`int_(0)^(pi//2) sin 2x log tan x dx` is equal to

Options

A. `pi`
B. `(pi)/(2)`
C. 0
D. 1

Explanation

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \), we can use the property of definite integrals that states:

\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]

In this case, we have \( a = \frac{\pi}{2} \). Therefore, we can rewrite the integral as follows:

### Step 1: Rewrite the Integral
Let’s express \( I \) in terms of \( \frac{\pi}{2} - x \):

\[ I = \int_0^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \]

Using the substitution \( x = \frac{\pi}{2} - t \), we have:

\[ dx = -dt \]

Changing the limits accordingly, when \( x = 0 \), \( t = \frac{\pi}{2} \) and when \( x = \frac{\pi}{2} \), \( t = 0 \). Therefore, we can rewrite \( I \):

\[ I = \int_{\frac{\pi}{2}}^0 \sin(2(\frac{\pi}{2} - t)) \log(\tan(\frac{\pi}{2} - t)) (-dt) \]

This simplifies to:

\[ I = \int_0^{\frac{\pi}{2}} \sin(\pi - 2t) \log(\cot t) \, dt \]

### Step 2: Simplify the Integral
Since \( \sin(\pi - 2t) = \sin(2t) \) and \( \log(\cot t) = \log\left(\frac{1}{\tan t}\right) = -\log(\tan t) \):

\[ I = \int_0^{\frac{\pi}{2}} \sin(2t) (-\log(\tan t)) \, dt = -\int_0^{\frac{\pi}{2}} \sin(2t) \log(\tan t) \, dt \]

This means:

\[ I = -I \]

### Step 3: Solve for \( I \)
Adding \( I \) to both sides gives:

\[ 2I = 0 \]

Thus, we find:

\[ I = 0 \]

### Conclusion
The value of the integral \( \int_0^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \) is:

\[ \boxed{0} \]