Question

Evaluate the integral: \[ I = \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \,dx. \]

• A. \( \pi \log 2 + 1 \)
• B. \( \frac{\pi}{2} \log 2 + 1 \)
• C. \( \frac{\pi}{4} \log 2 \)
• D. \( \frac{\pi}{8} \log 2 \)

Hint:

For definite integrals involving logarithms and trigonometric functions, use the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \) to simplify calculations.

Answer 1

The Correct Option is D

Step 1: Using the property of definite integrals We use the standard property: \[ \int_0^a f(x) \,dx = \int_0^a f(a - x) \,dx. \] Setting \( f(x) = \log(1 + \tan x) \), we substitute \( x \to \frac{\pi}{4} - x \): \[ I = \int_0^{\frac{\pi}{4}} \log(1 + \tan (\frac{\pi}{4} - x)) \,dx. \] Using the identity: \[ \tan (\frac{\pi}{4} - x) = \frac{1 - \tan x}{1 + \tan x}, \] we rewrite the integral as: \[ I = \int_0^{\frac{\pi}{4}} \log \left( 1 + \frac{1 - \tan x}{1 + \tan x} \right) \,dx. \] Simplifying: \[ 1 + \frac{1 - \tan x}{1 + \tan x} = \frac{2}{1 + \tan x}. \] Thus, \[ I = \int_0^{\frac{\pi}{4}} \log \frac{2}{1 + \tan x} \,dx. \]
Step 2: Splitting the Integral Expanding the logarithm: \[ I = \int_0^{\frac{\pi}{4}} \left[ \log 2 - \log(1 + \tan x) \right] dx. \] Splitting the integral: \[ I = \int_0^{\frac{\pi}{4}} \log 2 \,dx - \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \,dx. \] Since both integrals are equal: \[ 2I = \int_0^{\frac{\pi}{4}} \log 2 \,dx. \] Evaluating the first integral: \[ \int_0^{\frac{\pi}{4}} \log 2 \,dx = \log 2 \times \frac{\pi}{4}. \] Thus, \[ 2I = \frac{\pi}{4} \log 2. \] \[ I = \frac{\pi}{8} \log 2. \] % Final Answer Thus, the correct answer is option (4).

Answer 2

To evaluate the integral \( I = \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \,dx \), we will use symmetry properties and substitution to simplify the computation. First, note the substitution \( u = \frac{\pi}{4} - x \). Consequently, \( du = -dx \), and when \( x = 0 \), \( u = \frac{\pi}{4} \); when \( x = \frac{\pi}{4} \), \( u = 0 \). Therefore, the integral becomes: \[ I = \int_{\frac{\pi}{4}}^{0} \log(1 + \tan(\frac{\pi}{4} - u)) (-du) = \int_0^{\frac{\pi}{4}} \log(1 + \tan(\frac{\pi}{4} - u)) \,du. \] Using the identity \( \tan(\frac{\pi}{4} - u) = \frac{1 - \tan u}{1 + \tan u} \), we have: \[ 1 + \tan(\frac{\pi}{4} - u) = \frac{1 + \tan u + 1 - \tan u}{1 + \tan u} = \frac{2}{1 + \tan u}. \] Thus the integral becomes: \[ I = \int_0^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan u}\right) \,du = \int_0^{\frac{\pi}{4}} \left(\log 2 - \log(1 + \tan u)\right) \,du. \] Separate the integral: \[ I = \int_0^{\frac{\pi}{4}} \log 2 \,du - \int_0^{\frac{\pi}{4}} \log(1 + \tan u) \,du. \] The first part simplifies to: \[ \int_0^{\frac{\pi}{4}} \log 2 \,du = \log 2 \cdot \left[\frac{\pi}{4} - 0\right] = \frac{\pi}{4} \log 2. \] The second integral, due to symmetry, is also equal to \( I \). Hence, it follows: \[ I = \frac{\pi}{4} \log 2 - I. \] Solving for \( I \): \[ 2I = \frac{\pi}{4} \log 2 \quad \Rightarrow \quad I = \frac{\pi}{8} \log 2. \] Thus, the value of the integral is \(\boxed{\frac{\pi}{8} \log 2}\).