If a0,then int_{-pi}^pi frac{sin ^2 x}{1+a^x | vedclass

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(A) Let $I = \int_{-\pi}^\pi \frac{\sin ^2 x}{1+a^x} dx$  $(i)$ Using the property $\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$ is not directly

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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 $\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$ is not directly applicable here,so we use $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Replacing $x$ with $-x$: $I = \int_{-\pi}^\pi \frac{\sin ^2(-x)}{1+a^{-x}} dx = \int_{-\pi}^\pi \frac{\sin ^2 x}{1+\frac{1}{a^x}} dx = \int_{-\pi}^\pi \frac{a^x \sin ^2 x}{a^x+1} dx$  $(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_{-\pi}^\pi \frac{\sin ^2 x + a^x \sin ^2 x}{1+a^x} dx = \int_{-\pi}^\pi \frac{\sin ^2 x(1+a^x)}{1+a^x} dx$ $2I = \int_{-\pi}^\pi \sin ^2 x dx$ Since $\sin ^2 x$ is an even function,$2I = 2 \int_0^\pi \sin ^2 x dx$ $I = \int_0^\pi \frac{1-\cos 2x}{2} dx = \frac{1}{2} [x - \frac{\sin 2x}{2}]_0^\pi$ $I = \frac{1}{2} [(\pi - 0) - (0 - 0)] = \frac{\pi}{2}$

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

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