The value of `int_(0)^(-pi//4)(1+tanx)/(1-tanx)dx` is

Options

A. `-1/2log2`
B. `1/4log2`
C. `1/3log2`
D. `(-1)/3log2`

Answer

To solve the integral

\[ I = \int_{0}^{-\frac{\pi}{4}} \frac{1 + \tan x}{1 - \tan x} \, dx, \]

we can use a trigonometric identity to simplify the integrand.

### Step 1: Change of Variables
First, we can change the limits of integration by reversing them, which introduces a negative sign:

\[ I = -\int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx. \]

### Step 2: Use the Trigonometric Identity
Using the identity

\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x}, \]

we can rewrite the integral:

\[ I = -\int_{-\frac{\pi}{4}}^{0} \tan\left(\frac{\pi}{4} + x\right) \, dx. \]

### Step 3: Substitute
Next, we can substitute \( u = \frac{\pi}{4} + x \), which gives \( du = dx \). The limits change as follows:
- When \( x = -\frac{\pi}{4} \), \( u = 0 \).
- When \( x = 0 \), \( u = \frac{\pi}{4} \).

Thus, the integral becomes:

\[ I = -\int_{0}^{\frac{\pi}{4}} \tan u \, du. \]

### Step 4: Integrate
The integral of \( \tan u \) is:

\[ \int \tan u \, du = -\log|\cos u| + C. \]

So we have:

\[ I = -\left[-\log|\cos u|\right]_{0}^{\frac{\pi}{4}} = \log|\cos 0| - \log|\cos \frac{\pi}{4}|. \]

### Step 5: Evaluate the Limits
Now we evaluate the limits:

\[ I = \log(1) - \log\left(\frac{1}{\sqrt{2}}\right) = 0 - \left(-\frac{1}{2} \log 2\right) = \frac{1}{2} \log 2. \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{1}{2} \log 2. \]