If `int (dx)/(5+4 cos x) = k tan^(-1) (m tan ((x)/(2))) + C` then :

Options

A. ` k = (1)/(3), m = (1)/(3)`
B. `k = (1)/(3), m = (2)/(3)`
C. `k = (2)/(3), m = (2)/(3)`
D. `k = (2)/(3), = (2)/(3)`

Answer

To solve the integral \(\int \frac{dx}{5 + 4 \cos x}\), we will follow these steps:

### Step 1: Rewrite \(\cos x\) in terms of \(\tan\)
We can use the identity for \(\cos x\) in terms of \(\tan\):
\[ \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]
Let \(t = \tan\left(\frac{x}{2}\right)\). Then, we can substitute \(\cos x\) in the integral.

### Step 2: Substitute \(\cos x\) in the integral
Substituting the expression for \(\cos x\):
\[ \int \frac{dx}{5 + 4 \left(\frac{1 - t^2}{1 + t^2}\right)} = \int \frac{dx}{5 + \frac{4(1 - t^2)}{1 + t^2}} \]
This simplifies to:
\[ \int \frac{dx}{\frac{5(1 + t^2) + 4(1 - t^2)}{1 + t^2}} = \int \frac{(1 + t^2)dx}{9 + t^2} \]

### Step 3: Change of variable
Now, we need to express \(dx\) in terms of \(dt\):
\[ dx = \frac{2}{1 + t^2} dt \]
Substituting this into the integral gives:
\[ \int \frac{(1 + t^2) \cdot \frac{2}{1 + t^2} dt}{9 + t^2} = 2 \int \frac{dt}{9 + t^2} \]

### Step 4: Solve the integral
The integral \(\int \frac{dt}{9 + t^2}\) can be solved using the formula:
\[ \int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C \]
Here, \(a = 3\):
\[ 2 \int \frac{dt}{9 + t^2} = 2 \cdot \frac{1}{3} \tan^{-1}\left(\frac{t}{3}\right) + C = \frac{2}{3} \tan^{-1}\left(\frac{t}{3}\right) + C \]

### Step 5: Substitute back for \(t\)
Since \(t = \tan\left(\frac{x}{2}\right)\), we substitute back:
\[ \frac{2}{3} \tan^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)}{3}\right) + C \]

### Final Result
Thus, we have:
\[ \int \frac{dx}{5 + 4 \cos x} = \frac{2}{3} \tan^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)}{3}\right) + C \]

### Comparison with the given form
We compare this with the given form \(k \tan^{-1}(m \tan\left(\frac{x}{2}\right)) + C\):
- Here, \(k = \frac{2}{3}\) and \(m = \frac{1}{3}\).