`int_(0)^(pi//2) log | tan x | dx` is equal to

Options

A. `pi log 2`
B. `-pi log 2`
C. 0
D. none of these

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \log |\tan x| \, dx \), we will use a property of integrals along with some logarithmic identities. Here's the step-by-step solution:

### Step 1: Define the Integral
Let \( I = \int_0^{\frac{\pi}{2}} \log |\tan x| \, dx \).

### Step 2: Use the Property of Integrals
We can use the property of integrals that states:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
In our case, \( a = \frac{\pi}{2} \). Therefore, we can write:
\[ I = \int_0^{\frac{\pi}{2}} \log |\tan(\frac{\pi}{2} - x)| \, dx \]

### Step 3: Simplify the Integral
We know that:
\[ \tan\left(\frac{\pi}{2} - x\right) = \cot x \]
Thus, we can rewrite the integral as:
\[ I = \int_0^{\frac{\pi}{2}} \log |\cot x| \, dx \]

### Step 4: Combine the Two Integrals
Now we have two expressions for \( I \):
1. \( I = \int_0^{\frac{\pi}{2}} \log |\tan x| \, dx \)
2. \( I = \int_0^{\frac{\pi}{2}} \log |\cot x| \, dx \)

Adding these two equations gives:
\[ 2I = \int_0^{\frac{\pi}{2}} \left( \log |\tan x| + \log |\cot x| \right) \, dx \]

### Step 5: Use Logarithmic Properties
Using the property of logarithms \( \log a + \log b = \log(ab) \), we can combine the logs:
\[ \log |\tan x| + \log |\cot x| = \log |\tan x \cdot \cot x| = \log |1| = 0 \]

### Step 6: Evaluate the Integral
Thus, we have:
\[ 2I = \int_0^{\frac{\pi}{2}} 0 \, dx = 0 \]

### Step 7: Solve for \( I \)
This implies:
\[ 2I = 0 \implies I = 0 \]

### Final Answer
Therefore, the value of the integral is:
\[ \int_0^{\frac{\pi}{2}} \log |\tan x| \, dx = 0 \]

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