int_{0}^{pi} x f(sin x) dx = | vedclass

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(D) माना $I = \int_{0}^{\pi} x f(\sin x) dx$  $(1)$गुणधर्म $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ का उपयोग करने पर,हमें प्राप्त होता है:$I =

Vedclass · Accepted Answer

$\pi \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$

Vedclass · Answer

(D) माना $I = \int_{0}^{\pi} x f(\sin x) dx$  $(1)$गुणधर्म $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ का उपयोग करने पर,हमें प्राप्त होता है:$I = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x)) dx$चूँकि $\sin(\pi - x) = \sin x$,इसलिए:$I = \int_{0}^{\pi} (\pi - x) f(\sin x) dx$$I = \pi \int_{0}^{\pi} f(\sin x) dx - \int_{0}^{\pi} x f(\sin x) dx$$I = \pi \int_{0}^{\pi} f(\sin x) dx - I$$2I = \pi \int_{0}^{\pi} f(\sin x) dx$$I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) dx$गुणधर्म $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ का उपयोग करने पर यदि $f(2a-x) = f(x)$ हो:यहाँ $f(\sin(\pi - x)) = f(\sin x)$,इसलिए $\int_{0}^{\pi} f(\sin x) dx = 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$$I = \frac{\pi}{2} 	imes 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) dx = \pi \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ का उपयोग करने पर,हमें प्राप्त होता है $\int_{0}^{\frac{\pi}{2}} f(\sin x) dx = \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$अतः,$I = \pi \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.

$\int_{0}^{\pi} x f(\sin x) dx = $

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