left|frac{120}{pi^3} int_0^pi frac{x^2 sin x | vedclass

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(A) Let $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have $I = \in

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$15$

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(A) Let $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have $I = \int_0^\pi \frac{(\pi-x)^2 \sin(\pi-x) \cos(\pi-x)}{\sin^4(\pi-x) + \cos^4(\pi-x)} dx = \int_0^\pi \frac{(\pi-x)^2 \sin x (-\cos x)}{\sin^4 x + \cos^4 x} dx = -\int_0^\pi \frac{(\pi^2 - 2\pi x + x^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Adding the two expressions for $I$: $2I = \int_0^\pi \frac{(x^2 - (\pi^2 - 2\pi x + x^2)) \sin x \cos x}{\sin^4 x + \cos^4 x} dx = \int_0^\pi \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. $2I = 2\pi \int_0^\pi \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx - \pi^2 \int_0^\pi \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Using the property $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ if $f(2a-x) = f(x)$,we find the second integral is $0$ because $\sin(\pi-x)\cos(\pi-x) = -\sin x \cos x$. Thus,$2I = 2\pi \cdot \frac{\pi}{2} \int_0^\pi \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx = 0$. Wait,re-evaluating: $I = \int_0^{\pi/2} \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx + \int_{\pi/2}^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Let $x = \pi - t$ in the second integral: $\int_0^{\pi/2} \frac{(\pi-t)^2 \sin t (-\cos t)}{\sin^4 t + \cos^4 t} dt = -\int_0^{\pi/2} \frac{(\pi^2 - 2\pi t + t^2) \sin t \cos t}{\sin^4 t + \cos^4 t} dt$. $I = \int_0^{\pi/2} \frac{(t^2 - \pi^2 + 2\pi t - t^2) \sin t \cos t}{\sin^4 t + \cos^4 t} dt = \int_0^{\pi/2} \frac{(2\pi t - \pi^2) \sin t \cos t}{\sin^4 t + \cos^4 t} dt$. $= 2\pi \int_0^{\pi/2} \frac{t \sin t \cos t}{\sin^4 t + \cos^4 t} dt - \pi^2 \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt$. Using $\int_0^{\pi/2} f(\sin x, \cos x) dx = \int_0^{\pi/2} f(\cos x, \sin x) dx$,the first integral is $\frac{\pi}{4} \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt$. $I = (2\pi \cdot \frac{\pi}{4} - \pi^2) \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt = -\frac{\pi^2}{2} \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt$. Let $u = \sin^2 t$,$du = 2 \sin t \cos t dt$. Integral becomes $-\frac{\pi^2}{4} \int_0^1 \frac{du}{u^2 + (1-u)^2} = -\frac{\pi^2}{4} \int_0^1 \frac{du}{2u^2 - 2u + 1} = -\frac{\pi^2}{8} \int_0^1 \frac{du}{(u-1/2)^2 + 1/4} = -\frac{\pi^2}{8} [2 	an^{-1}(2u-1)]_0^1 = -\frac{\pi^2}{4} (\frac{\pi}{4} - (-\frac{\pi}{4})) = -\frac{\pi^3}{8}$. Taking absolute value: $|-\frac{\pi^3}{8}| = \frac{\pi^3}{8}$. Result: $\frac{120}{\pi^3} \cdot \frac{\pi^3}{8} = 15$.

$\left|\frac{120}{\pi^3} \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx\right|$ is equal to:

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