If the integal `int (5 tan x dx )/(tan x-2) =x +a ln |sin x-2 cos x | +C,` then 'a' is equal to :

Options

A. 1
B. 2
C. `-1`
D. `-2`

Answer

To solve the integral

\[ \int \frac{5 \tan x \, dx}{\tan x - 2} = x + a \ln |\sin x - 2 \cos x| + C, \]

we need to find the value of \( a \).

### Step-by-Step Solution:

1. **Rewrite the Integral**:
Start with the integral:
\[ \int \frac{5 \tan x \, dx}{\tan x - 2}. \]
We can express \( 5 \tan x \) as \( \tan x - 2 + 4 \tan x + 2 \) to facilitate the integration.

2. **Split the Integral**:
Rewrite the integral as:
\[ \int \left( \frac{\tan x - 2}{\tan x - 2} + \frac{4 \tan x + 2}{\tan x - 2} \right) dx. \]
This simplifies to:
\[ \int 1 \, dx + \int \frac{4 \tan x + 2}{\tan x - 2} \, dx. \]

3. **Integrate the First Part**:
The first integral is straightforward:
\[ \int 1 \, dx = x. \]

4. **Focus on the Second Integral**:
Now, we need to evaluate:
\[ \int \frac{4 \tan x + 2}{\tan x - 2} \, dx. \]
We can split this further:
\[ \int \frac{4 \tan x}{\tan x - 2} \, dx + \int \frac{2}{\tan x - 2} \, dx. \]

5. **Substitution for the Integral**:
Let \( t = \sin x - 2 \cos x \). Then, differentiate:
\[ dt = (\cos x + 2 \sin x) \, dx. \]
Rearranging gives:
\[ dx = \frac{dt}{\cos x + 2 \sin x}. \]

6. **Substituting in the Integral**:
Substitute \( t \) into the integral:
\[ \int \frac{dt}{t}. \]
The integral of \( \frac{1}{t} \) is:
\[ \ln |t| + C = \ln |\sin x - 2 \cos x| + C. \]

7. **Combine the Results**:
Now, we have:
\[ \int \frac{5 \tan x \, dx}{\tan x - 2} = x + 2 \ln |\sin x - 2 \cos x| + C. \]

8. **Comparing with the Given Expression**:
From the original equation, we have:
\[ x + a \ln |\sin x - 2 \cos x| + C. \]
Comparing both sides, we see that \( a = 2 \).

### Final Answer:
Thus, the value of \( a \) is:
\[ \boxed{2}. \]