The value of `int _(0)^(4//pi) (3x ^(2) sin ""(1)/(x)-x cos ""(1)/(x )) dx` is:

Options

A. `(8 sqrt2)/(pi ^(3))`
B. `(24sqrt2)/(pi ^(3))`
C. `(32sqrt2)/(pi ^(3))`
D. None of these

Answer

To solve the integral

\[ I = \int_{0}^{\frac{4}{\pi}} \left( 3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right) \right) dx, \]

we can break it down into two parts:

\[ I = \int_{0}^{\frac{4}{\pi}} 3x^2 \sin\left(\frac{1}{x}\right) dx - \int_{0}^{\frac{4}{\pi}} x \cos\left(\frac{1}{x}\right) dx. \]

Let's denote the first integral as \( I_1 \) and the second integral as \( I_2 \).

### Step 1: Solve \( I_1 = \int_{0}^{\frac{4}{\pi}} 3x^2 \sin\left(\frac{1}{x}\right) dx \)

We will use integration by parts for \( I_1 \). Let:

- \( u = \sin\left(\frac{1}{x}\right) \) (first function)
- \( dv = 3x^2 dx \) (second function)

Then, we differentiate and integrate:

- \( du = \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) dx = -\frac{\cos\left(\frac{1}{x}\right)}{x^2} dx \)
- \( v = x^3 \)

Using integration by parts formula \( \int u \, dv = uv - \int v \, du \):

\[ I_1 = \left[ x^3 \sin\left(\frac{1}{x}\right) \right]_{0}^{\frac{4}{\pi}} - \int_{0}^{\frac{4}{\pi}} x^3 \left(-\frac{\cos\left(\frac{1}{x}\right)}{x^2}\right) dx \]

This simplifies to:

\[ I_1 = \left[ x^3 \sin\left(\frac{1}{x}\right) \right]_{0}^{\frac{4}{\pi}} + \int_{0}^{\frac{4}{\pi}} x \cos\left(\frac{1}{x}\right) dx \]

### Step 2: Evaluate the boundary term

Now we need to evaluate \( \left[ x^3 \sin\left(\frac{1}{x}\right) \right]_{0}^{\frac{4}{\pi}} \):

1. At \( x = \frac{4}{\pi} \):

\[ \left( \frac{4}{\pi} \right)^3 \sin\left(\frac{\pi}{4}\right) = \frac{64}{\pi^3} \cdot \frac{1}{\sqrt{2}} = \frac{64}{\sqrt{2} \pi^3} \]

2. At \( x = 0 \):

As \( x \to 0 \), \( x^3 \sin\left(\frac{1}{x}\right) \) oscillates between \( -x^3 \) and \( x^3 \), thus the limit approaches \( 0 \).

So,

\[ \left[ x^3 \sin\left(\frac{1}{x}\right) \right]_{0}^{\frac{4}{\pi}} = \frac{64}{\sqrt{2} \pi^3} - 0 = \frac{64}{\sqrt{2} \pi^3} \]

### Step 3: Substitute back into \( I \)

Now substituting back into \( I \):

\[ I = \frac{64}{\sqrt{2} \pi^3} + I_2 \]

Since \( I_2 = \int_{0}^{\frac{4}{\pi}} x \cos\left(\frac{1}{x}\right) dx \), we notice that this term cancels out with the corresponding term in \( I \).

### Step 4: Final Calculation

Thus, we have:

\[ I = \frac{64}{\sqrt{2} \pi^3} \]

To rationalize:

\[ I = \frac{64 \sqrt{2}}{2 \pi^3} = \frac{32 \sqrt{2}}{\pi^3} \]

### Final Answer

The final value of the integral is:

\[ \frac{32 \sqrt{2}}{\pi^3} \]