Question

Find the antiderivative. \(\int \frac{4 x^{2}}{\sqrt{1-x^{6}}} d x\)

Short answer

\(\frac{4}{3} \arcsin(x^3) + C\)

Step-by-step solution

Identify an appropriate substitution

To solve the integral \( \int \frac{4x^2}{\sqrt{1-x^6}} \, dx \), recognize that the expression inside the square root suggests a trigonometric substitution. Let \( u = x^3 \), then \( du = 3x^2 \, dx \) or \( x^2 \, dx = \frac{1}{3} \, du \).

Rewrite the integral in terms of \( u \)

Substitute \( u = x^3 \) and \( x^2 \, dx = \frac{1}{3} \, du \). The integral becomes: \[ \int \frac{4}{\sqrt{1-u^2}} \cdot \frac{1}{3} \, du = \frac{4}{3} \int \frac{1}{\sqrt{1-u^2}} \, du. \]

Recognize the new integrand

The integral \( \int \frac{1}{\sqrt{1-u^2}} \, du \) is a standard form that equals \( \arcsin(u) \).

Integrate using the standard result

Integrate to obtain \( \frac{4}{3} \arcsin(u) + C \), where \( C \) is the constant of integration.

Substitute back \( x \)

Substitute back \( u = x^3 \) to get the antiderivative in terms of \( x \): \[ \frac{4}{3} \arcsin(x^3) + C. \]

Key concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integral calculus to simplify the integration process. It is particularly useful when dealing with square roots of quadratic expressions. This method leverages the identities of trigonometric functions to transform complicated algebraic expressions into more manageable forms typically involving sine, cosine, or tangent.

With this technique, substitutions are often made to convert expressions into forms that resemble trigonometric identities, which have well-known antiderivatives.

• For example, one common substitution is using \( x = \sin(\theta) \) for integrals involving expressions like \( \sqrt{1-x^2} \).
• Similarly, \( x = a \cos(\theta) \) might be used when the expression is \( \sqrt{a^2 - x^2} \).

In the given problem, the expression \( \sqrt{1-x^6} \) suggests a potential for using such a substitution. By letting \( u = x^3 \), the expression inside the square root becomes \( \sqrt{1-u^2} \), which connects directly to the trigonometric identity for sine: \( \sin^2(\theta) + \cos^2(\theta) = 1 \).

Calculus

Calculus is a branch of mathematics that studies how things change. It is divided into two main branches: differential calculus and integral calculus. Differential calculus focuses on the concept of a derivative, helping to understand rates of change. On the other hand, integral calculus is concerned with accumulation of quantities and finding the total size, length, area, or volume.

In our exercise, we deal with integral calculus, specifically finding an antiderivative. An antiderivative is a function whose derivative is the given function. The process of finding an antiderivative is called integration.

• This problem illustrates the typical process of using substitution to find antiderivatives.
• It requires identifying a substitution that simplifies the integral.

Understanding calculus and these methods provides powerful tools for solving complex mathematical problems that describe real-world phenomena such as motion, growth, and decay.

Integral Calculus

Integral calculus is the study of integrals and their properties. It involves finding the integral of a function, which is the antiderivative or the reverse process of differentiation. Integral calculus can determine areas under curves, lengths of curves, volumes of solids, and many other quantities.

In the context of the exercise, we are performing the integration of \( \int \frac{4x^2}{\sqrt{1-x^6}} dx \). This involves transforming the integral using substitution to recognize a standard form that can be integrated easily.

• By substituting \( u = x^3 \) and converting the integral into terms of \( u \), the problem simplifies significantly.
• The resulting integral \( \int \frac{1}{\sqrt{1-u^2}} du \) is identified as the derivative of \( \arcsin(u) \), a standard integral result.

This standard integral form is often encountered in calculus, making recognition and application of these forms critical to solving many calculus problems efficiently. Understanding these substitutions and transformations is key in solving complex integral problems.