Question

Evaluate: \[ I = \int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \, dx \]

Hint:

Use the identity \(\sin 2x = 2 \sin x \cos x\) to simplify trigonometric expressions in integrals involving products of sine and cosine.

Answer 1

Recall the identity \[\sin 2x = 2 \sin x \cos x\]. This implies \[\sin x \cos x = \frac{1}{2} \sin 2x\]. The integral transforms to \[I = \int_0^{\frac{\pi}{4}} \frac{\frac{1}{2} \sin 2x}{\cos^4 x + \sin^4 x} \, dx\]. Simplify the denominator: \[\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\cos^2 x \sin^2 x\]. Given \(\cos^2 x + \sin^2 x = 1\), the denominator becomes \[\cos^4 x + \sin^4 x = 1 - 2 \cos^2 x \sin^2 x\]. Substituting this back into the integral yields \[I = \int_0^{\frac{\pi}{4}} \frac{\frac{1}{2} \sin 2x}{1 - 2 \cos^2 x \sin^2 x} \, dx\]. This integral is solvable using standard or numerical methods to obtain the final result.