Evaluating Definite Integrals with Logarithms and Trigonometry

We are asked to find the value of the definite integral \( \int_{0}^{\frac{\pi}{4}} \log_e(1 + \tan x)dx \).

Let the given integral be denoted by \( I \). So, we have:

\( I = \int_{0}^{\frac{\pi}{4}} \log_e(1 + \tan x)dx \quad \dots (1) \)

To evaluate this integral, we can use a standard property of definite integrals:

Property: \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \)

In our case, \( a = \frac{\pi}{4} \) and \( f(x) = \log_e(1 + \tan x) \). Applying the property, we get:

\( I = \int_{0}^{\frac{\pi}{4}} \log_e(1 + \tan(\frac{\pi}{4} - x))dx \quad \dots (2) \)

Now, let's simplify the term \( \tan(\frac{\pi}{4} - x) \) using the trigonometric identity for \( \tan(A - B) \):

\( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)

Substituting \( A = \frac{\pi}{4} \) and \( B = x \), we get:

\( \tan(\frac{\pi}{4} - x) = \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} \)

Since \( \tan \frac{\pi}{4} = 1 \), the expression becomes:

\( \tan(\frac{\pi}{4} - x) = \frac{1 - \tan x}{1 + \tan x} \)

Now substitute this back into the expression inside the logarithm in integral (2):

\( 1 + \tan(\frac{\pi}{4} - x) = 1 + \frac{1 - \tan x}{1 + \tan x} \)

To combine the terms, find a common denominator:

\( 1 + \frac{1 - \tan x}{1 + \tan x} = \frac{1 \cdot (1 + \tan x)}{1 + \tan x} + \frac{1 - \tan x}{1 + \tan x} \)

\( = \frac{(1 + \tan x) + (1 - \tan x)}{1 + \tan x} \)

\( = \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} \)

\( = \frac{2}{1 + \tan x} \)

So, integral (2) becomes:

\( I = \int_{0}^{\frac{\pi}{4}} \log_e\left(\frac{2}{1 + \tan x}\right)dx \)

Using the logarithm property \( \log_e\left(\frac{a}{b}\right) = \log_e a - \log_e b \), we can split the logarithm:

\( I = \int_{0}^{\frac{\pi}{4}} (\log_e 2 - \log_e(1 + \tan x))dx \)

Now, we can split the integral into two parts:

\( I = \int_{0}^{\frac{\pi}{4}} \log_e 2 \, dx - \int_{0}^{\frac{\pi}{4}} \log_e(1 + \tan x)dx \)

Notice that the second integral on the right side is the original integral \( I \) from equation (1). So we have:

\( I = \int_{0}^{\frac{\pi}{4}} \log_e 2 \, dx - I \)

Now, we need to solve for \( I \). Add \( I \) to both sides of the equation:

\( I + I = \int_{0}^{\frac{\pi}{4}} \log_e 2 \, dx \)

\( 2I = \int_{0}^{\frac{\pi}{4}} \log_e 2 \, dx \)

Since \( \log_e 2 \) is a constant with respect to \( x \), we can take it out of the integral or integrate directly:

\( \int \log_e 2 \, dx = (\log_e 2) \cdot x + C \)

Now, evaluate the definite integral:

\( 2I = [\log_e 2 \cdot x]_{0}^{\frac{\pi}{4}} \)

\( 2I = (\log_e 2) \cdot \frac{\pi}{4} - (\log_e 2) \cdot 0 \)

\( 2I = \frac{\pi}{4} \log_e 2 - 0 \)

\( 2I = \frac{\pi}{4} \log_e 2 \)

Finally, divide both sides by 2 to find the value of \( I \):

\( I = \frac{1}{2} \cdot \frac{\pi}{4} \log_e 2 \)

\( I = \frac{\pi}{8} \log_e 2 \)

Thus, the value of the integral \( \int_{0}^{\frac{\pi}{4}} \log_e(1 + \tan x)dx \) is \( \frac{\pi}{8} \log_e 2 \).

Revision Table: Key Steps for Integral Evaluation

Additional Information on Definite Integral Properties

The property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \) used in this solution is a very powerful tool for evaluating certain types of definite integrals, especially those involving trigonometric functions over symmetric intervals like \( [0, \frac{\pi}{2}] \) or \( [0, \frac{\pi}{4}] \). Here's why it works:

• Consider the substitution \( u = a - x \). Then \( du = -dx \).
• When \( x = 0 \), \( u = a - 0 = a \).
• When \( x = a \), \( u = a - a = 0 \).
• So, \( \int_{0}^{a} f(x) dx = \int_{a}^{0} f(a-u) (-du) \).
• Using the property \( \int_{b}^{a} g(u) du = - \int_{a}^{b} g(u) du \), we get \( - \int_{a}^{0} f(a-u) du = \int_{0}^{a} f(a-u) du \).
• Since the variable of integration is a dummy variable, we can replace \( u \) with \( x \). Thus, \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \).

This property is particularly useful when \( f(x) + f(a-x) \) simplifies nicely, often to a constant or a form that makes the integration straightforward. In this problem, applying the property allowed us to express the original integral I in terms of itself and a simple integral of a constant, which could then be easily solved.