The value of `int_(0)^(pi//2)ln|tanx+cotx|` dx is equal to :

Options

A. `pi ln2`
B. `-pi ln 2`
C. `(pi)/(2) ln 2`
D. `-(pi)/(2)ln 2`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \ln |\tan x + \cot x| \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start with the expression inside the logarithm:
\[ \tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \]
This can be combined into a single fraction:
\[ \tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \]
Since \(\sin^2 x + \cos^2 x = 1\), we have:
\[ \tan x + \cot x = \frac{1}{\sin x \cos x} \]

### Step 2: Substitute into the integral
Now substituting this back into the integral, we get:
\[ I = \int_{0}^{\frac{\pi}{2}} \ln \left| \frac{1}{\sin x \cos x} \right| \, dx \]
Using the property of logarithms, this can be simplified to:
\[ I = \int_{0}^{\frac{\pi}{2}} \ln(1) - \ln(\sin x \cos x) \, dx \]
Since \(\ln(1) = 0\), we have:
\[ I = -\int_{0}^{\frac{\pi}{2}} \ln(\sin x \cos x) \, dx \]

### Step 3: Use logarithm properties
Using the property of logarithms, we can separate the integral:
\[ I = -\int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx - \int_{0}^{\frac{\pi}{2}} \ln(\cos x) \, dx \]
Let \( J = \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx \). By symmetry, we know:
\[ \int_{0}^{\frac{\pi}{2}} \ln(\cos x) \, dx = J \]
Thus, we can write:
\[ I = -2J \]

### Step 4: Evaluate \( J \)
It is known that:
\[ J = \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx = -\frac{\pi}{2} \ln(2) \]
So substituting this back, we find:
\[ I = -2 \left(-\frac{\pi}{2} \ln(2)\right) = \pi \ln(2) \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\pi \ln(2)} \]