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 equa 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 \( \tan x \) in the interval \( \left[ \frac{\pi}{4}, \frac{\pi}{3} \right] \)

To find which function is greater in the interval, we can evaluate the values of \( \sin x \) and \( \tan x \) at the endpoints:

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

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

Since \( \tan x \) is greater than \( \sin x \) at both endpoints and is continuous, we conclude that \( \tan x \) is the maximum function in the interval \( \left[ \frac{\pi}{4}, \frac{\pi}{3} \right] \).

### Step 2: Set up the integral

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

### Step 3: Integrate \( \tan x \)

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

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

### Step 4: Calculate the values of \( \cos \)

Now, we calculate:
- \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)
- \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)

Substituting these values into the integral:
\[ I = -\ln\left(\frac{1}{2}\right) + \ln\left(\frac{1}{\sqrt{2}}\right) \]
\[ I = \ln(2) + \ln\left(\frac{1}{\sqrt{2}}\right) \]
Using the property of logarithms:
\[ I = \ln(2) - \frac{1}{2}\ln(2) = \ln\left(2^{1} \cdot 2^{-\frac{1}{2}}\right) = \ln\left(2^{\frac{1}{2}}\right) = \ln\left(\sqrt{2}\right) \]

### Step 5: Relate \( I \) to \( \ln k \)

Since we have \( I = \ln k \), we can equate:
\[ \ln k = \ln\left(\sqrt{2}\right) \]
Thus, \( k = \sqrt{2} \).

### Step 6: Calculate \( k^2 \)

Finally, we calculate \( k^2 \):
\[ k^2 = (\sqrt{2})^2 = 2 \]

### Final Answer

The value of \( k^2 \) is \( \boxed{2} \).