` int(1)/(5+4 cos x)dx`

Answer

To solve the integral \( \int \frac{1}{5 + 4 \cos x} \, dx \), we can use the half-angle formula for cosine. Here’s a step-by-step solution:

### Step 1: Rewrite the integral using the half-angle formula
The half-angle formula for cosine is given by:
\[ \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]
Substituting this into the integral, we have:
\[ \int \frac{1}{5 + 4 \left( \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \right)} \, dx \]

### Step 2: Simplify the expression
This simplifies to:
\[ \int \frac{1 + \tan^2\left(\frac{x}{2}\right)}{5(1 + \tan^2\left(\frac{x}{2}\right)) + 4(1 - \tan^2\left(\frac{x}{2}\right))} \, dx \]
\[ = \int \frac{1 + \tan^2\left(\frac{x}{2}\right)}{(5 + 4) + (5 - 4)\tan^2\left(\frac{x}{2}\right)} \, dx \]
\[ = \int \frac{1 + \tan^2\left(\frac{x}{2}\right)}{9 + \tan^2\left(\frac{x}{2}\right)} \, dx \]

### Step 3: Change of variable
Let \( t = \tan\left(\frac{x}{2}\right) \). Then, the differential \( dx \) can be expressed as:
\[ dx = \frac{2}{1 + t^2} \, dt \]
Substituting this into the integral gives:
\[ \int \frac{1 + t^2}{9 + t^2} \cdot \frac{2}{1 + t^2} \, dt = 2 \int \frac{1}{9 + t^2} \, dt \]

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

### Step 5: Substitute back
Now, substituting \( t = \tan\left(\frac{x}{2}\right) \) back into the equation:
\[ \frac{2}{3} \tan^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)}{3}\right) + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ \int \frac{1}{5 + 4 \cos x} \, dx = \frac{2}{3} \tan^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)}{3}\right) + C \]