Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)`
Making the substitution `"tan"1/2x=t`, we have
`I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0`
The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.

Explanation

To solve the integral \( I = \int_{0}^{2\pi} \frac{dx}{5 - 2\cos x} \) and identify the mistake in the substitution \( \tan \frac{x}{2} = t \), we will proceed step by step.

### Step 1: Write down the integral
We start with the integral:
\[ I = \int_{0}^{2\pi} \frac{dx}{5 - 2\cos x} \]

### Step 2: Make the substitution
Using the substitution \( t = \tan \frac{x}{2} \), we know the following relationships:
- \( \cos x = \frac{1 - t^2}{1 + t^2} \)
- \( dx = \frac{2}{1 + t^2} dt \)

### Step 3: Change the limits of integration
When \( x = 0 \), \( t = \tan(0) = 0 \).
When \( x = 2\pi \), \( t = \tan(\pi) = 0 \).
Thus, the limits of integration change from \( 0 \) to \( 0 \).

### Step 4: Substitute in the integral
Substituting these into the integral gives:
\[ I = \int_{0}^{0} \frac{2}{1 + t^2} \cdot \frac{1}{5 - 2\left(\frac{1 - t^2}{1 + t^2}\right)} dt \]

### Step 5: Simplify the integrand
Now, simplify the integrand:
\[ 5 - 2\cos x = 5 - 2\left(\frac{1 - t^2}{1 + t^2}\right) = 5 - \frac{2(1 - t^2)}{1 + t^2} = \frac{5(1 + t^2) - 2(1 - t^2)}{1 + t^2} = \frac{5 + 5t^2 - 2 + 2t^2}{1 + t^2} = \frac{3 + 7t^2}{1 + t^2} \]

Thus, the integral becomes:
\[ I = \int_{0}^{0} \frac{2}{1 + t^2} \cdot \frac{1 + t^2}{3 + 7t^2} dt = \int_{0}^{0} \frac{2}{3 + 7t^2} dt \]

### Step 6: Evaluate the integral
Since the limits of integration are the same (from \( 0 \) to \( 0 \)), we find that:
\[ I = 0 \]

### Step 7: Identify the mistake
The mistake here lies in the substitution \( t = \tan \frac{x}{2} \) because this substitution is discontinuous at \( x = \pi \), which is within the interval of integration \( [0, 2\pi] \). This discontinuity causes the integral to evaluate incorrectly to zero when it should yield a positive value.

### Conclusion
The integral \( I = \int_{0}^{2\pi} \frac{dx}{5 - 2\cos x} \) should be evaluated using a different method or by correctly handling the discontinuity in the substitution.