Question

$\int\limits_{-\pi/4}^{\pi/4}\frac{e^x\,\sec^2x\,dx}{e^{2x}-1}$ is equal to

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

Answer 1

Answer: (A)

0

Answer 2

Let $f\left(x\right) =\frac{e^{x} sec^{2}\,x}{e^{2x}-1}$ $\therefore f \left(-x\right)=\frac{e^{-x} sec^{2}\left(-x\right)}{e^{-2x}-1}=\frac{\frac{1}{e^{x}}}{\frac{1}{e^{2x}}-1}sec^{2}\, x$ $=\frac{e^{2} x}{e^{x}\left(1-e^{2x}\right)} sec^{2}\,x=-\frac{e^{x}}{e^{2x}-1}sec^{2}\, x=f \left(x\right)$ $\therefore$ f i s an odd function $\therefore \int\limits_{-\pi 4}^{\pi 4}\frac{e^{x}sec^{2}\,x}{e^{2x}-1} dx=0$