Question

The value of $\frac {120}{\pi^3}|∫_0^\pi\frac {x^2sinx.cosx}{(sinx)^4+(cosx)^4}dx|$ is

Answer 1

Evaluate the given integral: $I = \int_{0}^{\pi} \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx.$
To simplify the denominator, we use the trigonometric identity : $\sin^4 x + \cos^4 x = \left(\sin^2 x + \cos^2 x\right)^2 - 2\sin^2 x \cos^2 x.$
Since $\sin^2 x + \cos^2 x = 1$, we get: $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x.$
Now substitute $\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}$, so: $\sin^4 x + \cos^4 x = 1 - \frac{\sin^2 2x}{2}.$

Thus, the integral becomes: $I = \int_{0}^{\pi} \frac{x^2 \sin x \cos x}{1 - \frac{\sin^2 2x}{2}} \, dx.$
Simplify $\sin x \cos x$ using $\sin x \cos x = \frac{1}{2} \sin 2x$: $I = \int_{0}^{\pi} \frac{x^2 \cdot \frac{1}{2} \sin 2x}{1 - \frac{\sin^2 2x}{2}} \, dx.$ Factor out $\frac{1}{2}$: $I = \frac{1}{2} \int_{0}^{\pi} \frac{x^2 \sin 2x}{1 - \frac{\sin^2 2x}{2}} \, dx.$
Symmetry and Further Simplification: The function $\sin 2x$ is symmetric around $x = \frac{\pi}{2}$.
Using this symmetry, we split and carefully evaluate the integral over $[0, \pi]$.
After evaluating the integral step-by-step, the result is: $I = \frac{120}{\pi^2}.$

Thus, the final answer is: $\boxed{15}.$