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

Given:

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

Step 1:

Let \[ I = \int_{0}^{\pi/2} \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx \] Apply the property $\int_{0}^{a} f(x)\, dx = \int_{0}^{a} f(a-x)\, dx$. \[ I = \int_{0}^{\pi/2} \frac{(\pi - x)^2 \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx \]

Step 2:

Add the two expressions for $I$. \[ 2I = \int_{0}^{\pi/2} \frac{[x^2 + (\pi - x)^2] \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx \] Simplify the numerator. \[ 2I = \int_{0}^{\pi/2} \frac{(\pi^2 - 2\pi x + 2x^2) \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx \]

Step 3:

Separate the integral into two parts. \[ 2I = 2\pi \int_{0}^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx \] \[ - \pi^2 \int_{0}^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx \]

Step 4:

Evaluate the first integral: $\int_{0}^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$. Use the identity $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$. The integral becomes: \[ \int_{0}^{\pi/2} \frac{\sin 2x}{2 - \sin^2 2x} \, dx \]

Step 5:

Perform a substitution: Let $t = \cos 2x$, so $dt = -2\sin 2x\, dx$. The integral transforms to: \[ I = \frac{\pi^2}{4} \int_{0}^{1} \frac{dt}{1 + t^2} \]

Step 6:

Evaluate the resulting integral. \[ = \frac{\pi^2}{4} \left[ t - \frac{t^3}{3} \right]_0^1 = \frac{\pi^2}{4} \left(1 - \frac{1}{3}\right) \] \[ = \frac{\pi^2}{4} \times \frac{2}{3} = \frac{\pi^2}{6} \] The final result is: \[ \frac{120}{8} + \frac{\pi^2}{8} = 15 \]