The value of the integral `int_0^(pi/4) frac{xdx}{sin^4 2x+cos^4 2x}` equals:

Options

A. `frac{sqrt 2 pi^2}{8}`
B. `frac{sqrt 2 pi^2}{16}`
C. `frac{sqrt 2 pi^2}{32}`
D. `frac{sqrt 2 pi^2}{64}`

Answer

To solve the integral

\[ I = \int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}, \]

we can use the property of definite integrals:

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

Here, we will set \( a = \frac{\pi}{4} \). Thus, we have:

\[ I = \int_0^{\frac{\pi}{4}} \frac{\left(\frac{\pi}{4} - x\right) \, dx}{\sin^4(2(\frac{\pi}{4} - x)) + \cos^4(2(\frac{\pi}{4} - x))}. \]

Now, simplifying the argument of the sine and cosine functions:

\[ \sin(2(\frac{\pi}{4} - x)) = \sin\left(\frac{\pi}{2} - 2x\right) = \cos(2x), \]
\[ \cos(2(\frac{\pi}{4} - x)) = \cos\left(\frac{\pi}{2} - 2x\right) = \sin(2x). \]

Thus, we can rewrite the integral as:

\[ I = \int_0^{\frac{\pi}{4}} \frac{\left(\frac{\pi}{4} - x\right) \, dx}{\cos^4(2x) + \sin^4(2x)}. \]

Now, we can express \( I \) as:

\[ I = \int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)} + \int_0^{\frac{\pi}{4}} \frac{\left(\frac{\pi}{4} - x\right) \, dx}{\sin^4(2x) + \cos^4(2x)}. \]

Combining both integrals gives:

\[ 2I = \int_0^{\frac{\pi}{4}} \frac{\frac{\pi}{4} \, dx}{\sin^4(2x) + \cos^4(2x)}. \]

Thus, we have:

\[ I = \frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{\frac{\pi}{4} \, dx}{\sin^4(2x) + \cos^4(2x)} = \frac{\pi}{8} \int_0^{\frac{\pi}{4}} \frac{dx}{\sin^4(2x) + \cos^4(2x)}. \]

Next, we need to simplify the denominator:

\[ \sin^4(2x) + \cos^4(2x) = (\sin^2(2x) + \cos^2(2x))^2 - 2\sin^2(2x)\cos^2(2x) = 1 - 2\sin^2(2x)\cos^2(2x). \]

Using the identity \( \sin^2(2x) = \frac{1 - \cos(4x)}{2} \) and \( \cos^2(2x) = \frac{1 + \cos(4x)}{2} \), we can express:

\[ \sin^2(2x)\cos^2(2x) = \frac{1}{4}(1 - \cos(4x))(1 + \cos(4x)) = \frac{1}{4}(1 - \cos^2(4x)) = \frac{1}{4}\sin^2(4x). \]

Thus, we have:

\[ \sin^4(2x) + \cos^4(2x) = 1 - \frac{1}{2}\sin^2(4x). \]

Now, substituting this back into the integral, we get:

\[ I = \frac{\pi}{8} \int_0^{\frac{\pi}{4}} \frac{dx}{1 - \frac{1}{2}\sin^2(4x)}. \]

This integral can be evaluated using a standard integral formula, leading to the final result.

### Final Result:
After evaluating, we find:

\[ I = \frac{\pi^2}{16\sqrt{2}}. \]