Evaluate The Integral:${ \int \frac{1}{\sqrt{169} - X^2} , Dx }$

Mar 10, 2025 by ADMIN 66 views

**Evaluating the Integral of a Rational Function**

Introduction

In this article, we will delve into the evaluation of a specific integral, which involves a rational function. The integral in question is ${ \int \frac{1}{\sqrt{169} - x^2} , dx }$. This type of integral is commonly encountered in calculus and requires a thorough understanding of various techniques and strategies for integration.

Understanding the Integral

The given integral is a rational function, which means it is the ratio of two polynomials. In this case, the numerator is a constant (1), and the denominator is a quadratic expression, $\sqrt{169} - x^2$ . The integral can be rewritten as ${ \int \frac{1}{\sqrt{169} - x^2} , dx = \int \frac{1}{13 - x^2} , dx }$. This simplification is possible because $\sqrt{169} = 13$ .

Applying the Pythagorean Identity

To evaluate this integral, we can use the Pythagorean identity, which states that $a^2 + b^2 = c^2$ . In this case, we can rewrite the denominator as $13 - x^2 = (3x)^2 + (4)^2$ . This allows us to apply the Pythagorean identity, which states that $\sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2}$ .

Using Trigonometric Substitution

Using the Pythagorean identity, we can rewrite the integral as ${ \int \frac{1}{13 - x^2} , dx = \int \frac{1}{(3x)^2 + 4^2} , dx }$. This is a classic example of a trigonometric substitution problem. We can substitute $x = 2\tan(\theta)$ , which implies $dx = 2\sec^2(\theta) d\theta$ . This substitution allows us to rewrite the integral in terms of $\theta$ .

Rewriting the Integral in Terms of θ

Using the substitution $x = 2\tan(\theta)$ , we can rewrite the integral as ${ \int \frac{1}{13 - x^2} , dx = \int \frac{1}{(3(2\tan(\theta)))^2 + 4^2} , dx }$. Simplifying this expression, we get ${ \int \frac{1}{13 - x^2} , dx = \int \frac{1}{(6\tan(\theta))^2 + 4^2} , dx }$. This can be further simplified to ${ \int \frac{1}{13 - x^2} , dx = \int \frac{1}{36\tan^2(\theta) + 16} , dx }$.

Evaluating the Integral

Using the substitution $x = 2\tan(\theta)$ , we can rewrite the integral as ${ \int \frac{1}{36\tan^2(\theta) + 16} , dx = \int \frac{1}{36\tan^2(\theta) + 16} \cdot 2\sec^2(\theta) d\theta }$. This can be simplified to ${ \int \frac{1}{36\tan^2(\theta) + 16} , dx = \frac{1}{18} \int \frac{1}{\tan^2(\theta) + \left(\frac{4}{6}\right)^2} , dx }$. This is a standard integral that can be evaluated using the arctangent function.

Final Evaluation

Evaluating the integral, we get ${ \int \frac{1}{\sqrt{169} - x^2} , dx = \frac{1}{18} \arctan\left(\frac{6x}{4}\right) + C }$. This is the final answer to the given integral.

Conclusion

In this article, we evaluated the integral ${ \int \frac{1}{\sqrt{169} - x^2} , dx }$. We used various techniques and strategies for integration, including the Pythagorean identity and trigonometric substitution. The final answer to the integral is ${ \frac{1}{18} \arctan\left(\frac{6x}{4}\right) + C }$. This result can be used to solve a wide range of problems in calculus and other areas of mathematics.

Future Directions

In future articles, we can explore other techniques and strategies for integration, including the use of partial fractions and the evaluation of improper integrals. We can also apply these techniques to solve a wide range of problems in calculus and other areas of mathematics.

References

[1] "Calculus" by Michael Spivak
[2] "Calculus: Early Transcendentals" by James Stewart
[3] "A First Course in Calculus" by Serge Lang

Glossary

Pythagorean identity: A mathematical identity that states $a^2 + b^2 = c^2$ .
Trigonometric substitution: A technique used to evaluate integrals by substituting a trigonometric function for a variable.
Arctangent function: A mathematical function that returns the angle whose tangent is a given value.

Keywords

Integral: A mathematical expression that represents the area under a curve.
Rational function: A mathematical function that is the ratio of two polynomials.
Pythagorean identity: A mathematical identity that states $a^2 + b^2 = c^2$ .
Trigonometric substitution: A technique used to evaluate integrals by substituting a trigonometric function for a variable.
Arctangent function: A mathematical function that returns the angle whose tangent is a given value.

Introduction

In our previous article, we evaluated the integral ${ \int \frac{1}{\sqrt{169} - x^2} , dx }$. In this article, we will provide a Q&A guide to help you better understand the concepts and techniques used in evaluating this integral.

Q: What is the Pythagorean identity?

A: The Pythagorean identity is a mathematical identity that states $a^2 + b^2 = c^2$ . This identity is used to rewrite the denominator of the integral in a more manageable form.

Q: How is the Pythagorean identity used in evaluating the integral?

A: The Pythagorean identity is used to rewrite the denominator of the integral as $13 - x^2 = (3x)^2 + 4^2$ . This allows us to apply the Pythagorean identity and simplify the integral.

Q: What is trigonometric substitution?

A: Trigonometric substitution is a technique used to evaluate integrals by substituting a trigonometric function for a variable. In this case, we used the substitution $x = 2\tan(\theta)$ to rewrite the integral in terms of $\theta$ .

Q: How is trigonometric substitution used in evaluating the integral?

A: Trigonometric substitution is used to rewrite the integral in terms of $\theta$ . This allows us to simplify the integral and evaluate it using the arctangent function.

Q: What is the arctangent function?

A: The arctangent function is a mathematical function that returns the angle whose tangent is a given value. In this case, we used the arctangent function to evaluate the integral.

Q: How is the arctangent function used in evaluating the integral?

A: The arctangent function is used to evaluate the integral by returning the angle whose tangent is a given value. In this case, we used the arctangent function to evaluate the integral and obtain the final answer.

Q: What is the final answer to the integral?

A: The final answer to the integral is ${ \frac{1}{18} \arctan\left(\frac{6x}{4}\right) + C }$. This is the result of evaluating the integral using the techniques and strategies discussed in this article.

Q: Can you provide more examples of integrals that can be evaluated using trigonometric substitution?

A: Yes, there are many examples of integrals that can be evaluated using trigonometric substitution. Some examples include:

\int \frac{1}{x^2 + 1} , dx }$

\int \frac{1}{x^2 - 1} , dx }$

\int \frac{1}{x^2 + 4} , dx }$

These integrals can be evaluated using trigonometric substitution and the arctangent function.

Q: Can you provide more information on the Pythagorean identity?

A: Yes, the Pythagorean identity is a fundamental concept in mathematics that states $a^2 + b^2 = c^2$ . This identity is used to rewrite the denominator of the integral in a more manageable form.

Q: Can you provide more information on trigonometric substitution?

A: Yes, trigonometric substitution is a technique used to evaluate integrals by substituting a trigonometric function for a variable. This technique is used to simplify the integral and evaluate it using the arctangent function.

Q: Can you provide more information on the arctangent function?

A: Yes, the arctangent function is a mathematical function that returns the angle whose tangent is a given value. This function is used to evaluate the integral and obtain the final answer.