int_0^{pi} x f(sin x) , dx ની કિંમત શોધો. | vedclass

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(D) ધારો કે $I = \int_0^{\pi} x f(\sin x) \, dx$. ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા: $I = \int_0^{\pi} (\pi - x) f(\

Vedclass · Accepted Answer

$\pi \int_0^{\frac{\pi}{2}} f(\sin x) \, dx$

Vedclass · Answer

(D) ધારો કે $I = \int_0^{\pi} x f(\sin x) \, dx$. ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા: $I = \int_0^{\pi} (\pi - x) f(\sin(\pi - x)) \, dx$. કારણ કે $\sin(\pi - x) = \sin x$, તેથી: $I = \int_0^{\pi} (\pi - x) f(\sin x) \, dx = \pi \int_0^{\pi} f(\sin x) \, dx - \int_0^{\pi} x f(\sin x) \, dx$. $I = \pi \int_0^{\pi} f(\sin x) \, dx - I$. $2I = \pi \int_0^{\pi} f(\sin x) \, dx$. ગુણધર્મ $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ નો ઉપયોગ કરતા, જો $f(2a-x) = f(x)$ હોય, અહીં $2a = \pi$, તેથી $a = \frac{\pi}{2}$. કારણ કે $f(\sin(\pi - x)) = f(\sin x)$, તેથી: $2I = \pi \cdot 2 \int_0^{\frac{\pi}{2}} f(\sin x) \, dx$. $I = \pi \int_0^{\frac{\pi}{2}} f(\sin x) \, dx$.

$\int_0^{\pi} x f(\sin x) \, dx$ ની કિંમત શોધો.

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