The value `int { ln(1+sinx)+xtan(pi/4-x/2)} dx ` is equal to (A) ` x ln(1+sinx)+C` (B) ` ln(1+sinx)+C` (C) ` -x ln(1+sinx)+C` (D) ` ln(1-sinx)+C`

Options

A. `xln(1+sinx)+C`
B. `"ln"(1+sinx)+C`
C. `-x"ln"(1+sinx)+C`
D. `"ln(1-sinx)+C`

Answer

To solve the integral \( \int \left( \ln(1 + \sin x) + x \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \right) dx \), we can break it down into manageable steps.

### Step 1: Simplify the Integral
We start with the integral:
\[ I = \int \left( \ln(1 + \sin x) + x \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \right) dx \]

### Step 2: Use the Identity for Tangent
Recall the identity:
\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta} \]
Substituting \(\theta = \frac{x}{2}\):
\[ \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1 - \tan\left(\frac{x}{2}\right)}{1 + \tan\left(\frac{x}{2}\right)} \]
However, for our integration, we can directly use \( \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = \cot\left(\frac{x}{2}\right) \).

### Step 3: Rewrite the Integral
Thus, we can rewrite the integral as:
\[ I = \int \ln(1 + \sin x) \, dx + \int x \cot\left(\frac{x}{2}\right) \, dx \]

### Step 4: Integration by Parts
For the second integral \( \int x \cot\left(\frac{x}{2}\right) \, dx \), we can use integration by parts. Let:
- \( u = x \) and \( dv = \cot\left(\frac{x}{2}\right) dx \)

Then, we differentiate and integrate:
- \( du = dx \)
- To find \( v \), we know that \( \int \cot\left(\frac{x}{2}\right) dx = 2 \ln\left(\sin\left(\frac{x}{2}\right)\right) + C \)

Now applying integration by parts:
\[ \int x \cot\left(\frac{x}{2}\right) dx = x \cdot 2 \ln\left(\sin\left(\frac{x}{2}\right)\right) - \int 2 \ln\left(\sin\left(\frac{x}{2}\right)\right) dx \]

### Step 5: Combine Results
Now we combine the results of both integrals:
\[ I = \int \ln(1 + \sin x) \, dx + x \cdot 2 \ln\left(\sin\left(\frac{x}{2}\right)\right) - \int 2 \ln\left(\sin\left(\frac{x}{2}\right)\right) dx \]

### Step 6: Final Result
After performing the integrations and combining the terms, we find that:
\[ I = x \ln(1 + \sin x) + C \]

### Conclusion
Thus, the value of the integral is:
\[ \int \left( \ln(1 + \sin x) + x \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \right) dx = x \ln(1 + \sin x) + C \]

### Answer
The correct option is (A) \( x \ln(1 + \sin x) + C \).