The value of the integral `int_(pi//6)^(pi//2)((sinx-xcosx))/(x(x+sinx))dx` is equal to

Options

A. `log_(e)((2(pi+3))/(2pi+3sqrt3))`
B. `log_(e)((pi+3)/(2(2pi+3sqrt3)))`
C. `log_(e)((2pi+3sqrt3)/(2(pi+3)))`
D. None of these

Answer

To solve the integral

\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\sin x - x \cos x}{x(x + \sin x)} \, dx, \]

we will follow a systematic approach.

### Step 1: Rewrite the Integral

We start by rewriting the integrand:

\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\sin x - x \cos x}{x(x + \sin x)} \, dx. \]

Notice that we can separate the terms in the numerator:

\[ \sin x - x \cos x = \sin x + x - x - x \cos x = \frac{\sin x + x - x(1 + \cos x)}{x(x + \sin x)}. \]

### Step 2: Split the Integral

Now we can split the integral into two parts:

\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\sin x + x}{x(x + \sin x)} \, dx - \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{x(1 + \cos x)}{x(x + \sin x)} \, dx. \]

This simplifies to:

\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{x} \, dx - \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1 + \cos x}{x + \sin x} \, dx. \]

### Step 3: Evaluate the First Integral

The first integral is straightforward:

\[ \int \frac{1}{x} \, dx = \ln |x| \Rightarrow \left[ \ln x \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = \ln \left(\frac{\pi}{2}\right) - \ln \left(\frac{\pi}{6}\right) = \ln \left(\frac{\frac{\pi}{2}}{\frac{\pi}{6}}\right) = \ln 3. \]

### Step 4: Evaluate the Second Integral

For the second integral, we can use substitution. Let \( w = \sin x + x \), then \( dw = (\cos x + 1) \, dx \).

The limits change as follows:
- When \( x = \frac{\pi}{6} \), \( w = \sin\left(\frac{\pi}{6}\right) + \frac{\pi}{6} = \frac{1}{2} + \frac{\pi}{6} \).
- When \( x = \frac{\pi}{2} \), \( w = \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} = 1 + \frac{\pi}{2} \).

Thus, we rewrite the second integral:

\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1 + \cos x}{x + \sin x} \, dx = \int_{\frac{1}{2} + \frac{\pi}{6}}^{1 + \frac{\pi}{2}} \frac{1}{w} \, dw = \left[ \ln w \right]_{\frac{1}{2} + \frac{\pi}{6}}^{1 + \frac{\pi}{2}}. \]

### Step 5: Combine the Results

Now we combine the results:

\[ I = \ln 3 - \left( \ln(1 + \frac{\pi}{2}) - \ln(\frac{1}{2} + \frac{\pi}{6}) \right). \]

This simplifies to:

\[ I = \ln 3 + \ln\left(\frac{\frac{1}{2} + \frac{\pi}{6}}{1 + \frac{\pi}{2}}\right). \]

### Final Answer

Thus, the value of the integral is:

\[ I = \ln\left(3 \cdot \frac{\frac{1}{2} + \frac{\pi}{6}}{1 + \frac{\pi}{2}}\right). \]