`int_(lnpi-ln2)^(lnpi) (e^(x))/(1-cos(2/3e^(x)))` dx is equal to

Answer

To solve the integral

\[ \int_{\ln \pi - \ln 2}^{\ln \pi} \frac{e^x}{1 - \cos\left(\frac{2}{3} e^x\right)} \, dx, \]

we will follow these steps:

### Step 1: Change of Variables
Let \( t = \frac{2}{3} e^x \). Then, we can express \( e^x \) in terms of \( t \):

\[ e^x = \frac{3}{2} t. \]

Now, we need to find \( dx \) in terms of \( dt \):

\[ dx = \frac{dt}{\frac{2}{3} e^x} = \frac{dt}{\frac{2}{3} \cdot \frac{3}{2} t} = \frac{dt}{t}. \]

### Step 2: Change the Limits of Integration
Next, we need to change the limits of integration. When \( x = \ln \pi - \ln 2 \):

\[ t = \frac{2}{3} e^{\ln \pi - \ln 2} = \frac{2}{3} \cdot \frac{\pi}{2} = \frac{\pi}{3}. \]

When \( x = \ln \pi \):

\[ t = \frac{2}{3} e^{\ln \pi} = \frac{2}{3} \cdot \pi. \]

Thus, the new limits are from \( \frac{\pi}{3} \) to \( \frac{2\pi}{3} \).

### Step 3: Substitute into the Integral
Now we can substitute into the integral:

\[ \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\frac{3}{2} t}{1 - \cos(t)} \cdot \frac{dt}{t} = \frac{3}{2} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{1}{1 - \cos(t)} \, dt. \]

### Step 4: Simplify the Integral
Using the identity \( 1 - \cos(t) = 2 \sin^2\left(\frac{t}{2}\right) \):

\[ \frac{3}{2} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{1}{1 - \cos(t)} \, dt = \frac{3}{2} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{1}{2 \sin^2\left(\frac{t}{2}\right)} \, dt = \frac{3}{4} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \csc^2\left(\frac{t}{2}\right) \, dt. \]

### Step 5: Further Change of Variables
Let \( u = \frac{t}{2} \), then \( dt = 2 du \). The limits change accordingly:

- When \( t = \frac{\pi}{3} \), \( u = \frac{\pi}{6} \).
- When \( t = \frac{2\pi}{3} \), \( u = \frac{\pi}{3} \).

Thus, the integral becomes:

\[ \frac{3}{4} \cdot 2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc^2(u) \, du = \frac{3}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc^2(u) \, du. \]

### Step 6: Evaluate the Integral
The integral of \( \csc^2(u) \) is \( -\cot(u) \):

\[ \frac{3}{2} \left[-\cot(u)\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{3}{2} \left[-\cot\left(\frac{\pi}{3}\right) + \cot\left(\frac{\pi}{6}\right)\right]. \]

Calculating the cotangent values:

\[ -\cot\left(\frac{\pi}{3}\right) = -\frac{1}{\sqrt{3}}, \quad \cot\left(\frac{\pi}{6}\right) = \sqrt{3}. \]

Thus, we have:

\[ \frac{3}{2} \left[-\frac{1}{\sqrt{3}} + \sqrt{3}\right] = \frac{3}{2} \left[\sqrt{3} - \frac{1}{\sqrt{3}}\right] = \frac{3}{2} \left[\frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}}\right] = \frac{3}{2} \cdot \frac{2}{\sqrt{3}} = \sqrt{3}. \]

### Final Answer
The value of the integral is

\[ \sqrt{3}. \]