`int_(1//3)^(3) (1)/(x)sin ((1)/(x)-x)dx` is equal to

Options

A. `(sqrt(3))/(2)`
B. `(sqrt(3)pi)/(2)`
C. 0
D. none of these

Answer

To solve the integral \( I = \int_{\frac{1}{3}}^{3} \frac{1}{x} \sin\left(\frac{1}{x} - x\right) \, dx \), we will use a substitution method. Here’s a step-by-step breakdown of the solution:

### Step 1: Substitution
Let \( t = \frac{1}{x} \). Then, we have:
\[ x = \frac{1}{t} \quad \text{and} \quad dx = -\frac{1}{t^2} \, dt \]

### Step 2: Change the Limits
When \( x = \frac{1}{3} \):
\[ t = \frac{1}{\frac{1}{3}} = 3 \]
When \( x = 3 \):
\[ t = \frac{1}{3} \]
Thus, the limits change from \( x = \frac{1}{3} \) to \( x = 3 \) into \( t = 3 \) to \( t = \frac{1}{3} \).

### Step 3: Substitute into the Integral
Now substituting \( t \) and \( dx \) into the integral:
\[ I = \int_{3}^{\frac{1}{3}} \frac{1}{\frac{1}{t}} \sin\left(t - \frac{1}{t}\right) \left(-\frac{1}{t^2}\right) dt \]
This simplifies to:
\[ I = \int_{3}^{\frac{1}{3}} -\frac{t}{1} \sin\left(t - \frac{1}{t}\right) dt \]
Reversing the limits changes the sign:
\[ I = \int_{\frac{1}{3}}^{3} \frac{1}{t} \sin\left(t - \frac{1}{t}\right) dt \]

### Step 4: Recognizing the Integral
Now we have:
\[ I = \int_{\frac{1}{3}}^{3} \frac{1}{t} \sin\left(t - \frac{1}{t}\right) dt \]
This integral is the same as the original integral \( I \) but with \( t \) instead of \( x \). Therefore, we can write:
\[ I = -I \]

### Step 5: Solving for \( I \)
Adding \( I \) to both sides gives:
\[ 2I = 0 \]
Thus,
\[ I = 0 \]

### Conclusion
Therefore, the value of the integral is:
\[ \int_{\frac{1}{3}}^{3} \frac{1}{x} \sin\left(\frac{1}{x} - x\right) \, dx = 0 \]