Q. 57 - 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.)

$\int \frac{\sin (1/x)}{x^2}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{\sin (1/x)}{x^2}\mathrm{d}x=\cos (1/\mathrm{x})+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals.

$\int \frac{\sin (1/x)}{x^2}\mathrm{d}x$

Step 2. Using the substitution method.

Let

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

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \frac{\sin (1/x)}{x^2}\mathrm{d}x=-\int \sin u\mathrm{du} \\ \int \frac{\sin (1/\mathrm{x})}{{\mathrm{x}}^2}\mathrm{dx}=-\int \mathrm{sinudu} \\ \int \frac{\sin (1/\mathrm{x})}{{\mathrm{x}}^2}\mathrm{dx}=-(-\mathrm{cosu})+\mathrm{C} \\ \int \frac{\sin (1/\mathrm{x})}{{\mathrm{x}}^2}\mathrm{dx}=\mathrm{cosu}+\mathrm{C} \\ \int \frac{\sin (1/\mathrm{x})}{{\mathrm{x}}^2}\mathrm{dx}=\cos (1/\mathrm{x})+\mathrm{C} \end{gathered}$$