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).