`int_(0)^(pi//2) log (tan x ) dx=`

Options

A. 0
B. `pi`
C. `pi/2`
D. `pi/4`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx \), we can use the properties of definite integrals. 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 definite integrals
We can use the property of definite integrals which states that:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \frac{\pi}{2} \). Therefore, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\tan(\frac{\pi}{2} - x)) \, dx \]

### Step 3: Simplify the expression
Using the identity \( \tan(\frac{\pi}{2} - x) = \cot x \), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\cot x) \, dx \]

### Step 4: Express \( \log(\cot x) \)
We know that:
\[ \cot x = \frac{1}{\tan x} \]
Thus,
\[ \log(\cot x) = \log\left(\frac{1}{\tan x}\right) = -\log(\tan x) \]
So we can rewrite \( I \):
\[ I = \int_{0}^{\frac{\pi}{2}} -\log(\tan x) \, dx = -I \]

### Step 5: Solve for \( I \)
Now, we have:
\[ I = -I \]
Adding \( I \) to both sides gives:
\[ 2I = 0 \]
Thus,
\[ I = 0 \]

### Conclusion
The value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx = 0 \]