If the area bounded by `f(x)=tan^(3)x+tanx` from x = 0 to `x=(pi)/(4)`is k square units, then the maximum value of `g(x)=k sin x` is `(AA x in [0, (pi)/(4)])`

Options

A. 2
B. `(1)/(2sqrt2)`
C. 4
D. 8

Answer

To solve the problem, we need to find the area bounded by the function \( f(x) = \tan^3 x + \tan x \) from \( x = 0 \) to \( x = \frac{\pi}{4} \), and then use this area to find the maximum value of \( g(x) = k \sin x \).

### Step-by-Step Solution:

1. **Identify the area to be calculated**:
We need to calculate the area under the curve \( f(x) \) from \( x = 0 \) to \( x = \frac{\pi}{4} \):
\[ k = \int_{0}^{\frac{\pi}{4}} f(x) \, dx = \int_{0}^{\frac{\pi}{4}} (\tan^3 x + \tan x) \, dx \]

2. **Separate the integral**:
We can split the integral into two parts:
\[ k = \int_{0}^{\frac{\pi}{4}} \tan^3 x \, dx + \int_{0}^{\frac{\pi}{4}} \tan x \, dx \]

3. **Calculate \( \int \tan x \, dx \)**:
The integral of \( \tan x \) is:
\[ \int \tan x \, dx = -\ln |\cos x| + C \]
Evaluating from \( 0 \) to \( \frac{\pi}{4} \):
\[ \int_{0}^{\frac{\pi}{4}} \tan x \, dx = \left[-\ln |\cos x|\right]_{0}^{\frac{\pi}{4}} = -\ln \left(\frac{1}{\sqrt{2}}\right) - (-\ln(1)) = \ln(\sqrt{2}) = \frac{1}{2} \ln(2) \]

4. **Calculate \( \int \tan^3 x \, dx \)**:
We can use the identity \( \tan^3 x = \tan x \cdot (1 + \tan^2 x) \):
\[ \int \tan^3 x \, dx = \int \tan x \, dx + \int \tan x \cdot \tan^2 x \, dx \]
The second integral can be solved using the substitution \( u = \tan x \):
\[ \int \tan^2 x \, dx = \int (sec^2 x - 1) \, dx = \int sec^2 x \, dx - \int 1 \, dx = \tan x - x + C \]
Thus,
\[ \int_{0}^{\frac{\pi}{4}} \tan^3 x \, dx = \int_{0}^{\frac{\pi}{4}} \tan x \, dx + \int_{0}^{\frac{\pi}{4}} \tan^2 x \, dx \]
Evaluating \( \int_{0}^{\frac{\pi}{4}} \tan^2 x \, dx \):
\[ = \left[ \tan x - x \right]_{0}^{\frac{\pi}{4}} = \left(1 - \frac{\pi}{4}\right) - (0 - 0) = 1 - \frac{\pi}{4} \]

5. **Combine the results**:
Now we can find \( k \):
\[ k = \left(1 - \frac{\pi}{4}\right) + \frac{1}{2} \ln(2) \]

6. **Define \( g(x) \)**:
The function \( g(x) \) is given by:
\[ g(x) = k \sin x \]

7. **Find the maximum value of \( g(x) \)**:
The maximum value of \( \sin x \) in the interval \( [0, \frac{\pi}{4}] \) is \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \).
Thus, the maximum value of \( g(x) \) is:
\[ g\left(\frac{\pi}{4}\right) = k \cdot \frac{1}{\sqrt{2}} = \left(1 - \frac{\pi}{4} + \frac{1}{2} \ln(2)\right) \cdot \frac{1}{\sqrt{2}} \]

### Final Answer:
The maximum value of \( g(x) \) is:
\[ \frac{1 - \frac{\pi}{4} + \frac{1}{2} \ln(2)}{\sqrt{2}} \]