Q. 64 - 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.)

$\int \frac{1}{\sqrt{\mathrm{x}}\sec \sqrt{\mathrm{x}}}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{1}{\sqrt{\mathrm{x}}\sec \sqrt{\mathrm{x}}}\mathrm{d}x=\frac{1}{2}\sin \sqrt{\mathrm{x}}+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals. 

$\int \frac{1}{\sqrt{\mathrm{x}}\sec \sqrt{\mathrm{x}}}\mathrm{d}x$

Step 2. Using the substitution method.

Let

$$\begin{gathered} u=\sqrt{\mathrm{x}} \\ u=x^{1/2} \\ \frac{du}{dx}=\frac{1}{2}{x}^{1/2-1} \\ \frac{du}{dx}=\frac{1}{2}{x}^{-1/2} \\ \frac{du}{dx}=\frac{1}{2}·\frac{1}{\sqrt{x}} \\ \frac{1}{2}du=\frac{1}{\sqrt{x}}dx \end{gathered}$$

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \frac{1}{\sqrt{\mathrm{x}}\sec \sqrt{\mathrm{x}}}\mathrm{d}x=\frac{1}{2}\int \frac{1}{\sec u}\mathrm{du} \\ \int \frac{1}{\sqrt{\mathrm{x}}\sec \sqrt{\mathrm{x}}}\mathrm{dx}=\frac{1}{2}\int \cos \mathrm{u}\mathrm{du} \\ \int \frac{1}{\sqrt{\mathrm{x}}\sec \sqrt{\mathrm{x}}}\mathrm{dx}=\frac{1}{2}\mathrm{sinu}+\mathrm{C} \\ \int \frac{1}{\sqrt{\mathrm{x}}\sec \sqrt{\mathrm{x}}}\mathrm{dx}=\frac{1}{2}\sin \sqrt{\mathrm{x}}+\mathrm{C} \end{gathered}$$