Let `f (x) = x cos x, x in [(3pi)/(2), 2pi] and g (x) ` be its inverse. If `int _(0)^(2pi)g (x) dx = api ^(2) + beta pi+ gamma,` where ` alpha, beta and gamma in R,` then find the value of `2 (alpha + beta+ gamma).`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Define the Functions
Let \( f(x) = x \cos x \) for \( x \in \left[\frac{3\pi}{2}, 2\pi\right] \). We need to find the integral of its inverse function \( g(x) \), which is defined as \( g(x) = f^{-1}(x) \).

### Step 2: Determine the Limits of Integration
First, we evaluate \( f(x) \) at the endpoints of the interval:
- \( f\left(\frac{3\pi}{2}\right) = \frac{3\pi}{2} \cos\left(\frac{3\pi}{2}\right) = \frac{3\pi}{2} \cdot 0 = 0 \)
- \( f(2\pi) = 2\pi \cos(2\pi) = 2\pi \cdot 1 = 2\pi \)

Thus, \( f(x) \) maps \( \left[\frac{3\pi}{2}, 2\pi\right] \) to \( [0, 2\pi] \).

### Step 3: Change of Variables in the Integral
We need to evaluate the integral:
\[ \int_0^{2\pi} g(x) \, dx \]
Using the property of inverse functions, we can change the variable:
Let \( x = f(t) \), then \( dx = f'(t) dt \). The limits change accordingly:
- When \( x = 0 \), \( t = \frac{3\pi}{2} \)
- When \( x = 2\pi \), \( t = 2\pi \)

Thus, we have:
\[ \int_0^{2\pi} g(x) \, dx = \int_{\frac{3\pi}{2}}^{2\pi} t f'(t) \, dt \]

### Step 4: Compute the Derivative
Next, we compute the derivative \( f'(x) \):
\[ f'(x) = \cos x - x \sin x \]

### Step 5: Substitute into the Integral
Now substituting \( f'(t) \) into the integral:
\[ \int_{\frac{3\pi}{2}}^{2\pi} t (\cos t - t \sin t) \, dt \]
This can be separated into two integrals:
\[ \int_{\frac{3\pi}{2}}^{2\pi} t \cos t \, dt - \int_{\frac{3\pi}{2}}^{2\pi} t^2 \sin t \, dt \]

### Step 6: Evaluate the Integrals
1. **First Integral**: \( \int t \cos t \, dt \)
- Using integration by parts, let \( u = t \) and \( dv = \cos t \, dt \):
\[ \int t \cos t \, dt = t \sin t - \int \sin t \, dt = t \sin t + \cos t \]
Evaluating from \( \frac{3\pi}{2} \) to \( 2\pi \):
\[ \left[ t \sin t + \cos t \right]_{\frac{3\pi}{2}}^{2\pi} = \left[ 2\pi \cdot 0 + 1 \right] - \left[ \frac{3\pi}{2} \cdot (-1) + 0 \right] = 1 + \frac{3\pi}{2} \]

2. **Second Integral**: \( \int t^2 \sin t \, dt \)
- Using integration by parts again, let \( u = t^2 \) and \( dv = \sin t \, dt \):
\[ \int t^2 \sin t \, dt = -t^2 \cos t + 2 \int t \cos t \, dt \]
Substitute the result from the first integral:
\[ = -t^2 \cos t + 2 \left( t \sin t + \cos t \right) \]
Evaluating this from \( \frac{3\pi}{2} \) to \( 2\pi \):
\[ \text{Evaluate at } 2\pi: -4\pi^2 \cdot 1 + 2(2\pi \cdot 0 + 1) = -4\pi^2 + 2 \]
\[ \text{Evaluate at } \frac{3\pi}{2}: -\frac{9\pi^2}{4} \cdot 0 + 2\left(\frac{3\pi}{2} \cdot (-1) + 0\right) = -3\pi \]
Thus, the second integral evaluates to:
\[ \left(-4\pi^2 + 2\right) - \left(-3\pi\right) = -4\pi^2 + 2 + 3\pi \]

### Step 7: Combine Results
Combining both integrals:
\[ I = \left(1 + \frac{3\pi}{2}\right) - \left(-4\pi^2 + 2 + 3\pi\right) \]
Simplifying gives:
\[ I = 4\pi^2 - \frac{3\pi}{2} - 1 \]

### Step 8: Compare with Given Expression
We know:
\[ I = \alpha \pi^2 + \beta \pi + \gamma \]
From our result:
- \( \alpha = 4 \)
- \( \beta = -\frac{3}{2} \)
- \( \gamma = -1 \)

### Step 9: Calculate \( 2(\alpha + \beta + \gamma) \)
Now we calculate:
\[ \alpha + \beta + \gamma = 4 - \frac{3}{2} - 1 = 4 - 1.5 - 1 = 1.5 \]
Thus,
\[ 2(\alpha + \beta + \gamma) = 2 \cdot 1.5 = 3 \]

### Final Answer
The value of \( 2(\alpha + \beta + \gamma) \) is \( \boxed{3} \).