If the value of the integral `I=int_((pi)/(4))^((pi)/(3))" max "(sinx, tanx)dx` is equal to ln k, then the value of `k^(2)` is equal to

Answer

To solve the integral \( I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \max(\sin x, \tan x) \, dx \) and find the value of \( k^2 \) such that \( I = \ln k \), we will follow these steps:

### Step 1: Determine the maximum of \( \sin x \) and \( \tan x \)

We need to find the maximum of \( \sin x \) and \( \tan x \) in the interval \( \left[\frac{\pi}{4}, \frac{\pi}{3}\right] \).

- At \( x = \frac{\pi}{4} \):
\[ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \tan\left(\frac{\pi}{4}\right) = 1 \]

- At \( x = \frac{\pi}{3} \):
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

Now, we check the values of \( \sin x \) and \( \tan x \) at some points in the interval to see which function is greater.

- At \( x = \frac{\pi}{4} \), \( \tan x \) is greater.
- At \( x = \frac{\pi}{3} \), \( \tan x \) is also greater.

Since \( \tan x \) is increasing and \( \sin x \) is less than \( \tan x \) in this interval, we conclude:
\[ \max(\sin x, \tan x) = \tan x \text{ for } x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right] \]

### Step 2: Rewrite the integral

Now we can rewrite the integral:
\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \tan x \, dx \]

### Step 3: Calculate the integral of \( \tan x \)

The integral of \( \tan x \) is:
\[ \int \tan x \, dx = -\ln(\cos x) + C \]

Thus, we evaluate:
\[ I = \left[-\ln(\cos x)\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} \]

### Step 4: Substitute the limits

Now we substitute the limits:
\[ I = -\ln(\cos(\frac{\pi}{3})) + \ln(\cos(\frac{\pi}{4})) \]

Calculating the cosine values:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

Thus:
\[ I = -\ln\left(\frac{1}{2}\right) + \ln\left(\frac{1}{\sqrt{2}}\right) \]

### Step 5: Simplify the expression

Using properties of logarithms:
\[ I = \ln(2) - \ln(\sqrt{2}) = \ln(2) - \ln(2^{1/2}) = \ln(2) - \frac{1}{2} \ln(2) = \frac{1}{2} \ln(2) \]

### Step 6: Relate to \( \ln k \)

We have:
\[ I = \frac{1}{2} \ln(2) = \ln(k) \]
This implies:
\[ k = 2^{1/2} = \sqrt{2} \]

### Step 7: Find \( k^2 \)

Now we compute \( k^2 \):
\[ k^2 = (\sqrt{2})^2 = 2 \]

### Final Answer

Thus, the value of \( k^2 \) is:
\[ \boxed{2} \]