The value of `int_0^(pi/2)sin2x.log tanxdx `is

Options

A. 0
B. `pi`
C. `pi/2`
D. none of these

Answer

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

### Step 1: Set up 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:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
In our case, \( a = \frac{\pi}{2} \). Therefore, we can rewrite the integral as:
\[ I = \int_0^{\frac{\pi}{2}} \sin(2\left(\frac{\pi}{2} - x\right)) \log(\tan\left(\frac{\pi}{2} - x\right)) \, dx \]

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

Thus, we have:
\[ I = \int_0^{\frac{\pi}{2}} \sin(2x) \log(\cot x) \, dx \]

### Step 4: Rewrite \( \log(\cot x) \)
We know that:
\[ \log(\cot x) = \log\left(\frac{1}{\tan x}\right) = -\log(\tan x) \]
So we can rewrite the integral as:
\[ I = \int_0^{\frac{\pi}{2}} \sin(2x) (-\log(\tan x)) \, dx = -\int_0^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \]

### Step 5: Combine the integrals
Now we have:
\[ I = -I \]
Adding \( I \) to both sides gives:
\[ 2I = 0 \]
Thus, we find:
\[ I = 0 \]

### Final Answer
The value of the integral is:
\[ \int_0^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx = 0 \]