The value of the integral `int_(0)^(pi//2) sin 2x log tan x dx` equals

Options

A. 0
B. `(pi)/(8) log 2`
C. `(pi)/(4) log 2`
D. `(pi)/(6) log 2`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin 2x \log \tan x \, dx \), we will use a property of definite integrals and some logarithmic identities. Let's go through the solution step by step.

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{\frac{\pi}{2}} \sin 2x \log \tan x \, dx \]

### Step 2: Use the Property of Definite Integrals
We can use the property of definite integrals that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we will set \( a = \frac{\pi}{2} \) and replace \( x \) with \( \frac{\pi}{2} - x \):
\[ I = \int_{0}^{\frac{\pi}{2}} \sin 2\left(\frac{\pi}{2} - x\right) \log \tan\left(\frac{\pi}{2} - x\right) \, dx \]

### Step 3: Simplify the Integral
Using the identities:
- \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \)
- \( \tan\left(\frac{\pi}{2} - x\right) = \cot x \)

We can rewrite \( \sin 2\left(\frac{\pi}{2} - x\right) \) as:
\[ \sin\left(\pi - 2x\right) = \sin 2x \]
Thus, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin 2x \log \cot x \, dx \]

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

### Step 5: Solve for \( I \)
Adding \( I \) to both sides gives:
\[ I + I = 0 \implies 2I = 0 \implies I = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \sin 2x \log \tan x \, dx = 0 \]