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 Journals 33, No. 4 (September 2004))

Short answer

In summary, to evaluate the integral \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\), we first proved the property that if \(f\) is integrable on \([0, b],\) then \(\int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x\). Then, we used this property along with the given identity \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) to transform the given integral. After making the necessary substitutions, we found that \(\int_{0}^{\pi/4} \ln (1+\tan x)dx = 2\int_{0}^{\pi/8} \ln (1+\tan y)dy\).

Step-by-step solution

Proving the necessary property

Let \(f\) be integrable on the interval \([0, b]\). We want to show that $$\int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x.$$ To prove this, we will use the substitution method. Let \(u = b - x\). Notice, when \(x = 0\), \(u = b\) and when \(x = b\), \(u = 0\). Also, \(-du = dx\). Thus, we can rewrite the integral as follows: $$\int_{0}^{b} f(x) d x = \int_{b}^{0} f(b-u)(-du) = \int_{0}^{b} f(b-u) du$$ Now, let's divide the interval \([0, b]\) into two equal parts by setting \(x = \frac{b}{2}\). Then, using the additivity property of the integral, we can rewrite the integral in the following way: $$\int_{0}^{b} f(x) d x = \int_{0}^{b/2} f(x) dx + \int_{b/2}^{b} f(x) dx$$ For the second part of the integral, we will use the substitution \(u = b - x\). When \(x = \frac{b}{2}\), we have \(u = \frac{b}{2}\) and when \(x = b\), we have \(u = 0\). Thus, we can rewrite the second part of the integral as follows: $$\int_{b/2}^{b} f(x) dx = \int_{b/2}^{0} f(b - u)(-du) = \int_{0}^{b/2} f(b - u) du$$ Now, combining the two parts of the integral, we have: $$\int_{0}^{b} f(x) d x = \int_{0}^{b/2} f(x) dx + \int_{0}^{b/2} f(b - u) du$$ Let's make the substitution \(u = b - x\): $$\int_{0}^{b} f(x) d x = \int_{0}^{b/2} [f(x) + f(b - x)] dx$$ This proves part (a) of the exercise.

Evaluating the given integral

Now, we will use part (a) and the given identity \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) to evaluate the integral \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\): From the preceding result, we know that $$\int_{0}^{\pi / 4} \ln (1+\tan x)d x = \int_{0}^{\pi / 8} [\ln(1+\tan x)+\ln(1+\tan(\pi / 4-x))] d x$$ Using the given identity \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\), set \(\alpha = x\) and \(\beta = \pi / 4 - x\). Then, we have $$\tan x = \frac{\tan x+\tan (\pi / 4-x)}{1-\tan x \tan (\pi / 4-x)}$$ Solving for \(\tan (\pi / 4-x)\), we get $$\tan (\pi / 4-x) = \frac{1-\tan x}{\tan x}$$ Now, substitute this expression into the second logarithm: $$\ln (1+ \frac{1-\tan x}{\tan x}) = \ln (\frac{1}{\tan x})$$ Now, we can rewrite the integral as $$\int_{0}^{\pi / 4} \ln(1+\tan x)dx = \int_{0}^{\pi / 8} [\ln(1+\tan x)+ \ln(\frac{1}{\tan x})]dx$$ Using the logarithmic property, \(\ln (a) + \ln (b) = \ln (ab),\) we have $$\int_{0}^{\pi / 4} \ln(1+\tan x)dx = \int_{0}^{\pi / 8} [\ln((1+\tan x) \cdot (\frac{1}{\tan x}))]dx$$ Simplifying the expression inside the logarithm, we get $$\int_{0}^{\pi / 4} \ln(1+\tan x)dx = \int_{0}^{\pi / 8} [\ln(1+\cot x)]dx$$ Since \(\cot x=\frac{1}{\tan x}\), we can change the variable as \(y=\pi/4-x\). Thus, the integral becomes $$\int_{0}^{\pi / 4} \ln(1+\tan x)dx = \int_{0}^{\pi / 8} [\ln(1+\cot (\pi/4-y))]dy$$ Now, we have $$\int_{0}^{\pi / 4} \ln(1+\tan x)dx = \int_{0}^{\pi / 8} [\ln(1+\tan y)]dy$$ Since both integrals are equal, we can equate them and get $$I=2 \int_{0}^{\pi / 8} [\ln(1+\tan y)]dy$$ The integral value is doubled because we made a change of variables, effectively "folding" the integral in half. Therefore, the final answer is $$\int_{0}^{\pi / 4} \ln (1+\tan x) d x = 2\int_{0}^{\pi / 8} \ln (1+\tan y) d y.$$

