The value of `int_(0)^(pi//4) log (1+ tan theta ) d theta` is equal to

Options

A. `(pi)/(2)log2`
B. `-(pi)/(4)log 2`
C. `(pi)/(8) log2`
D. None of these

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan \theta) \, d\theta \), we can use a symmetry property of definite integrals. Here’s a step-by-step solution:

### Step 1: Use the property of definite integrals
We can use the property of definite integrals that states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \frac{\pi}{4} \). Therefore, we can write:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan(\frac{\pi}{4} - \theta)) \, d\theta \]

### Step 2: Simplify \( \tan(\frac{\pi}{4} - \theta) \)
Using the tangent subtraction formula:
\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{\tan\frac{\pi}{4} - \tan\theta}{1 + \tan\frac{\pi}{4} \tan\theta} = \frac{1 - \tan\theta}{1 + \tan\theta} \]
Thus, we have:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan\theta}{1 + \tan\theta}\right) \, d\theta \]

### Step 3: Simplify the logarithmic expression
Now we simplify the logarithmic expression:
\[ 1 + \frac{1 - \tan\theta}{1 + \tan\theta} = \frac{(1 + \tan\theta) + (1 - \tan\theta)}{1 + \tan\theta} = \frac{2}{1 + \tan\theta} \]
So, we can rewrite \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan\theta}\right) \, d\theta \]

### Step 4: Split the logarithm
Using the property of logarithms, we can split the integral:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(2) \, d\theta - \int_{0}^{\frac{\pi}{4}} \log(1 + \tan\theta) \, d\theta \]
This simplifies to:
\[ I = \log(2) \cdot \int_{0}^{\frac{\pi}{4}} d\theta - I \]

### Step 5: Evaluate the integral
The integral \( \int_{0}^{\frac{\pi}{4}} d\theta \) evaluates to \( \frac{\pi}{4} \):
\[ I = \log(2) \cdot \frac{\pi}{4} - I \]

### Step 6: Solve for \( I \)
Now, we can solve for \( I \):
\[ 2I = \log(2) \cdot \frac{\pi}{4} \]
\[ I = \frac{\log(2) \cdot \frac{\pi}{4}}{2} = \frac{\log(2) \cdot \pi}{8} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{\log(2) \cdot \pi}{8} \]

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