Find the value of `int_(0)^(pi//2)sin2xlogtanxdx`.

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \), we can use a symmetry property of definite integrals.

### Step-by-Step Solution:

1. **Define the Integral:**
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \]

2. **Use the Property of Definite Integrals:**
We can use the property that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
Here, \( a = \frac{\pi}{2} \). Thus, we can write:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2(\frac{\pi}{2} - x)) \log(\tan(\frac{\pi}{2} - x)) \, dx \]

3. **Simplify the Integral:**
Using the identities:
- \( \sin(2(\frac{\pi}{2} - x)) = \sin(\pi - 2x) = \sin(2x) \)
- \( \tan(\frac{\pi}{2} - x) = \cot x \)

We can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\cot x) \, dx \]

4. **Express \(\log(\cot x)\):**
We know that:
\[ \log(\cot x) = \log\left(\frac{1}{\tan x}\right) = -\log(\tan x) \]
Therefore, we can rewrite \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) (-\log(\tan x)) \, dx = -\int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \]

5. **Combine the Integrals:**
Now we have two expressions for \( I \):
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \]
and
\[ I = -\int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \]
Adding these two equations gives:
\[ I + I = 0 \implies 2I = 0 \implies I = 0 \]

### Final Answer:
\[ \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx = 0 \]