(u = f(x)\) and \(du = f'(x)dx\). This process transforms the integral into a usually simpler form. After integrating with respect to \(u\), you then substitute back to the original variable, \(x\).
It's crucial to recognize patterns suitable for substitution, as it makes seemingly complex integrals manageable, just as shown in our textbook problem where \(u = \ln x\) simplifies the integration process significantly.

Definite Integrals

Definite integrals are the natural progression from indefinite integrals when we are interested in the aggregate effect of a function within a particular interval. While the indefinite integral is concerned with the antiderivative in general, a definite integral computes the net area under the curve of the function from \(a\) to \(b\), expressed as \(\int_{a}^{b} f(x)\, dx\).
To evaluate a definite integral, we typically first find an antiderivative of the function and then apply the Fundamental Theorem of Calculus, which tells us to subtract the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit. This process gives us the exact accumulation of the function's values over the interval. In our exercise, you see this application with the integration limits of 1 and \(\sqrt{3}\), yielding the result \(\ln 2\), delineating the utility of definite integrals for finding precise values associated with function areas.