intlimits_{ - pi /2}^{pi /2} {frac{{{{sin }^2 | vedclass

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(D) माना $I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \quad ......(1)$गुणधर्म $\int\limits_a^b f(x)dx = \int\limits_a^

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$\frac{\pi}{4}$

Vedclass · Answer

(D) माना $I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \quad ......(1)$गुणधर्म $\int\limits_a^b f(x)dx = \int\limits_a^b f(a+b-x)dx$ का उपयोग करते हुए,हम $x$ को $-\frac{\pi}{2} + \frac{\pi}{2} - x = -x$ से प्रतिस्थापित करते हैं:$I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}(-x)}}{{1 + {2^{-x}}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + \frac{1}{2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{2^x}{{\sin }^2}x}}{{2^x + 1}}} dx \quad ......(2)$$(1)$ और $(2)$ को जोड़ने पर:$2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x + {2^x}{{\sin }^2}x}}{{1 + {2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x dx}$चूँकि $\sin^2 x$ एक सम फलन है,$2I = 2 \int\limits_0^{\pi /2} {\sin^2 x dx} = 2 \int\limits_0^{\pi /2} {\frac{1 - \cos 2x}{2} dx}$$2I = \int\limits_0^{\pi /2} {(1 - \cos 2x) dx} = [x - \frac{\sin 2x}{2}]_0^{\pi /2} = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2}$अतः,$I = \frac{\pi}{4}$.

$\int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $ का मान क्या है?

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