Question

\(\left| \frac{120}{\pi^3} \int_0^{\frac{\pi}{2}} \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx \right| \text{ is equal to } \underline{\hspace{2cm}}.\)

Answer 1

Correct Answer: 15

The integral to be evaluated is:

\[ \int_{0}^{\pi} \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx. \]

The denominator can be simplified using the identity:

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

Given that \(\sin^2 x + \cos^2 x = 1\), the denominator becomes:

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

Using the identity \(\sin^2 x \cos^2 x = \left(\frac{\sin 2x}{2}\right)^2 = \frac{\sin^2 2x}{4}\), we get:

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

The integral can now be written as:

\[ \int_{0}^{\pi} \frac{x^2 \sin x \cos x}{1 - \frac{\sin^2 2x}{2}} dx. \]

Substitute \(\sin x \cos x = \frac{1}{2} \sin 2x\) into the integral:

\[ \int_{0}^{\pi} \frac{x^2 \cdot \frac{1}{2} \sin 2x}{1 - \frac{\sin^2 2x}{2}} dx = \frac{1}{2} \int_{0}^{\pi} \frac{x^2 \sin 2x}{1 - \frac{\sin^2 2x}{2}} dx. \]

Utilize the symmetry of \(\sin 2x\) around \(x = \frac{\pi}{2}\) to evaluate the integral over \([0, \pi]\). Split the integral and evaluate each part.

The evaluation of the integral yields:

\[ \frac{120}{\pi^2} \int_{0}^{\pi} \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx = 15. \]

Therefore, the final answer is:

\[ 15. \]