Question

$\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{e^x\left(x\sin x\right)}{e^{2x}-1}dx$ is equal to

• A. $0$
• B. $2$
• C. $e$
• D. none of these

Answer 1

Answer: (A)

$0$

Solution

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{e^x\left(x\sin x\right)}{e^{2x}-1}dx.$

let $f\left(x\right)=\frac{e^x\left(x\sin x\right)}{e^{2x}-1}$ $f(-x)=\frac{e^{-x}(-x\sin -x)}{\frac{1-e^{2x}}{e^{2x}}}$

$f(-x)=-\frac{e^x\left(x\sin x\right)}{e^{2x}-1}=-f\left(x\right)$

$f\left(x\right)$ is an odd function

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}f\left(x\right)=0$ property of define integral.

$A$ is correct