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Found in: Page 289

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the integrals.$$\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx$$

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

See the step by step solution

## Step 1 Definition of the integral

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

## Step 2 : 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}

## Step 3 : 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.