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

Answer

To solve the integral \( \int \frac{1}{4 + 5 \cos x} \, dx \), we can use the Weierstrass substitution, which involves substituting \( t = \tan\left(\frac{x}{2}\right) \). This substitution is useful because it simplifies the trigonometric functions.

### Step-by-Step Solution:

1. **Substitution**:
Let \( t = \tan\left(\frac{x}{2}\right) \). Then, we have:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \]
and the differential \( dx \) can be expressed as:
\[ dx = \frac{2}{1 + t^2} \, dt \]

2. **Rewrite the Integral**:
Substitute \( \cos x \) and \( dx \) into the integral:
\[ \int \frac{1}{4 + 5 \cos x} \, dx = \int \frac{1}{4 + 5 \left(\frac{1 - t^2}{1 + t^2}\right)} \cdot \frac{2}{1 + t^2} \, dt \]

3. **Simplify the Denominator**:
The expression in the denominator becomes:
\[ 4 + 5 \left(\frac{1 - t^2}{1 + t^2}\right) = \frac{(4 + 5) + (5)(-t^2)}{1 + t^2} = \frac{9 - 5t^2}{1 + t^2} \]
Thus, the integral now looks like:
\[ \int \frac{2(1 + t^2)}{9 - 5t^2} \, dt \]

4. **Separate the Integral**:
We can separate the integral:
\[ \int \frac{2}{9 - 5t^2} \, dt + \int \frac{2t^2}{9 - 5t^2} \, dt \]

5. **Integrate Each Part**:
For the first part, we can use the formula for the integral of \( \frac{1}{a^2 - x^2} \):
\[ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \]
Here, \( a^2 = \frac{9}{5} \) or \( a = \frac{3}{\sqrt{5}} \):
\[ \int \frac{2}{9 - 5t^2} \, dt = \frac{2}{\sqrt{5} \cdot 3} \tan^{-1}\left(\frac{\sqrt{5}}{3} t\right) + C \]

For the second part, we can use the substitution \( u = 9 - 5t^2 \):
\[ \int \frac{2t^2}{9 - 5t^2} \, dt = -\frac{2}{5} \int du = -\frac{2}{5}u + C = -\frac{2}{5}(9 - 5t^2) + C \]

6. **Combine Results**:
Combine the results from both integrals and substitute back \( t = \tan\left(\frac{x}{2}\right) \):
\[ \int \frac{1}{4 + 5 \cos x} \, dx = \frac{2}{3\sqrt{5}} \tan^{-1}\left(\frac{\sqrt{5}}{3} \tan\left(\frac{x}{2}\right)\right) - \frac{2}{5}(9 - 5\tan^2\left(\frac{x}{2}\right)) + C \]

### Final Answer:
\[ \int \frac{1}{4 + 5 \cos x} \, dx = \frac{2}{3\sqrt{5}} \tan^{-1}\left(\frac{\sqrt{5}}{3} \tan\left(\frac{x}{2}\right)\right) - \frac{18}{5} + \frac{2}{5} \tan^2\left(\frac{x}{2}\right) + C \]