`"If " int(5tanx)/(tanx-2)dx=x+a " In"(sinx-2cosx)+k " then " a= `

Options

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

Explanation

To solve the integral \(\int \frac{5 \tan x}{\tan x - 2} \, dx = x + a \ln(\sin x - 2 \cos x) + k\), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{5 \tan x}{\tan x - 2} \, dx \]
We can express \(\tan x\) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus, we rewrite the integral as:
\[ I = \int \frac{5 \frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x} - 2} \, dx \]

### Step 2: Simplify the Integral
Now, we simplify the expression:
\[ I = \int \frac{5 \sin x}{\sin x - 2 \cos x} \, dx \]
Next, we can factor out \(\cos x\) from the denominator:
\[ I = \int \frac{5 \sin x}{\sin x - 2 \cos x} \, dx \]

### Step 3: Separate Terms
We can separate the terms in the numerator:
\[ 5 \sin x = (\sin x - 2 \cos x) + 2 \sin x + 2 \cos x \]
Thus, we rewrite the integral:
\[ I = \int \left(1 + \frac{2 \sin x + 2 \cos x}{\sin x - 2 \cos x}\right) \, dx \]

### Step 4: Split the Integral
Now we can split the integral into two parts:
\[ I = \int 1 \, dx + 2 \int \frac{\sin x + \cos x}{\sin x - 2 \cos x} \, dx \]
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 5: Substitution for the Second Integral
For the second integral, we can use the substitution:
Let \(t = \sin x - 2 \cos x\). Then, the derivative \(dt\) is:
\[ dt = (\cos x + 2 \sin x) \, dx \]
Thus, we can express \(dx\) in terms of \(dt\):
\[ dx = \frac{dt}{\cos x + 2 \sin x} \]

### Step 6: Solve the Second Integral
Now we rewrite the second integral:
\[ 2 \int \frac{\sin x + \cos x}{t} \cdot \frac{dt}{\cos x + 2 \sin x} \]
This simplifies to:
\[ 2 \int \frac{1}{t} \, dt = 2 \ln |t| + C = 2 \ln |\sin x - 2 \cos x| + C \]

### Step 7: Combine Results
Combining the results, we have:
\[ I = x + 2 \ln |\sin x - 2 \cos x| + C \]
Comparing this with the given expression \(x + a \ln(\sin x - 2 \cos x) + k\), we find:
\[ a = 2 \]

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