Question

The value of the integral $\int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}$

• 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 1

Answer: (C)

$\frac{\sqrt{2} \pi^2}{32}$

Answer 2

$I = \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}$
Let $2x = t$, then $dx = \frac{1}{2} dt$, $I = \frac{1}{4} \int_{0}^{\pi} \frac{t \, dt}{\sin^4 t + \cos^4 t}$ Using symmetry: $I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{t + \frac{\pi}{2} - t}{\sin^4 t + \cos^4 t} \, dt$ $I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2}}{\sin^4 t + \cos^4 t} \, dt - I$ $2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}$ Let $\tan t = y$, then $\sec^2 t \, dt = dy$: $2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{(1 + y^2) \, dy}{1 + y^4}$ $2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{dy}{y^2 + 1}$ Let $y = p$, then: $I = \frac{\pi}{16} \int_{0}^{\infty} \frac{dp}{p^2 + (\sqrt{2})^2}$ Using the standard integral formula: $I = \frac{\pi}{16 \sqrt{2}} \left[ \tan^{-1} \left( \frac{p}{\sqrt{2}} \right) \right]_{0}^{\infty}$ $I = \frac{\pi}{16 \sqrt{2}} \cdot \frac{\pi}{2}$ $I = \frac{\pi^2}{16 \sqrt{2}}$