`int_(0)^(pi) (1)/(5+2 cos x)dx`

Answer

To solve the integral \( I = \int_{0}^{\pi} \frac{1}{5 + 2 \cos x} \, dx \), we will use the substitution method and properties of trigonometric functions. Here’s a step-by-step solution:

### Step 1: Use the Weierstrass Substitution
Let \( t = \tan\left(\frac{x}{2}\right) \). Then, we have:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \]
and
\[ dx = \frac{2}{1 + t^2} \, dt. \]

### Step 2: Change the Limits of Integration
When \( x = 0 \):
\[ t = \tan(0) = 0. \]
When \( x = \pi \):
\[ t = \tan\left(\frac{\pi}{2}\right) \to \infty. \]
Thus, the limits change from \( 0 \) to \( \infty \).

### Step 3: Substitute into the Integral
Now substituting \( \cos x \) and \( dx \) into the integral:
\[ I = \int_{0}^{\infty} \frac{1}{5 + 2 \left(\frac{1 - t^2}{1 + t^2}\right)} \cdot \frac{2}{1 + t^2} \, dt. \]
This simplifies to:
\[ I = \int_{0}^{\infty} \frac{2}{(5(1 + t^2) + 2(1 - t^2))} \, dt = \int_{0}^{\infty} \frac{2}{(5 + 2) + (5 - 2)t^2} \, dt. \]
\[ = \int_{0}^{\infty} \frac{2}{7 + 3t^2} \, dt. \]

### Step 4: Factor Out Constants
We can factor out constants from the integral:
\[ I = \frac{2}{3} \int_{0}^{\infty} \frac{1}{\frac{7}{3} + t^2} \, dt. \]

### Step 5: Use the Standard Integral Formula
We know that:
\[ \int_{0}^{\infty} \frac{1}{a^2 + x^2} \, dx = \frac{\pi}{2a}. \]
In our case, \( a^2 = \frac{7}{3} \), so \( a = \sqrt{\frac{7}{3}} \).

### Step 6: Calculate the Integral
Thus,
\[ \int_{0}^{\infty} \frac{1}{\frac{7}{3} + t^2} \, dt = \frac{\pi}{2\sqrt{\frac{7}{3}}} = \frac{\pi \sqrt{3}}{2\sqrt{7}}. \]
Now substituting this back into our expression for \( I \):
\[ I = \frac{2}{3} \cdot \frac{\pi \sqrt{3}}{2\sqrt{7}} = \frac{\pi \sqrt{3}}{3\sqrt{7}}. \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{\pi}{\sqrt{7}}. \]