The value of int_{-frac{pi}{2}}^{frac{pi}{2}} | vedclass

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(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \,d x$ ... $(i)$ Using the property $\int_{a}^{b} f(x) \,d x = \int_{a}^{b}

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$\frac{\pi}{4}$

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(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \,d x$ ... $(i)$ Using the property $\int_{a}^{b} f(x) \,d x = \int_{a}^{b} f(a+b-x) \,d x$, we get: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 (-x)}{1+2^{-x}} \,d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+\frac{1}{2^x}} \,d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2^x \sin ^2 x}{2^x+1} \,d x$ ... (ii) Adding $(i)$ and (ii): $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x (1+2^x)}{1+2^x} \,d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \,d x$ Since $\sin ^2 x$ is an even function, $2I = 2 \int_{0}^{\frac{\pi}{2}} \sin ^2 x \,d x$, so $I = \int_{0}^{\frac{\pi}{2}} \sin ^2 x \,d x$. Using $\sin ^2 x = \frac{1-\cos 2x}{2}$: $I = \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2x}{2} \,d x = \frac{1}{2} [x - \frac{\sin 2x}{2}]_{0}^{\frac{\pi}{2}} = \frac{1}{2} [(\frac{\pi}{2} - 0) - (0 - 0)] = \frac{\pi}{4}$.

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \,d x$ is

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