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

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + 2 \cos x} \), we can follow these steps:

### Step 1: Rewrite the integral using the identity for cosine
We know that \( \cos x = \frac{1 - \tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})} \). We will use this identity to rewrite the integral in terms of \( t = \tan(\frac{x}{2}) \).

### Step 2: Change of variable
Using the substitution \( t = \tan\left(\frac{x}{2}\right) \), we have:
\[ dx = \frac{2}{1 + t^2} dt \]
When \( x = 0 \), \( t = 0 \) and when \( x = \frac{\pi}{2} \), \( t = 1 \).

### Step 3: Substitute in the integral
Substituting \( \cos x \) and \( dx \) into the integral, we get:
\[ I = \int_{0}^{1} \frac{2}{1 + 2\left(\frac{1 - t^2}{1 + t^2}\right)} \cdot \frac{1}{1 + t^2} dt \]

### Step 4: Simplify the integrand
Simplifying the integrand:
\[ 1 + 2\cos x = 1 + 2\left(\frac{1 - t^2}{1 + t^2}\right) = \frac{(1 + t^2) + 2(1 - t^2)}{1 + t^2} = \frac{3 - t^2}{1 + t^2} \]
Thus,
\[ I = \int_{0}^{1} \frac{2}{\frac{3 - t^2}{1 + t^2}} \cdot \frac{1}{1 + t^2} dt = \int_{0}^{1} \frac{2(1 + t^2)}{3 - t^2} dt \]

### Step 5: Split the integral
Now we can split the integral:
\[ I = \int_{0}^{1} \frac{2}{3 - t^2} dt + \int_{0}^{1} \frac{2t^2}{3 - t^2} dt \]

### Step 6: Evaluate the first integral
The first integral can be evaluated as:
\[ \int \frac{2}{3 - t^2} dt = -\frac{2}{\sqrt{3}} \log\left| \sqrt{3} + t \right| + C \]
Evaluating from 0 to 1:
\[ \left[-\frac{2}{\sqrt{3}} \log(\sqrt{3} + 1) + \frac{2}{\sqrt{3}} \log(\sqrt{3})\right] \]

### Step 7: Evaluate the second integral
For the second integral:
\[ \int \frac{2t^2}{3 - t^2} dt \]
We can use the substitution \( u = 3 - t^2 \) to evaluate this integral.

### Step 8: Combine results
After evaluating both integrals, we combine the results to find the value of \( I \).

### Final Result
After performing the calculations, we find:
\[ I = \frac{1}{\sqrt{3}} \log\left( \frac{2 + \sqrt{3}}{1} \right) \]