intlimits_{frac{-pi}{2}}^{frac{pi}{2}} frac{s | vedclass

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Question

(A) माना $I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + (2017)^x} \, dx$. गुणधर्म $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) माना $I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + (2017)^x} \, dx$. गुणधर्म $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$ का उपयोग करने पर,हमें प्राप्त होता है $I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2(-x)}{1 + (2017)^{-x}} \, dx$. चूंकि $\sin^2(-x) = \sin^2 x$,इसलिए $I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + \frac{1}{(2017)^x}} \, dx = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{(2017)^x \sin^2 x}{(2017)^x + 1} \, dx$. $I$ के दोनों समीकरणों को जोड़ने पर: $2I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x + (2017)^x \sin^2 x}{1 + (2017)^x} \, dx = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx$. चूंकि $\sin^2 x$ एक सम फलन है,इसलिए $2I = 2 \int\limits_{0}^{\frac{\pi}{2}} \sin^2 x \, dx$. $I = \int\limits_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \int\limits_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

$\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + (2017)^x} \, dx$ का मान ज्ञात कीजिए।

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