Evaluate the following integral.

{eq}\displaystyle \int e^x \cosh x \ dx {/eq}

Question:

Evaluate the following integral.

{eq}\displaystyle \int e^x \cosh x \ dx {/eq}

Cosine Hyperbolic Function as Integrand:

The cosine hyperbolic function is defined as the ratio of the sum of exponential functions {eq}e^x {/eq} and {eq}e^{-x} {/eq} over the integer value {eq}2 {/eq}. The product rule of exponential expression for the simplification of the integral expression is:

{eq}e^a\cdot e^b=e^{a+b} {/eq}

Answer and Explanation: 1

Given integral:

{eq}\displaystyle \int e^x \cosh x \ dx=?\\[2ex] {/eq}

The cosine hyperbolic function in exponential form is:

{eq}\cosh x=\dfrac{e^x+e^{-x}}{2}\\[2ex] {/eq}

Substitute the above exponential formula in the integral expression and simplify it.

{eq}\begin{align*} \int e^x \cosh x \ dx&=\int e^x \left ( \frac{e^x+e^{-x}}{2} \right ) \ dx\\[2ex] &=\frac{1}{2}\cdot \int e^x (e^x+e^{-x})\ dx\\[2ex] &=\frac{1}{2}\cdot \int (e^x \cdot e^x+e^x \cdot e^{-x})\ dx\\[2ex] &=\frac{1}{2}\cdot \int (e^{x+x}+e^{x-x})\ dx&\left [ \because e^a \cdot e^b=e^{a+b} \right ]\\[2ex] &=\frac{1}{2}\cdot \int (e^{2x}+e^{0})\ dx\\[2ex] &=\frac{1}{2}\cdot \int (e^{2x}+1)\ dx&\left [ \because e^{0}=1 \right ]\\[2ex] \end{align*} {/eq}

Solve each anti-derivative using the common integral formulas.

{eq}\begin{align*} \frac{1}{2}\cdot \int (e^{2x}+1)\ dx&=\frac{1}{2}\cdot\left(\int e^{2x}\ dx+\int 1\ dx\right)\\[2ex] &=\frac{1}{2}\cdot\left(\frac{e^{2x}}{2}+x\right)+C\\[2ex] &=\frac{e^{2x}}{4}+\frac{x}{2}+C\\[2ex] \end{align*} {/eq}

Thus, the value of the integral expression is {eq}\boxed{\displaystyle \frac{e^{2x}}{4}+\frac{x}{2}+C} {/eq}.


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Integration Problems in Calculus: Solutions & Examples

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Chapter 13 / Lesson 13
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