Question

Evaluate the integrals.

\(\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx\)

Short answer

The value of \(\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx\)is \(40\).

Step-by-step solution

Definition of the integral

Integral calculus is an area of mathematics that deals with the calculation, properties, and applications of integrals.

Evaluating the integral.

The given data is,

\(\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx\)

Given that:\(\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}\)and\(\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\)

Therefore

\(\begin{aligned}{c}\sinh x + \cosh x = \frac{{{e^x} - {e^{ - x}}}}{2} + \frac{{{e^x} + {e^{ - x}}}}{2}\\ = \frac{{2{e^x}}}{2}\\ = {e^x}\\ = \int_{ - 10}^{10} {\frac{{2{e^x}}}{{{e^x}}}} dx\\ = \int_{ - 10}^{10} 2 dx\\ = \int_{ - 10}^{10} 2 \cdot {x^0}dx\end{aligned}\)

Using the power rule of integration, which states that for \(n \ne  - 1\) 

Then it is expressed as,

\(\begin{aligned}{c}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C\\ = \left( {2 \times \frac{{{x^{0 + 1}}}}{{0 + 1}}} \right)_{ - 10}^{10}\\ = (2x)_{ - 10}^{10}\\ = (2 \times 10 - 2 \times ( - 10))\\ = 20 + 20\\ = 40\end{aligned}\)

Hence the value of \(\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx\) is 40.