Question

Evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) using the following steps. a. If \(f\) is integrable on \([0, b],\) use substitution to show that $$\int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x$$ b. Use part (a) and the identity tan \((\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) to evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) (Source: The College Mathematics Journal, \(33,4,\) Sep 2004 )

Short answer

By showing that $\int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x$ and using the identity $\tan(\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$, we can evaluate the integral $\int_{0}^{\pi / 4} \ln (1+\tan x) d x$. The final result for the given integral is $\frac{\pi}{8} \ln(2)$.

Step-by-step solution

Substitution to show the equality

Let \(f\) be an integrable function on the interval \([0, b]\). Define \(g(x) = f(x) + f(b - x)\). We will apply a substitution to the integral \(\int_0^b g(x) dx\) to prove the result. Let \(u = b - x\). Then, \(du = -dx\), and when \(x=0\), \(u=b\), and when \(x=b\), \(u=0\). Consider the integral: $$\int_0^b g(x)dx = \int_0^b (f(x)+f(b-x))dx$$ Applying the substitution, we get $$\int_0^b g(x)dx =\int_b^0 (f(b-u) + f(u))(-du)$$ Since the limits of integration are reversed, we can rewrite this as $$\int_0^b g(x)dx = \int_0^b (f(u)+f(b-u))du$$ Comparing this to the original integral, we can see that it has the same form, but with a different variable. It follows that $$\int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x$$

Using the identity to evaluate the integral

Given the identity \(\tan(\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) , we have: $$\tan \left( x+\frac{\pi}{4} \right) = \frac{\tan x + \tan \frac{\pi}{4}}{1 - \tan x \tan \frac{\pi}{4}} = \frac{\tan x + 1}{1 - \tan x}$$ Now, let's choose \(f(x) = \ln (1 + \tan x)\) for the integral. Clearly, \([0, \frac{\pi}{4}] \subseteq [0, b]\). We want to evaluate $$\int_{0}^{\pi / 4} \ln (1+\tan x) d x$$ Using the result from Step 1, we have: $$\int_{0}^{\pi / 4} \ln (1+\tan x) d x = \int_{0}^{\pi / 8}(\ln(1+\tan x) + \ln(1+\tan(\tfrac{\pi}{4}-x)) ) d x$$ Using the identity, we can rewrite the second term in the sum as: $$\ln(1+\tan(\tfrac{\pi}{4}-x)) = \ln \left( 1 + \frac{\tan \tfrac{\pi}{4} - \tan x}{1 + \tan x} \right) = \ln \left( \frac{1 + \tan \tfrac{\pi}{4}}{1 + \tan x} \right)$$ Now, we can simplify the integral by combining the logarithmic terms: $$\int_{0}^{\pi / 4} \ln (1+\tan x) d x = \int_{0}^{\pi / 8}(\ln(1+\tan x) + \ln \left( \frac{1 + \tan \tfrac{\pi}{4}}{1 + \tan x} \right) ) d x = \int_{0}^{\pi / 8} \ln \left( \frac{1+\tan x}{1+\tan x} \cdot \frac{1 + \tan \tfrac{\pi}{4}}{1 + \tan x} \right) d x$$ Simplifying and using the properties of logarithms, we have: $$\int_{0}^{\pi / 4} \ln (1+\tan x) d x = \int_{0}^{\pi / 8} \ln \left( \frac{1 + \tan \tfrac{\pi}{4}}{1 + \tan x} \right) d x = \int_{0}^{\pi / 8} \ln(2) d x$$ The last integral is easily evaluated as $$\int_{0}^{\pi / 4} \ln (1+\tan x) d x = \Big[ x \ln(2) \Big]_{0}^{\pi/8} = \frac{\pi}{8} \ln(2)$$ Thus, the answer for the given exercise is $$\int_{0}^{\pi / 4} \ln (1+\tan x) d x = \frac{\pi}{8} \ln(2)$$

Key concepts

Integration Techniques

Integral calculus is a fundamental part of mathematics that deals with finding the area under a curve, among many other applications. To evaluate a given integral, mathematicians have developed various integration techniques. These techniques include but are not limited to the power rule, integration by parts, trigonometric substitution, partial fractions, and the substitution method.

Each technique serves a specific type of function or scenario. The power rule is straightforward when dealing with polynomial functions. Integration by parts is useful when an integral is the product of two functions. Trigonometric substitution works well with integrals involving square roots and trigonometric functions, while partial fractions deal with rational functions.

The use of these techniques simplifies complex integrals into more manageable forms that can often be integrated directly. It's essential for students to understand when and how to use each of these techniques to correctly evaluate integrals.

Substitution Method

The substitution method is a powerful tool in integration, particularly suitable when an integral includes a composite function—a function inside another function. It simplifies the integral by substituting a part of the integrand with a new variable, usually denoted as u. The idea is to transform the integral into a form that is easier to evaluate.

The choice of u is crucial and often guided by parts of the function that have a noticeable derivative present elsewhere in the integrand. After substituting, we find du, which allows us to replace dx in the integral. This process often simplifies the integrand and changes the limits of integration if they are defined. Once the integral is evaluated, the substitution is reversed to express the antiderivative in terms of the original variable.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all acceptable values of the involved variables. These identities are indispensable in simplifying and evaluating expressions and integrals involving trigonometric functions.

Commonly used trigonometric identities include the Pythagorean identities, angle sum and difference identities, double angle formulas, and half-angle formulas. For example, the angle sum identity for tangent, which states that \( \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \), is vital in integrating functions where a simple substitution does not suffice.

By transforming the integrand using these identities, one can often convert a complex trigonometric integrand into a simpler form that is easier to integrate, leveraging the structural patterns in trigonometry.

Properties of Logarithms

Logarithms are another critical mathematical function often encountered in calculus. The properties of logarithms help to manipulate and simplify logarithmic expressions before integration. Understanding these properties can make a significant difference in evaluating integrals that involve logarithmic functions.

Some fundamental properties include the product property \( \ln(ab) = \ln(a) + \ln(b) \) and the quotient property \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \). There's also the power property \( \ln(a^k) = k\ln(a) \), which is useful when the logarithm is raised to a power.

These properties allow the simplification of the integrand into a form that may be straightforward to integrate, as seen in the \( \ln(1 + \tan(x)) \) integration problem where the logarithm of a quotient is converted into the difference of two logarithms, ultimately simplifying the integration process.