Q. 65 - 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.)

$\int \frac{e^x}{\csc e^x}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{e^x}{\csc e^x}\mathrm{d}x=-{\mathrm{cose}}^{\mathrm{x}}+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals.

$\int \frac{e^x}{\csc e^x}\mathrm{d}x$

Step 2. Using the substitution method.

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

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \frac{e^x}{\csc e^x}\mathrm{d}x=\int \frac{1}{\csc u}\mathrm{du} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{csce}}^{\mathrm{x}}}\mathrm{dx}=\int \sin \mathrm{u}\mathrm{du} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{csce}}^{\mathrm{x}}}\mathrm{dx}=-\mathrm{cosu}+\mathrm{C} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{csce}}^{\mathrm{x}}}\mathrm{dx}=-{\mathrm{cose}}^{\mathrm{x}}+\mathrm{C} \end{gathered}$$