If f(x) = frac{sin^2 pi x}{1+pi^x},then int ( | vedclass

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(B) We are given $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$. \ We need to evaluate $I = \int (f(x) + f(-x)) \, dx$. \ First,calculate $f(-x)$: \ $f(-x)

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$\frac{1}{2} x - \frac{\sin 2 \pi x}{4 \pi} + c$,(where $c$ is a constant of integration)

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(B) We are given $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$. \ We need to evaluate $I = \int (f(x) + f(-x)) \, dx$. \ First,calculate $f(-x)$: \ $f(-x) = \frac{\sin^2(-\pi x)}{1+\pi^{-x}} = \frac{\sin^2(\pi x)}{1+\frac{1}{\pi^x}} = \frac{\pi^x \sin^2 \pi x}{\pi^x + 1}$. \ Now,add $f(x)$ and $f(-x)$: \ $f(x) + f(-x) = \frac{\sin^2 \pi x}{1+\pi^x} + \frac{\pi^x \sin^2 \pi x}{1+\pi^x} = \frac{\sin^2 \pi x (1+\pi^x)}{1+\pi^x} = \sin^2 \pi x$. \ Now,integrate: \ $I = \int \sin^2 \pi x \, dx = \int \frac{1 - \cos 2 \pi x}{2} \, dx$. \ $I = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos 2 \pi x \, dx$. \ $I = \frac{x}{2} - \frac{1}{2} \cdot \frac{\sin 2 \pi x}{2 \pi} + c = \frac{x}{2} - \frac{\sin 2 \pi x}{4 \pi} + c$.

If $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$,then $\int (f(x) + f(-x)) \, dx$ is equal to

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