To solve the integral \( \int_{0}^{\frac{\pi}{2}} \theta^2 \csc^2 \theta \, d\theta \) and show that it equals \( \pi \log 2 \), we will use integration by parts.
### Step-by-Step Solution:
1. **Identify the Functions for Integration by Parts**:
We will use the integration by parts formula:
\[
\int u \, dv = uv - \int v \, du
\]
Here, let:
- \( u = \theta^2 \) (algebraic function)
- \( dv = \csc^2 \theta \, d\theta \)
2. **Differentiate and Integrate**:
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[
du = 2\theta \, d\theta
\]
- Integrate \( dv \):
\[
v = -\cot \theta
\]
3. **Apply the Integration by Parts Formula**:
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:
\[
\int_{0}^{\frac{\pi}{2}} \theta^2 \csc^2 \theta \, d\theta = \left[ -\theta^2 \cot \theta \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} 2\theta \cot \theta \, d\theta
\]
4. **Evaluate the Boundary Term**:
Evaluate \( \left[ -\theta^2 \cot \theta \right]_{0}^{\frac{\pi}{2}} \):
- At \( \theta = \frac{\pi}{2} \):
\[
-\left(\frac{\pi}{2}\right)^2 \cot\left(\frac{\pi}{2}\right) = 0 \quad (\text{since } \cot\left(\frac{\pi}{2}\right) = 0)
\]
- At \( \theta = 0 \):
\[
-\left(0\right)^2 \cot(0) = 0 \quad (\text{since } \cot(0) \text{ is undefined but the term is } 0)
\]
Thus, the boundary term evaluates to \( 0 \).
5. **Now Focus on the Remaining Integral**:
We have:
\[
\int_{0}^{\frac{\pi}{2}} \theta^2 \csc^2 \theta \, d\theta = 0 + \int_{0}^{\frac{\pi}{2}} 2\theta \cot \theta \, d\theta
\]
So we need to evaluate:
\[
\int_{0}^{\frac{\pi}{2}} 2\theta \cot \theta \, d\theta
\]
6. **Use Integration by Parts Again**:
For \( \int \theta \cot \theta \, d\theta \), let:
- \( u = \theta \)
- \( dv = \cot \theta \, d\theta \)
Then:
- \( du = d\theta \)
- \( v = \log(\sin \theta) \)
Applying integration by parts again:
\[
\int \theta \cot \theta \, d\theta = \theta \log(\sin \theta) - \int \log(\sin \theta) \, d\theta
\]
7. **Evaluate the New Integral**:
The integral \( \int \log(\sin \theta) \, d\theta \) is known from the problem statement:
\[
\int_{0}^{\frac{\pi}{2}} \log(\sin \theta) \, d\theta = -\frac{\pi}{2} \log 2
\]
8. **Putting It All Together**:
Now we can substitute back:
\[
\int_{0}^{\frac{\pi}{2}} 2\theta \cot \theta \, d\theta = 2\left[ \theta \log(\sin \theta) \right]_{0}^{\frac{\pi}{2}} - 2\left(-\frac{\pi}{2} \log 2\right)
\]
Evaluating the boundary term gives \( 0 \) again, so:
\[
= 0 + \pi \log 2
\]
9. **Final Result**:
Therefore, we conclude:
\[
\int_{0}^{\frac{\pi}{2}} \theta^2 \csc^2 \theta \, d\theta = \pi \log 2
\]
