int_0^pi x (sin^2(sin x) + cos^2(cos x)) dx = | vedclass

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(B) Let $I = \int_0^\pi x (\sin^2(\sin x) + \cos^2(\cos x)) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^\pi (

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$\frac{\pi^2}{2}$

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(B) Let $I = \int_0^\pi x (\sin^2(\sin x) + \cos^2(\cos x)) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^\pi (\pi - x) (\sin^2(\sin(\pi - x)) + \cos^2(\cos(\pi - x))) dx$ Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we get: $I = \int_0^\pi (\pi - x) (\sin^2(\sin x) + \cos^2(-\cos x)) dx$ $I = \int_0^\pi (\pi - x) (\sin^2(\sin x) + \cos^2(\cos x)) dx$ Adding the two expressions for $I$: $2I = \int_0^\pi \pi (\sin^2(\sin x) + \cos^2(\cos x)) dx$ $2I = \pi \int_0^\pi (\sin^2(\sin x) + \cos^2(\cos x)) dx$ Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$: $2I = 2\pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2(\cos x)) dx$ $I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2(\cos x)) dx . . . (i)$ Applying the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ again: $I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin(\frac{\pi}{2}-x)) + \cos^2(\cos(\frac{\pi}{2}-x))) dx$ $I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\cos x) + \cos^2(\sin x)) dx . . . (ii)$ Adding $(i)$ and $(ii)$: $2I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2(\cos x) + \sin^2(\cos x) + \cos^2(\sin x)) dx$ $2I = \pi \int_0^{\frac{\pi}{2}} ((\sin^2(\sin x) + \cos^2(\sin x)) + (\sin^2(\cos x) + \cos^2(\cos x))) dx$ Since $\sin^2 	heta + \cos^2 	heta = 1$: $2I = \pi \int_0^{\frac{\pi}{2}} (1 + 1) dx = 2\pi \int_0^{\frac{\pi}{2}} dx$ $2I = 2\pi [x]_0^{\frac{\pi}{2}} = 2\pi (\frac{\pi}{2}) = \pi^2$ $I = \frac{\pi^2}{2}$

$\int_0^\pi x (\sin^2(\sin x) + \cos^2(\cos x)) dx = $

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