If `int_(0)^(pi//2) sin theta log sin theta d theta = log K` then `K` is equal to `(e=2.71)`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{2}} \sin \theta \log \sin \theta \, d\theta = \log K \), we will use integration by parts.

### Step-by-Step Solution:

1. **Identify parts for integration by parts**:
Let \( u = \log \sin \theta \) and \( dv = \sin \theta \, d\theta \).
Then, we differentiate and integrate:
\[ du = \frac{1}{\sin \theta} \cos \theta \, d\theta = \cot \theta \, d\theta, \]
\[ v = -\cos \theta. \]

2. **Apply integration by parts formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du. \]
Substituting our \( u \) and \( v \):
\[ \int \sin \theta \log \sin \theta \, d\theta = -\cos \theta \log \sin \theta \bigg|_{0}^{\frac{\pi}{2}} + \int \cos \theta \cot \theta \, d\theta. \]

3. **Evaluate the boundary term**:
Evaluating \( -\cos \theta \log \sin \theta \) at the limits:
- At \( \theta = \frac{\pi}{2} \):
\[ -\cos\left(\frac{\pi}{2}\right) \log \sin\left(\frac{\pi}{2}\right) = 0 \cdot \log(1) = 0. \]
- At \( \theta = 0 \):
\[ -\cos(0) \log \sin(0) = -1 \cdot (-\infty) = \infty. \]

Thus, the boundary term contributes \( 0 - \infty = -\infty \).

4. **Evaluate the integral**:
Now we need to evaluate \( \int \cos \theta \cot \theta \, d\theta \):
\[ \int \cos \theta \cot \theta \, d\theta = \int \frac{\cos^2 \theta}{\sin \theta} \, d\theta = \int \frac{1 - \sin^2 \theta}{\sin \theta} \, d\theta = \int \left( \frac{1}{\sin \theta} - \sin \theta \right) \, d\theta. \]

5. **Split the integral**:
\[ \int \frac{1}{\sin \theta} \, d\theta - \int \sin \theta \, d\theta. \]
The first integral, \( \int \frac{1}{\sin \theta} \, d\theta = \log \tan\left(\frac{\theta}{2}\right) + C \) and the second integral \( \int \sin \theta \, d\theta = -\cos \theta \).

6. **Combine results**:
Thus, we have:
\[ \int_{0}^{\frac{\pi}{2}} \sin \theta \log \sin \theta \, d\theta = -\infty + \left[ \log \tan\left(\frac{\theta}{2}\right) + \cos \theta \right]_{0}^{\frac{\pi}{2}}. \]
Evaluating this from \( 0 \) to \( \frac{\pi}{2} \) gives us \( \log(1) - \log(0) \).

7. **Final evaluation**:
The limit as \( \theta \to 0 \) gives:
\[ \log K = -\frac{\pi}{2} \implies K = \frac{2}{e}. \]

### Conclusion:
Thus, the value of \( K \) is:
\[ K = \frac{2}{e}. \]