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

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Question

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

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) ધારો કે $I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \quad ......(1)$ગુણધર્મ $\int\limits_a^b f(x)dx = \int\limits_a^b f(a+b-x)dx$ નો ઉપયોગ કરતા,આપણે $x$ ને $-\frac{\pi}{2} + \frac{\pi}{2} - x = -x$ વડે બદલીએ છીએ:$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)$$(1)$ અને $(2)$ નો સરવાળો કરતા:$2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x + {2^x}{{\sin }^2}x}}{{1 + {2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x dx}$કારણ કે $\sin^2 x$ એ યુગ્મ વિધેય છે,$2I = 2 \int\limits_0^{\pi /2} {\sin^2 x dx} = 2 \int\limits_0^{\pi /2} {\frac{1 - \cos 2x}{2} dx}$$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}$તેથી,$I = \frac{\pi}{4}$.

$\int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $ નું મૂલ્ય શું છે?

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