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

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(B) આપણી પાસે $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ છે. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) આપણી પાસે $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ છે. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા: $I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin(\pi - x)) + \cos^2(\cos(\pi - x)) \right\} dx$ $I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin x) + \cos^2(-\cos x) \right\} dx$ $I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ $I$ માટેના આ બંને પદોનો સરવાળો કરતા: $2I = \int_0^\pi \pi \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ $I = \frac{\pi}{2} \int_0^\pi \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ ગુણધર્મ $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ નો ઉપયોગ કરતા: $I = \frac{\pi}{2} \cdot 2 \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ $I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ હવે ફરીથી $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા: $I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin(\pi/2 - x)) + \cos^2(\cos(\pi/2 - x)) \right\} dx$ $I = \pi \int_0^{\pi/2} \left\{ \sin^2(\cos x) + \cos^2(\sin x) \right\} dx$ $I$ ના આ બંને સ્વરૂપોનો સરવાળો કરતા: $2I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) + \sin^2(\cos x) + \cos^2(\sin x) \right\} dx$ $2I = \pi \int_0^{\pi/2} \left\{ (\sin^2(\sin x) + \cos^2(\sin x)) + (\cos^2(\cos x) + \sin^2(\cos x)) \right\} dx$ $2I = \pi \int_0^{\pi/2} (1 + 1) dx = 2\pi \int_0^{\pi/2} dx = 2\pi \cdot \frac{\pi}{2} = \pi^2$ આમ,$I = \frac{\pi^2}{2}$. કારણ કે $\pi^2 \approx 9.869$,તેથી $I \approx \frac{9.869}{2} = 4.9345$. તેથી,$[I] = [4.9345] = 4$.

જો $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$ હોય,તો $[I] = \ldots$ શોધો. અહીં,$[.]$ એ મહત્તમ પૂર્ણાંક વિધેય દર્શાવે છે.

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