Question

Prove that \(\int_{0}^{\pi/4} \log(1 + \tan x) \, dx = \frac{\pi}{8} \log 2\).

Hint:

For definite integrals of the form \(\int_{0}^{a} f(x) dx\), always check if applying the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx\) simplifies the integrand or relates the new integral back to the original one. This is a very common and powerful technique, especially with trigonometric and logarithmic functions.

Answer 1

Step 1: Understanding the Concept:
This problem involves evaluating a definite integral. The integrand \(\log(1 + \tan x)\) does not have a simple antiderivative. Therefore, we should look for properties of definite integrals that can simplify the problem. The "King Property" of definite integrals is particularly useful here.
Step 2: Key Formula or Approach:
We will use the following property of definite integrals, often called the King Property:
\[ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \] We also need the tangent subtraction formula:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \] And the logarithm properties:
\[ \log\left(\frac{m}{n}\right) = \log m - \log n \] Step 3: Detailed Explanation:
Let the given integral be \(I\).
\[ I = \int_{0}^{\pi/4} \log(1 + \tan x) dx \quad \cdots (1) \] Using the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx\), with \(a = \pi/4\):
\[ I = \int_{0}^{\pi/4} \log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right) dx \] Now, we use the tangent subtraction formula for \(\tan(\frac{\pi}{4} - x)\), knowing that \(\tan(\pi/4) = 1\):
\[ \tan\left(\frac{\pi}{4} - x\right) = \frac{\tan(\pi/4) - \tan x}{1 + \tan(\pi/4)\tan x} = \frac{1 - \tan x}{1 + \tan x} \] Substitute this back into the integral for \(I\):
\[ I = \int_{0}^{\pi/4} \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) dx \] Simplify the expression inside the logarithm:
\[ 1 + \frac{1 - \tan x}{1 + \tan x} = \frac{(1 + \tan x) + (1 - \tan x)}{1 + \tan x} = \frac{2}{1 + \tan x} \] So the integral becomes:
\[ I = \int_{0}^{\pi/4} \log\left(\frac{2}{1 + \tan x}\right) dx \] Using the logarithm property \(\log(m/n) = \log m - \log n\):
\[ I = \int_{0}^{\pi/4} (\log 2 - \log(1 + \tan x)) dx \] Split the integral into two parts:
\[ I = \int_{0}^{\pi/4} \log 2 dx - \int_{0}^{\pi/4} \log(1 + \tan x) dx \quad \cdots (2) \] We can see that the second term on the right-hand side is the original integral \(I\) from equation (1).
\[ I = \int_{0}^{\pi/4} \log 2 dx - I \] Now, we solve for \(I\):
\[ 2I = \int_{0}^{\pi/4} \log 2 dx \] Since \(\log 2\) is a constant, we can take it out of the integral:
\[ 2I = \log 2 \int_{0}^{\pi/4} 1 dx \] \[ 2I = \log 2 [x]_{0}^{\pi/4} \] \[ 2I = \log 2 \left(\frac{\pi}{4} - 0\right) \] \[ 2I = \frac{\pi}{4} \log 2 \] Finally, divide by 2 to find \(I\):
\[ I = \frac{\pi}{8} \log 2 \] Step 4: Final Answer:
We have shown that \(\int_{0}^{\pi/4} \log(1 + \tan x) dx = \frac{\pi}{8} \log 2\).