Evaluate The Integral:(viii) ∫ D X X 2 A 2 + X 2 \int \frac{d X}{x^2 \sqrt{a^2+x^2}} ∫ X 2 A 2 + X 2 ​ D X ​ ​

Introduction

In this article, we will delve into the evaluation of a specific integral, which involves a trigonometric function. The integral in question is $\int \frac{d x}{x^2 \sqrt{a^2+x^2}}$. This type of integral is commonly encountered in mathematics, particularly in the field of calculus. We will use various techniques and formulas to simplify and evaluate this integral.

Understanding the Integral

The given integral is $\int \frac{d x}{x^2 \sqrt{a^2+x^2}}$. To begin, let's break down the components of this integral. We have a denominator that consists of two terms: $x^2$ and $\sqrt{a^2+x^2}$. The numerator is simply $d x$. Our goal is to simplify this expression and find its indefinite integral.

Using Trigonometric Substitution

One of the most effective methods for evaluating this type of integral is through trigonometric substitution. We can substitute $x = a \tan \theta$, which will allow us to rewrite the integral in terms of trigonometric functions. This substitution is particularly useful when dealing with expressions involving $x^2$ and $\sqrt{a^2+x^2}$.

Let's perform the substitution:

$x = a \tan \theta$

We can then find the derivative of $x$ with respect to $\theta$:

$\frac{d x}{d \theta} = a \sec^2 \theta$

Now, we can rewrite the integral in terms of $\theta$:

$\int \frac{d x}{x^2 \sqrt{a^2+x^2}} = \int \frac{a \sec^2 \theta}{a^2 \tan^2 \theta \sqrt{a^2+a^2 \tan^2 \theta}} d \theta$

Simplifying the expression, we get:

$\int \frac{d \theta}{\tan^2 \theta \sqrt{1+\tan^2 \theta}}$

Using the Double-Angle Formula

We can simplify the expression further by using the double-angle formula for tangent:

$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$

Substituting this into the integral, we get:

$\int \frac{d \theta}{\frac{\sin^2 \theta}{\cos^2 \theta} \sqrt{1+\frac{\sin^2 \theta}{\cos^2 \theta}}}$

Simplifying the expression, we get:

$\int \frac{\cos^2 \theta}{\sin^2 \theta \sqrt{\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}}} d \theta$

Evaluating the Integral

We can now evaluate the integral by simplifying the expression:

$\int \frac{\cos^2 \theta}{\sin^2 \theta \sqrt{\frac{1}{\cos^2 \theta}}} d \theta$

Simplifying the expression, we get:

$\int \frac{\cos^2 \theta}{\sin^2 \theta \frac{1}{\cos \theta}} d \theta$

$\int \frac{\cos \theta}{\sin^2 \theta} d \theta$

Using the Power Rule

We can evaluate the integral by using the power rule:

$\int \frac{\cos \theta}{\sin^2 \theta} d \theta = \int \frac{1}{\sin^2 \theta} d (\sin \theta)$

Simplifying the expression, we get:

$\int \frac{1}{\sin^2 \theta} d (\sin \theta) = \int \csc^2 \theta d \theta$

Evaluating the Integral

We can now evaluate the integral by using the antiderivative of $\csc^2 \theta$:

$\int \csc^2 \theta d \theta = -\cot \theta + C$

Conclusion

In this article, we evaluated the integral $\int \frac{d x}{x^2 \sqrt{a^2+x^2}}$ using trigonometric substitution and the power rule. We simplified the expression and found its indefinite integral. This type of integral is commonly encountered in mathematics, particularly in the field of calculus. We hope that this article has provided a clear and concise explanation of how to evaluate this type of integral.

Final Answer

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Introduction

In our previous article, we evaluated the integral $\int \frac{d x}{x^2 \sqrt{a^2+x^2}}$ using trigonometric substitution and the power rule. In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have.

Q: What is the purpose of using trigonometric substitution in this integral?

A: Trigonometric substitution is a technique used to simplify expressions involving $x^2$ and $\sqrt{a^2+x^2}$. By substituting $x = a \tan \theta$, we can rewrite the integral in terms of trigonometric functions, which can be easier to evaluate.

Q: Why did we use the double-angle formula for tangent in the integral?

A: We used the double-angle formula for tangent to simplify the expression $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$. This allowed us to rewrite the integral in a more manageable form.

Q: How did we evaluate the integral $\int \csc^2 \theta d \theta$?

A: We evaluated the integral $\int \csc^2 \theta d \theta$ by using the antiderivative of $\csc^2 \theta$, which is $-\cot \theta + C$.

Q: What is the final answer to the integral $\int \frac{d x}{x^2 \sqrt{a^2+x^2}}$?

A: The final answer to the integral $\int \frac{d x}{x^2 \sqrt{a^2+x^2}}$ is $\boxed{-\frac{1}{a} \sec^{-1} \left( \frac{x}{a} \right) + C}$.

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

A: Yes, here are a few examples of integrals that can be evaluated using trigonometric substitution:

• $\int \frac{d x}{\sqrt{a^2+x^2}}$
• $\int \frac{d x}{x \sqrt{a^2+x^2}}$
• $\int \frac{d x}{x^2 \sqrt{a^2-x^2}}$

These are just a few examples, and there are many other integrals that can be evaluated using trigonometric substitution.

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

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

• Not substituting the correct trigonometric function
• Not simplifying the expression correctly
• Not using the correct antiderivative

It's also important to make sure that the substitution is valid and that the resulting expression is simplified correctly.

Conclusion

In this article, we provided a Q&A section to help clarify any doubts or questions that readers may have about evaluating the integral $\int \frac{d x}{x^2 \sqrt{a^2+x^2}}$ using trigonometric substitution. We hope that this article has been helpful in providing a clear and concise explanation of how to evaluate this type of integral.