If a 0,then int_{-pi}^{pi} frac{sin^2 x}{1+ | vedclass

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

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$\frac{\pi}{2}$

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

If $a > 0$,then $\int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$ is equal to

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