Q. 70 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{d}x={\cos }^{-1}{\mathrm{e}}^{\mathrm{x}}+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals.

$\int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{d}x$

Step 2. Using the substitution method.

Let

$$\begin{gathered} \mathrm{u}={\mathrm{e}}^{\mathrm{x}} \\ \frac{du}{dx}={\mathrm{e}}^{\mathrm{x}} \\ d\mathrm{u}={\mathrm{e}}^{\mathrm{x}}d\mathrm{x} \end{gathered}$$

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{d}x=\int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\left({\mathrm{e}}^{\mathrm{x}}\right)}^2-1}}\mathrm{d}x \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{d}x=\int \frac{1}{\sqrt{u^2-1}}\mathrm{du} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{dx}=-\int \frac{1}{\sqrt{1-{\mathrm{u}}^2}}\mathrm{du} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{dx}=-\left(-{\cos }^{-1}\mathrm{u}\right)+\mathrm{C} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{dx}={\cos }^{-1}\mathrm{u}+\mathrm{C} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{dx}={\cos }^{-1}{\mathrm{e}}^{\mathrm{x}}+\mathrm{C} \end{gathered}$$