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Chapter 7: Problem 22

Evaluate the following integrals. $$\int x \tan ^{-1} x^{2} d x$$

Short Answer

Expert verified
Question: Evaluate the following integral: $$\int{x \tan ^{-1} x^{2} d x}$$ Answer: The integral cannot be expressed using elementary functions. However, we can express it in terms of another integral: $$x \cdot \left(\frac{1}{2}\int \tan^{-1}(x^2) dx\right) - \int \left(\frac{1}{2}\int \tan^{-1}(x^2) dx\right) dx$$

Step by step solution

01

Choose u and dv

Let's choose the following: $$u = x \quad \Rightarrow \quad d u=d(x)$$ $$d v = \tan ^{-1} x^{2} d x$$
02

Determine du and v

Now, we will find 'du' and 'v': $$du = dx$$ For 'v', we need to integrate \(\tan ^{-1} x^{2}\): $$v=\int \tan ^{-1} x^{2} d x$$ However, this integration is also very complex and needs special techniques to solve. Therefore, we will use a substitution technique to solve this integral. Let \(u = x^2 \Rightarrow du = 2x dx\). In this case, $$v=\frac{1}{2}\int \tan ^{-1}u du$$ We still need to integrate \(\tan ^{-1}u\) which cannot be integrated using elementary functions. We need to stop here and use the substitution back into the integration by parts formula.
03

Apply Integration by Parts

Using the integration by parts formula, we have: $$\int x \tan ^{-1} x^{2} d x = uv - \int v du$$ Substituting the values of u, v, and du: $$= x \cdot \left(\frac{1}{2}\int \tan^{-1}(x^2) dx\right) - \int \left(\frac{1}{2}\int \tan^{-1}(x^2) dx\right) dx$$ Unfortunately, we can't proceed further, as the integral of \(\tan^{-1}(x^2)\) cannot be expressed using elementary functions. However, we have expressed the given integral in terms of the integral of a simpler function. This expression can be used for further analysis or numerical approximation methods, if necessary.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indefinite Integral
In the realm of calculus, an indefinite integral represents the collection of all antiderivatives of a function. It is the reverse process of differentiation, often considered as the act of finding a function whose derivative is given. When we write an indefinite integral, we use the symbol \( \int \), followed by the function and the differential (e.g., \( \int f(x)\,dx \)). The result of an indefinite integral includes an arbitrary constant (\( C \)), as the antiderivative is not unique.

The exercise given involved evaluating the indefinite integral \( \int x \tan^{-1} x^2\,dx \). The challenge arises due to the complexity of integrating inverse trigonometric functions multiplied by polynomials, which often requires advanced techniques such as integration by parts or substitution.
Inverse Trigonometric Functions
In calculus, inverse trigonometric functions are used to determine angles given the values of trigonometric ratios. They are the inverse operations of the usual trigonometric functions. Examples of inverse trigonometric functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), among others.

When integrating an inverse trigonometric function, such as \( \tan^{-1}(x) \), one must often employ integration techniques that can handle these non-algebraic functions. When these functions appear in more complex integrals, as in the current exercise, they may necessitate the use of integration by parts or substitution to resolve the integral into a more workable form.
Substitution Method
The substitution method, often referred to as u-substitution, is a technique for evaluating integrals. The primary idea is to substitute a part of the integral with a new variable, simplifying the integral into a more familiar form. The substitution is chosen such that the derivative of the new variable is present in the original integral.

In the exercise provided, substitution was used within the integration by parts method to simplify the integral of \( \tan^{-1}(x^2) \). However, it's worth noting that even after the substitution was made, the resulting integral proved too complex to express in terms of elementary functions. Even so, substitution remains a potent tool for transforming complicated expressions into more manageable ones.
Integral Calculus
Integral calculus is a significant branch of mathematics that deals with accumulation and area. While differential calculus focuses on rates of change, integral calculus is concerned with the total size or value, such as areas under curves and the accumulated quantity. It includes techniques and methodologies to tackle various types of integrals, both definite (with limits) and indefinite (without limits).

Our current problem is an application of integral calculus. We attempted to find the integral of a product involving a polynomial and an inverse trigonometric function using integration by parts鈥攁 fundamental technique in integral calculus. Despite the complexity of the integral, the language of calculus enables us to manipulate and analyze such expressions to advance towards a solution or understand their behavior.

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Most popular questions from this chapter

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