The value of `int_(1//2)^(1) (2 x sin (1)/(x) - cos (1)/(x) ) dx` is

Options

A. `sin 1`
B. `cos 1`
C. `2 sin 1`
D. none of these

Answer

To solve the integral \( \int_{\frac{1}{2}}^{1} \left( 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \right) dx \), we can break it down into two separate integrals:

1. **Split the Integral**:
\[ \int_{\frac{1}{2}}^{1} \left( 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \right) dx = \int_{\frac{1}{2}}^{1} 2x \sin\left(\frac{1}{x}\right) dx - \int_{\frac{1}{2}}^{1} \cos\left(\frac{1}{x}\right) dx \]

2. **Factor Out Constants**:
The first integral can be simplified by factoring out the constant:
\[ \int_{\frac{1}{2}}^{1} 2x \sin\left(\frac{1}{x}\right) dx = 2 \int_{\frac{1}{2}}^{1} x \sin\left(\frac{1}{x}\right) dx \]

3. **Substitution for the First Integral**:
For the integral \( \int x \sin\left(\frac{1}{x}\right) dx \), we can use the substitution \( u = \frac{1}{x} \), which gives \( du = -\frac{1}{x^2} dx \) or \( dx = -\frac{1}{u^2} du \). The limits change as follows:
- When \( x = \frac{1}{2} \), \( u = 2 \)
- When \( x = 1 \), \( u = 1 \)

Thus, we have:
\[ \int_{\frac{1}{2}}^{1} x \sin\left(\frac{1}{x}\right) dx = -\int_{2}^{1} \frac{\sin(u)}{u^2} du = \int_{1}^{2} \frac{\sin(u)}{u^2} du \]

4. **Evaluate the First Integral**:
Therefore, we have:
\[ 2 \int_{\frac{1}{2}}^{1} x \sin\left(\frac{1}{x}\right) dx = 2 \int_{1}^{2} \frac{\sin(u)}{u^2} du \]

5. **Evaluate the Second Integral**:
The second integral \( \int_{\frac{1}{2}}^{1} \cos\left(\frac{1}{x}\right) dx \) can also be evaluated using the same substitution \( u = \frac{1}{x} \):
\[ \int_{\frac{1}{2}}^{1} \cos\left(\frac{1}{x}\right) dx = \int_{2}^{1} \cos(u) \left(-\frac{1}{u^2}\right) du = \int_{1}^{2} \frac{\cos(u)}{u^2} du \]

6. **Combine the Results**:
Now we can combine both integrals:
\[ \int_{\frac{1}{2}}^{1} \left( 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \right) dx = 2 \int_{1}^{2} \frac{\sin(u)}{u^2} du - \int_{1}^{2} \frac{\cos(u)}{u^2} du \]

7. **Final Expression**:
Thus, the final expression becomes:
\[ \int_{\frac{1}{2}}^{1} \left( 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \right) dx = \int_{1}^{2} \left( 2 \frac{\sin(u)}{u^2} - \frac{\cos(u)}{u^2} \right) du \]

8. **Evaluate the Integral**:
This can be evaluated numerically or using tables of integrals.