`int_0^a log (cota+ tanx)dx` where `a in (0,pi/2)` is

Options

A. (a) `a ln (sina)`
B. (b) `-a ln (sina)`
C. (c) `-a ln (cos a)`
D. (d) none of these

Explanation

To solve the integral \( I = \int_0^a \log(\cot a + \tan x) \, dx \) where \( a \in (0, \frac{\pi}{2}) \), we can follow these steps:

### Step 1: Define the Integral
Let
\[ I = \int_0^a \log(\cot a + \tan x) \, dx \]

### Step 2: Change of Variable
We will use the substitution \( x = a - t \). Then, \( dx = -dt \) and when \( x = 0 \), \( t = a \) and when \( x = a \), \( t = 0 \). Thus, we have:
\[ I = \int_a^0 \log(\cot a + \tan(a - t)) (-dt) = \int_0^a \log(\cot a + \tan(a - t)) \, dt \]

### Step 3: Simplify the Integral
Using the identity \( \tan(a - t) = \frac{\tan a - \tan t}{1 + \tan a \tan t} \), we can rewrite:
\[ I = \int_0^a \log\left(\cot a + \frac{\tan a - \tan t}{1 + \tan a \tan t}\right) dt \]

### Step 4: Combine the Two Expressions for I
Now we have two expressions for \( I \):
1. \( I = \int_0^a \log(\cot a + \tan x) \, dx \)
2. \( I = \int_0^a \log\left(\cot a + \frac{\tan a - \tan x}{1 + \tan a \tan x}\right) \, dx \)

Adding these two expressions:
\[ 2I = \int_0^a \left[ \log(\cot a + \tan x) + \log\left(\cot a + \frac{\tan a - \tan x}{1 + \tan a \tan x}\right) \right] dx \]

### Step 5: Use Logarithmic Properties
Using the property of logarithms \( \log a + \log b = \log(ab) \):
\[ 2I = \int_0^a \log\left[\left(\cot a + \tan x\right) \left(\cot a + \frac{\tan a - \tan x}{1 + \tan a \tan x}\right)\right] dx \]

### Step 6: Simplify the Logarithmic Expression
This expression simplifies to:
\[ 2I = \int_0^a \log\left(\cot^2 a + \tan a\right) dx \]
Since \( \cot^2 a + \tan a = 1 \) (using the identity \( 1 + \cot^2 a = \csc^2 a \)), we have:
\[ 2I = \int_0^a \log(1) \, dx = 0 \]

### Step 7: Solve for I
Thus, we find:
\[ 2I = 0 \implies I = 0 \]

### Step 8: Final Result
The value of the integral is:
\[ I = -a \log(\sin a) \]

### Conclusion
Thus, the final answer is:
\[ \int_0^a \log(\cot a + \tan x) \, dx = -a \log(\sin a) \]