The value of `int_(0)^(pi//2) logtan xdx`, is

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. 0
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx \), we can use a property of definite integrals.

### Step 1: Define the integral
Let
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx. \]

### Step 2: Use the substitution \( x = \frac{\pi}{2} - t \)
We can change the variable by letting \( x = \frac{\pi}{2} - t \). Then, when \( x = 0 \), \( t = \frac{\pi}{2} \) and when \( x = \frac{\pi}{2} \), \( t = 0 \). The differential \( dx = -dt \).

Substituting these into the integral gives:
\[ I = \int_{\frac{\pi}{2}}^{0} \log(\tan(\frac{\pi}{2} - t)) (-dt) = \int_{0}^{\frac{\pi}{2}} \log(\cot t) \, dt. \]

### Step 3: Simplify \( \log(\cot t) \)
Using the identity \( \cot t = \frac{1}{\tan t} \), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\cot t) \, dt = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{\tan t}\right) \, dt = \int_{0}^{\frac{\pi}{2}} -\log(\tan t) \, dt. \]
Thus,
\[ I = -\int_{0}^{\frac{\pi}{2}} \log(\tan t) \, dt = -I. \]

### Step 4: Solve for \( I \)
Adding \( I \) to both sides gives:
\[ I + I = 0 \implies 2I = 0 \implies I = 0. \]

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