If I = int_0^pi x left{ sin^2(sin x) + cos^2( | vedclass

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(B) We have $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:

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$4$

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(B) We have $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin(\pi - x)) + \cos^2(\cos(\pi - x)) \right\} dx$ $I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin x) + \cos^2(-\cos x) \right\} dx$ $I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ Adding the two expressions for $I$: $2I = \int_0^\pi \pi \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ $I = \frac{\pi}{2} \int_0^\pi \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$: $I = \frac{\pi}{2} \cdot 2 \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ $I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ Now,using $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ again: $I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin(\pi/2 - x)) + \cos^2(\cos(\pi/2 - x)) \right\} dx$ $I = \pi \int_0^{\pi/2} \left\{ \sin^2(\cos x) + \cos^2(\sin x) \right\} dx$ Adding these two forms of $I$: $2I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) + \sin^2(\cos x) + \cos^2(\sin x) \right\} dx$ $2I = \pi \int_0^{\pi/2} \left\{ (\sin^2(\sin x) + \cos^2(\sin x)) + (\cos^2(\cos x) + \sin^2(\cos x)) \right\} dx$ $2I = \pi \int_0^{\pi/2} (1 + 1) dx = 2\pi \int_0^{\pi/2} dx = 2\pi \cdot \frac{\pi}{2} = \pi^2$ Thus,$I = \frac{\pi^2}{2}$. Since $\pi^2 \approx 9.869$,then $I \approx \frac{9.869}{2} = 4.9345$. Therefore,$[I] = [4.9345] = 4$.

If $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$,then $[I] = \ldots$. Here,$[.]$ denotes the greatest integer function.

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