int_{-frac{pi}{4}}^{frac{pi}{4}} frac{e^x(x s | vedclass

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Question

(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^x(x \sin x)}{e^{2x}-1} dx$. Consider the integrand $f(x) = \frac{e^x(x \sin x)}{e^{2x}-1}$.

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(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^x(x \sin x)}{e^{2x}-1} dx$. Consider the integrand $f(x) = \frac{e^x(x \sin x)}{e^{2x}-1}$. We can rewrite $f(x)$ as: $f(x) = \frac{e^x(x \sin x)}{e^x(e^x - e^{-x})} = \frac{x \sin x}{e^x - e^{-x}} = \frac{x \sin x}{2 \sinh x}$. Now,check if $f(x)$ is an even or odd function by evaluating $f(-x)$: $f(-x) = \frac{(-x) \sin(-x)}{e^{-x} - e^x} = \frac{(-x)(-\sin x)}{-(e^x - e^{-x})} = \frac{x \sin x}{-(e^x - e^{-x})} = -f(x)$. Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function. According to the property of definite integrals,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function. Therefore,$I = 0$.

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^x(x \sin x)}{e^{2x}-1} dx =$

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