The value of intlimits_{ - pi /2}^{pi /2} {fr | vedclass

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

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$\frac{\pi}{4}$

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(D) Let $I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \quad ......(1)$Using the property $\int\limits_a^b f(x)dx = \int\limits_a^b f(a+b-x)dx$,we replace $x$ with $-\frac{\pi}{2} + \frac{\pi}{2} - x = -x$:$I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}(-x)}}{{1 + {2^{-x}}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + \frac{1}{2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{2^x}{{\sin }^2}x}}{{2^x + 1}}} dx \quad ......(2)$Adding $(1)$ and $(2)$:$2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x + {2^x}{{\sin }^2}x}}{{1 + {2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x dx}$Since $\sin^2 x$ is an even function,$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}$Therefore,$I = \frac{\pi}{4}$.

The value of $\int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $ is

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