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

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(B) ધારો કે $I = \int_0^\pi x (\sin^2(\sin x) + \cos^2(\cos x)) dx$. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા: $I = \int_0^\pi (\

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

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(B) ધારો કે $I = \int_0^\pi x (\sin^2(\sin x) + \cos^2(\cos x)) dx$. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા: $I = \int_0^\pi (\pi - x) (\sin^2(\sin(\pi - x)) + \cos^2(\cos(\pi - x))) dx$ કારણ કે $\sin(\pi - x) = \sin x$ અને $\cos(\pi - x) = -\cos x$,આપણને મળે: $I = \int_0^\pi (\pi - x) (\sin^2(\sin x) + \cos^2(\cos x)) dx$ $I$ માટેના બંને સમીકરણોનો સરવાળો કરતા: $2I = \pi \int_0^\pi (\sin^2(\sin x) + \cos^2(\cos x)) dx$ ગુણધર્મ $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ નો ઉપયોગ કરતા: $2I = 2\pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2(\cos x)) dx$ $I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2(\cos x)) dx . . . (i)$ ફરીથી ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા: $I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\cos x) + \cos^2(\sin x)) dx . . . (ii)$ $(i)$ અને $(ii)$ નો સરવાળો કરતા: $2I = \pi \int_0^{\frac{\pi}{2}} ((\sin^2(\sin x) + \cos^2(\sin x)) + (\sin^2(\cos x) + \cos^2(\cos x))) dx$ $2I = \pi \int_0^{\frac{\pi}{2}} (1 + 1) dx = 2\pi [x]_0^{\frac{\pi}{2}} = \pi^2$ $I = \frac{\pi^2}{2}$

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

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