Key concepts

Definite Integral

A definite integral is a fundamental concept in calculus used to calculate the accumulated area under a curve, symbolized by the integral sign with limits of integration. It gives the net signed area, accounting for both areas above and below the x-axis between two bounds.
This integral is written with an integral symbol, \(\int\), followed by a function and the differential \(dx\). The numbers on the lower and upper sides of the integral sign are the limits of integration, denoted as \(a\) and \(b\). For example, \(\int_{a}^{b} f(x) \, dx\) calculates the area under the curve \(f(x)\) from \(x = a\) to \(x = b\).
In definite integrals, the integrand, which is the function under the integral sign, is evaluated over an interval, resulting in a numerical value rather than another function. This concept is crucial for solving problems in physics, engineering, and economics, as it allows for calculating quantities like total distance traveled, total area, or accumulated growth over a specific period.

Substitution Method

Integration by substitution is a powerful technique that simplifies certain integrals, making them easier to evaluate. It's akin to reverse chain rule differentiation. Instead of integrating directly, we change the original integral into a more manageable form.
The key is finding a substitution, \(u\), that simplifies the integrand. Commonly, \(u\) is a function within the integrand whose derivative also appears. Once the substitution is made, the integral in terms of \(x\) transforms into an integral in terms of \(u\). Next, solve the simpler integral and substitute back \(x\) afterward.
Let's consider a simple example: \(\int x \, e^{x^2} \, dx\). Here, set \(u = x^2\) so that \(du = 2x \, dx\) or \(\frac{1}{2} du = x \, dx\). The integral becomes \(\int e^{u} \frac{1}{2} \, du\), which is much simpler to solve. After integrating, \(e^u\), reverse the substitution to return to the variable \(x\).
This method is particularly valuable in definite integrals, allowing the computation of new limits of integration in \(u\) terms, ensuring all transformations and calculations remain consistent and complete.

Integration Properties

Integration properties help simplify complex calculations and are essential tools for calculus students. These properties ensure integration operations are consistent, allowing for creative problem solving.

• Linearity: \(\int c \, f(x) \, dx = c \, \int f(x) \, dx\) and \(\int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx\). This property allows pulling constants out of the integral and splitting sums, making it easier to evaluate.
• Additivity: For definite integrals, if \(a\) is less than \(b\) and \(b\) is less than \(c\), then \(\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx\). This property allows breaking complex intervals into smaller pieces or rearranging them sequentially so as to integrate separately.
• Value at a point: The integral of zero is zero. If the integrand is zero in any sub-interval, it doesn't contribute to the integral value — conveying the significance of regions entirely under the x-axis contributing negatively.

Using these properties wisely can transform complex integrals like changing integration limits, confirming the answer's integrity or facing unconventional integrals, ultimately making them much easier to tackle.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions such as sine, cosine, and tangent that are true for all values of the included variables. These identities simplify integration problems, especially when integrating trigonometric expressions over defined boundaries.
Some helpful trigonometric identities include:

• Pythagorean Identity: \(\sin^2 x + \cos^2 x = 1\), which often appears when substituting or simplifying integrals.
• Angle Addition Formulas: Example, \(\tan( heta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \, \tan \phi}\), assists in transforming integrals that involve sum angles, making them manageable for further operations and evaluations.
• Reciprocal Identities: These express relationships like \(\tan x = \frac{\sin x}{\cos x}\) and \(\cot x = \frac{1}{\tan x}\), useful when simplifying or changing the nature of integrands in an integral.

By exploiting these identities, it's possible to re-write integrands into more calculable forms or uncover hidden patterns, making direct evaluation feasible.