Mathematics

$$\int\limits_{\dfrac{{ - \pi }}{4}}^{\dfrac{\pi }{4}} {\dfrac{{{e^x}\left( {x\sin x} \right)}}{{{e^{2x}} - 1}}} \,dx$$ is equal to 


ANSWER

$$0$$


SOLUTION
Solution 
$$\displaystyle \int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}} \dfrac{e^{x}(x \sin x )}{e^{2x}-1} dx. $$

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

$$f(-x)=-\dfrac{e^{x}(x \sin x)}{e^{2x}-1}=-f(x)$$

$$f(x)$$ is an odd function 

$$\displaystyle \int _{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}} f(x)=0$$ property of define integral.   
$$A$$ is correct 





























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Single Correct Medium Published on 17th 09, 2020
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