Question

Prove that: \(\int_0^{\pi/4} \log_e(1 + \tan \theta) \, d\theta = \frac{\pi}{8} \log_e 2\).

Hint:

The "King Property" (\(\int_0^a f(x) dx = \int_0^a f(a-x) dx\)) is extremely useful for definite integrals, especially when the integrand involves trigonometric functions with limits like \(0\) to \(\pi/2\) or \(0\) to \(\pi/4\). Adding the original integral \(I\) to the transformed integral is a common trick that often leads to simplification.

Answer 1

Step 1: Understanding the Concept:
This is a problem of evaluating a definite integral. A common and powerful technique for integrals of the form \(\int_0^a f(x) dx\) is to use the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\). This property is sometimes called the "King Property".
Step 2: Key Formula or Approach:
1. Let the given integral be \(I\).
2. Apply the property \(\int_0^a f(\theta) d\theta = \int_0^a f(a-\theta) d\theta\), where \(a = \pi/4\).
3. Use the trigonometric identity \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\).
4. Simplify the resulting integral and add it to the original integral \(I\) to solve for \(I\).
Step 3: Detailed Explanation or Calculation:
Let the integral be \(I\): \[ I = \int_0^{\pi/4} \log_e(1 + \tan \theta) d\theta . (1) \] Using the property \(\int_0^a f(\theta) d\theta = \int_0^a f(a-\theta) d\theta\), with \(a = \pi/4\): \[ I = \int_0^{\pi/4} \log_e(1 + \tan(\pi/4 - \theta)) d\theta \] We know the identity \(\tan(\pi/4 - \theta) = \frac{\tan(\pi/4) - \tan \theta}{1 + \tan(\pi/4) \tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta}\).
Substituting this into the integral: \[ I = \int_0^{\pi/4} \log_e\left(1 + \frac{1 - \tan \theta}{1 + \tan \theta}\right) d\theta \] Simplify the term inside the logarithm: \[ I = \int_0^{\pi/4} \log_e\left(\frac{(1 + \tan \theta) + (1 - \tan \theta)}{1 + \tan \theta}\right) d\theta \] \[ I = \int_0^{\pi/4} \log_e\left(\frac{2}{1 + \tan \theta}\right) d\theta \] Using the logarithm property \(\log(a/b) = \log a - \log b\): \[ I = \int_0^{\pi/4} (\log_e 2 - \log_e(1 + \tan \theta)) d\theta \] Split the integral into two parts: \[ I = \int_0^{\pi/4} \log_e 2 \, d\theta - \int_0^{\pi/4} \log_e(1 + \tan \theta) d\theta . (2) \] From equation (1), we know that the second term is \(I\). So, we have: \[ I = \int_0^{\pi/4} \log_e 2 \, d\theta - I \] \[ 2I = \int_0^{\pi/4} \log_e 2 \, d\theta \] Since \(\log_e 2\) is a constant: \[ 2I = \log_e 2 \int_0^{\pi/4} 1 \, d\theta = \log_e 2 [\theta]_0^{\pi/4} \] \[ 2I = \log_e 2 (\pi/4 - 0) = \frac{\pi}{4} \log_e 2 \] Solving for \(I\): \[ I = \frac{\pi}{8} \log_e 2 \] Step 4: Final Answer:
We have successfully proven that \(\int_0^{\pi/4} \log_e(1 + \tan \theta) d\theta = \frac{\pi}{8} \log_e 2\).