`int(1)/(4+5 cosx)dx`

Answer

To solve the integral \( I = \int \frac{1}{4 + 5 \cos x} \, dx \), we will follow a systematic approach.

### Step 1: Rewrite the integral using a trigonometric identity
We know that \( \cos x = \frac{1 - \tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})} \). Let's substitute this into the integral.

\[ I = \int \frac{1}{4 + 5 \left( \frac{1 - \tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})} \right)} \, dx \]

### Step 2: Simplify the denominator
We can rewrite the denominator:

\[ 4 + 5 \left( \frac{1 - \tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})} \right) = \frac{(4 + 5) + (5 - 4)\tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})} = \frac{9 - \tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})} \]

Thus, we have:

\[ I = \int \frac{1 + \tan^2(\frac{x}{2})}{9 - \tan^2(\frac{x}{2})} \, dx \]

### Step 3: Substitute \( t = \tan(\frac{x}{2}) \)
Now, let \( t = \tan(\frac{x}{2}) \). Then, we have:

\[ dx = \frac{2}{1 + t^2} \, dt \]

### Step 4: Substitute in the integral
Substituting \( t \) and \( dx \) into the integral gives:

\[ I = \int \frac{1 + t^2}{9 - t^2} \cdot \frac{2}{1 + t^2} \, dt = 2 \int \frac{1}{9 - t^2} \, dt \]

### Step 5: Use the formula for integration
The integral \( \int \frac{1}{a^2 - x^2} \, dx \) is given by:

\[ \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right| + C \]

In our case, \( a = 3 \) and \( x = t \):

\[ I = 2 \cdot \frac{1}{2 \cdot 3} \ln \left| \frac{3 + t}{3 - t} \right| + C = \frac{1}{3} \ln \left| \frac{3 + t}{3 - t} \right| + C \]

### Step 6: Substitute back \( t = \tan(\frac{x}{2}) \)
Now, substituting back \( t = \tan(\frac{x}{2}) \):

\[ I = \frac{1}{3} \ln \left| \frac{3 + \tan(\frac{x}{2})}{3 - \tan(\frac{x}{2})} \right| + C \]

### Final Answer
Thus, the final result is:

\[ \int \frac{1}{4 + 5 \cos x} \, dx = \frac{1}{3} \ln \left| \frac{3 + \tan(\frac{x}{2})}{3 - \tan(\frac{x}{2})} \right| + C \]