Question

Evaluate :${\int }_{-1}^1\left[\frac{\cos hx}{\left(1+e^{2x}\right)}\right]dx$

• A. $0$
• B. $1$
• C. $\left[e^2-1\right]/2e$
• D. $\left[e^2+1\right]/2e$

Answer 1

The correct option is C

$\left[e^2-1\right]/2e$

Evaluating the integral :

Consider the given equation as:

$I={\int }_{-1}^1\frac{\cos hx}{1+e^{2x}}dx\dots \dots \left(1\right)$

We know that,

$\cos hx=\frac{e^x+e^{-x}}{2}$

Substitute the value for $\cos hx$ in Equation $\left(1\right)$

$\begin{array}{rcl}I& =& {\int }_{-1}^1\frac{{\displaystyle \frac{e^x+e^{-x}}{2}}}{1+e^{2x}}dx\\ I& =& \frac{1}{2}{\int }_{-1}^1\frac{{\displaystyle e^x+\frac{1}{e^{-x}}}}{1+e^{2x}}dx\\ I& =& \frac{1}{2}{\int }_{-1}^1\frac{{\displaystyle e^{2x}+1}}{e^x\left(1+e^{2x}\right)}dx\\ I& =& \frac{1}{2}{\int }_{-1}^1\frac{{\displaystyle \left(1+e^{2x}\right)}}{e^x\left(1+e^{2x}\right)}dx\\ I& =& \frac{1}{2}{\int }_{-1}^1\frac{{\displaystyle 1}}{e^x}dx\\ I& =& \frac{1}{2}{\int }_{-1}^1{e}^{-x}dx\\ I& =& \frac{1}{2}{\left[\frac{e^{-x}}{-1}\right]}_{-1}^1\\ I& =& -\frac{1}{2}\left[e^{-1}-e^{-\left(-1\right)}\right]\\ I& =& -\frac{1}{2}\left[e^{-1}-e^1\right]\\ I& =& \frac{e^1-e^{-1}}{2}\\ I& =& \frac{e^1-{\displaystyle \frac{1}{e^1}}}{2}\\ I& =& \frac{e^2-1}{2e}\end{array}$

Therefore, the correct answer is Option (C).