Question

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

Answer 1

Correct Answer: 15

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: \[ \boxed{15}. \]