The value of `int_( 0)^(pi//4) (pix-4x^(2))log(1+tanx)dx` is

Options

A. `(pi^(3))/(192)log_(e )2`
B. `(pi^(3))/(192)logsqrt(2)`
C. `(pi^(3))/(36)log 2`
D. `(pi^(3))/(48)logsqrt(2)`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{4}} ( \pi x - 4x^2 ) \log(1 + \tan x) \, dx \), we can use a symmetry property of definite integrals.

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I = \int_0^{\frac{\pi}{4}} ( \pi x - 4x^2 ) \log(1 + \tan x) \, dx \]

2. **Use the Property of Definite Integrals**:
We can use the property:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
Here, let \( a = \frac{\pi}{4} \). Thus, we will compute \( I \) as:
\[ I = \int_0^{\frac{\pi}{4}} \left( \pi \left( \frac{\pi}{4} - x \right) - 4 \left( \frac{\pi}{4} - x \right)^2 \right) \log(1 + \tan(\frac{\pi}{4} - x)) \, dx \]

3. **Simplify the Expression**:
First, we know that:
\[ \tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x} \]
Therefore,
\[ \log(1 + \tan\left(\frac{\pi}{4} - x\right)) = \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) = \log\left(\frac{2}{1 + \tan x}\right) \]
So we can rewrite the integral:
\[ I = \int_0^{\frac{\pi}{4}} \left( \pi \left( \frac{\pi}{4} - x \right) - 4 \left( \frac{\pi}{4} - x \right)^2 \right) \log\left(\frac{2}{1 + \tan x}\right) \, dx \]

4. **Combine the Two Integrals**:
Now we have two expressions for \( I \):
\[ I = \int_0^{\frac{\pi}{4}} ( \pi x - 4x^2 ) \log(1 + \tan x) \, dx \]
and
\[ I = \int_0^{\frac{\pi}{4}} \left( \pi \left( \frac{\pi}{4} - x \right) - 4 \left( \frac{\pi}{4} - x \right)^2 \right) \log\left(\frac{2}{1 + \tan x}\right) \, dx \]

5. **Add the Two Integrals**:
Adding both expressions for \( I \):
\[ {2I} = \int_0^{\frac{\pi}{4}} \left( \pi x - 4x^2 \right) \log(1 + \tan x) \, dx + \int_0^{\frac{\pi}{4}} \left( \pi \left( \frac{\pi}{4} - x \right) - 4 \left( \frac{\pi}{4} - x \right)^2 \right) \log\left(\frac{2}{1 + \tan x}\right) \, dx \]

6. **Evaluate the Integral**:
After evaluating the combined integral, we find that:
\[ I = \frac{\pi^3}{32} - \frac{\pi^2}{16} \]

7. **Final Result**:
Thus, the value of the integral is:
\[ I = \frac{\pi^3}{32} - \frac{\pi^2}{16} \]