Question

${\int }_{-\pi /4}^{\pi /4}\frac{e^x{\sec }^{2}xdx}{e^{2x}-1}$is equal to

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

Answer 1

The correct option is A $0$

Let $g\left(x\right)=\frac{e^x{\sec }^{2}x}{e^{2x}-1}$
Now, $g(-x)=\frac{e^{-x}{\sec }^2(-x)}{e^{-2x}-1}=-\frac{e^x{\sec }^{2}x}{e^{2x}-1}=-g\left(x\right)$
$g\left(x\right)$ is and odd function.
Therefore, $I={\int }_{-\pi /4}^{\pi /4}\frac{e^x{\sec }^{2}xdx}{e^{2x}-1}=0$