Evaluate The Integral: $\int \frac{\left(1-x^2\right)^{3 / 2}}{x^6} \, Dx$

Mar 1, 2025 by ADMIN 75 views

**Evaluating the Integral of a Trigonometric Function**

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Introduction

In this article, we will delve into the world of calculus and evaluate the integral of a trigonometric function. The given integral is $\int \frac{\left(1-x^2\right)^{3 / 2}}{x^6} \, dx$ . This integral appears to be complex, but with the right approach, we can simplify it and find its solution.

Understanding the Integral

The given integral involves a trigonometric function, specifically the expression $\left(1-x^2\right)^{3 / 2}$ . This expression can be rewritten in terms of trigonometric functions using the identity $\sin^2 \theta + \cos^2 \theta = 1$ . By substituting $x = \sin \theta$ , we can rewrite the integral as $\int \frac{\left(\cos^2 \theta\right)^{3 / 2}}{\sin^6 \theta} \, d\theta$ .

Applying Trigonometric Substitution

To simplify the integral, we can apply a trigonometric substitution. Let $x = \sin \theta$ , which implies $dx = \cos \theta \, d\theta$ . Substituting these expressions into the integral, we get:

\int \frac{\left(1-\sin^2 \theta\right)^{3 / 2}}{\sin^6 \theta} \, d\theta = \int \frac{\left(\cos^2 \theta\right)^{3 / 2}}{\sin^6 \theta} \, d\theta

Simplifying the Integral

Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$ , we can rewrite the integral as:

\int \frac{\left(1-\sin^2 \theta\right)^{3 / 2}}{\sin^6 \theta} \, d\theta = \int \frac{\left(1-\sin^2 \theta\right)^{3 / 2}}{\sin^6 \theta} \, d\theta

Evaluating the Integral

To evaluate the integral, we can use the substitution $u = \sin^2 \theta$ , which implies $du = 2 \sin \theta \cos \theta \, d\theta$ . Substituting these expressions into the integral, we get:

\int \frac{\left(1-\sin^2 \theta\right)^{3 / 2}}{\sin^6 \theta} \, d\theta = \int \frac{\left(1-u\right)^{3 / 2}}{u^3} \, du

Solving the Integral

To solve the integral, we can use the substitution $v = 1-u$ , which implies $dv = -du$ . Substituting these expressions into the integral, we get:

\int \frac{\left(1-u\right)^{3 / 2}}{u^3} \, du = -\int \frac{v^{3 / 2}}{v^3} \, dv

Evaluating the Integral

To evaluate the integral, we can use the power rule of integration, which states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ . Applying this rule to the integral, we get:

-\int \frac{v^{3 / 2}}{v^3} \, dv = -\int v^{-3/2} \, dv = -\frac{v^{-1/2}}{-1/2} + C

Simplifying the Result

Simplifying the result, we get:

-\frac{v^{-1/2}}{-1/2} + C = \frac{2}{\sqrt{v}} + C = \frac{2}{\sqrt{1-u}} + C

Substituting Back

Substituting back $u = \sin^2 \theta$ , we get:

\frac{2}{\sqrt{1-u}} + C = \frac{2}{\sqrt{1-\sin^2 \theta}} + C = \frac{2}{\sqrt{\cos^2 \theta}} + C = \frac{2}{\cos \theta} + C

Final Result

The final result is:

\int \frac{\left(1-x^2\right)^{3 / 2}}{x^6} \, dx = \frac{2}{\cos \theta} + C

Conclusion

In this article, we evaluated the integral of a trigonometric function using trigonometric substitution and the power rule of integration. The final result is a function of $\theta$ , which can be expressed in terms of $x$ using the substitution $x = \sin \theta$ . This result can be used to solve a variety of problems in calculus and physics.

Future Work

In future work, we can explore other methods for evaluating trigonometric integrals, such as using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ to rewrite the integral in terms of a single trigonometric function. We can also investigate the use of trigonometric substitution in other areas of mathematics, such as differential equations and group theory.

References

[1] "Calculus" by Michael Spivak
[2] "Trigonometry" by I.M. Gelfand
[3] "Differential Equations" by Morris Tenenbaum

Glossary

Trigonometric function: A function that involves the trigonometric functions sine, cosine, and tangent.
Trigonometric substitution: A method of simplifying an integral by substituting a trigonometric function for a variable.
Power rule of integration: A rule that states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ .

Appendix

Proof of the power rule of integration: The power rule of integration can be proven using the definition of the derivative and the fundamental theorem of calculus.

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Introduction

In our previous article, we evaluated the integral of a trigonometric function using trigonometric substitution and the power rule of integration. In this article, we will answer some common questions related to the evaluation of trigonometric integrals.

Q: What is trigonometric substitution?

A: Trigonometric substitution is a method of simplifying an integral by substituting a trigonometric function for a variable. This method is particularly useful when the integral involves trigonometric functions such as sine, cosine, and tangent.

Q: How do I choose the right trigonometric substitution?

A: To choose the right trigonometric substitution, you need to identify the trigonometric function that is involved in the integral. For example, if the integral involves the expression $\sin^2 \theta + \cos^2 \theta = 1$ , you can substitute $x = \sin \theta$ or $x = \cos \theta$ .

Q: What is the power rule of integration?

A: The power rule of integration is a rule that states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ . This rule is useful for integrating powers of $x$ .

Q: How do I apply the power rule of integration?

A: To apply the power rule of integration, you need to identify the power of $x$ in the integral. For example, if the integral is $\int x^2 \, dx$ , you can apply the power rule of integration by substituting $n = 2$ .

Q: What is the final result of the integral?

A: The final result of the integral is $\frac{2}{\cos \theta} + C$ . This result can be expressed in terms of $x$ using the substitution $x = \sin \theta$ .

Q: Can I use trigonometric substitution for other types of integrals?

A: Yes, you can use trigonometric substitution for other types of integrals that involve trigonometric functions. For example, you can use trigonometric substitution to evaluate integrals that involve the expressions $\sin^2 \theta + \cos^2 \theta = 1$ or $\tan^2 \theta + 1 = \sec^2 \theta$ .

Q: What are some common mistakes to avoid when using trigonometric substitution?

A: Some common mistakes to avoid when using trigonometric substitution include:

Not identifying the trigonometric function that is involved in the integral
Not choosing the right trigonometric substitution
Not applying the power rule of integration correctly
Not expressing the final result in terms of the original variable

Q: Can I use trigonometric substitution to solve differential equations?

A: Yes, you can use trigonometric substitution to solve differential equations that involve trigonometric functions. For example, you can use trigonometric substitution to solve differential equations that involve the expressions $\sin^2 \theta + \cos^2 \theta = 1$ or $\tan^2 \theta + 1 = \sec^2 \theta$ .