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

The Correct Option is C

\( I = \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}\)
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: \[ I = \frac{\pi}{16 \sqrt{2}} \left[ \tan^{-1} \left( \frac{p}{\sqrt{2}} \right) \right]_{0}^{\infty} \] \[ I = \frac{\pi}{16 \sqrt{2}} \cdot \frac{\pi}{2} \] \[ I = \frac{\pi^2}{16 \sqrt{2}} \]