The value of the integral $\int_{0}^{π / 4} \frac{x d x}{\sin^{4} (2 x) + \cos^{4} (2 x)}$ equals:

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$\frac{\sqrt{2} π^{2}}{16}$

$\frac{\sqrt{2} π^{2}}{32}$

$\frac{\sqrt{2} π^{2}}{64}$

$\frac{\sqrt{2} π^{2}}{8}$

answer is D.

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Detailed Solution

$\int_{0}^{\frac{π}{4}} \frac{x d x}{\sin^{4} (2 x) + \cos^{4} (2 x)} L e t 2 x = t t h e n d x = \frac{1}{2} d t I = \frac{1}{4} \int_{0}^{\frac{π}{2}} \frac{t d t}{\sin^{4} t + \cos^{4} t} I = \frac{1}{4} \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - t) d t}{\sin^{4} (\frac{π}{2} - t) + \cos^{4} (\frac{π}{2} - t)} I = \frac{1}{4} \int_{0}^{\frac{π}{2}} \frac{\frac{π}{2} d t}{\sin^{4} t + \cos^{4} t} - I 2 I = \frac{π}{8} \int_{0}^{\frac{π}{2}} \frac{d t}{\sin^{4} t + \cos^{4} t} 2 I = \frac{π}{8} \int_{0}^{\frac{π}{2}} \frac{\sec^{4} t d t}{\tan^{4} t + 1} L e t \tan t = y t h e n \sec^{2} t d t = d y 2 I = \frac{π}{8} \int_{0}^{\infty} \frac{(1 + y^{2}) d y}{1 + y^{4}} = \frac{π}{16} \int_{0}^{\infty} \frac{1 + \frac{1}{y^{2}}}{y^{2} + \frac{1}{y^{2}}} d y y - \frac{1}{y} = p I = \frac{π}{16} \int_{- \infty}^{\infty} \frac{d p}{p^{2} + {(\sqrt{2})}^{2}} = \frac{π}{16 \sqrt{2}} {[\tan^{- 1} (\frac{p}{\sqrt{2}})]}_{- \infty}^{\infty} I = \frac{π^{2}}{16 \sqrt{2}}$