Problem 29 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 29 through 43, evaluate the integrals. Not all require a trigonometric substitution. Choose the simplest method of integration. \(\int x^{3} \sqrt{4-x^{2}} d x\)

Step-by-Step Solution

Answer

The solution of the integral \(\int x^{3} \sqrt{4-x^{2}} d x\) is \(2*(\frac{x}{2})^4 - \frac{4}{3}*(\frac{x}{2})^6 + C\)

Step 1: Trigonometric Substitution

Substitute \(x = 2\sin{\theta}\) such that \(|\frac{dx}{d\theta}| = 2\cos{\theta}\). Remember, it was chosen because of the trigonometric identity \(\cos^2{\theta} = 1 - \sin^2{\theta}\) which will simplify our square root.

Step 2: Transferring the integral

The integral becomes \(\int (2\sin{\theta})^3 \sqrt{4 - (2\sin{\theta})^2} * 2\cos{\theta} d\theta = 8 \int \sin^3{\theta} \cos^2{\theta} d\theta\). We can further simplify it by using another identity, \(\cos^2{\theta} = 1 - \sin^2{\theta}\).

Step 3: New Integral and Substitution

The integral now transforms to \(8 \int \sin^3{\theta}(1 - \sin^2{\theta}) d\theta\). We can use regular substitution now. Let \(u = \sin{\theta}\) which gives \(du = \cos{\theta} d\theta\). Substituting back, it simplifies to \(8\int u^3(1 - u^2) du\).

Step 4: Solve the integral in \(u\)

Now we have a simple polynomial integral to calculate. We integrate to find \(8\int u^3 - 8\int u^5 du = 8*\frac{u^4}{4} - 8*\frac{u^6}{6} = 2u^4 - \frac{4u^6}{3}\).

Step 5: Substituting back in terms of \(x\)

Utilizing our initial substitution \(u = \sin{\theta}\), and remembering \(x = 2\sin{\theta}\), we have our final solution: \(2*(\frac{x}{2})^4 - \frac{4}{3}*(\frac{x}{2})^6 + C\), where \(C\) is the constant of integration.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a clever technique used in calculus to simplify integrals involving square roots. In this particular problem, we substitute \(x = 2\sin{\theta}\). This choice helps because of the identity \(\cos^2{\theta} = 1 - \sin^2{\theta}\), which neatly transforms the square root term \(\sqrt{4-x^2}\) into a trigonometric expression \(2\cos{\theta}\). This substitution simplifies the integral, making it easier to solve. Typically, trigonometric substitutions are used when dealing with expressions of the form \(\sqrt{a^2 - x^2}\), \(\sqrt{x^2 - a^2}\), or \(\sqrt{x^2 + a^2}\), where it’s beneficial to transform them into trigonometric identities.

Integrals

The integral of a function represents the area under the curve of that function. In calculus, evaluating an integral can help you solve many complex problems, from finding areas to solving differential equations.

In this exercise, the integral initially presented is \(\int x^3 \sqrt{4-x^2} \, dx\). By using particular methods such as substitution, we transform this into a more manageable form. Understanding the properties and rules of integration can provide insights into calculating these areas accurately.

Integration Methods

Integration has several techniques, each suited for different types of problems. The key is to identify the most efficient method. In this exercise, we used a combination of trigonometric substitution and simple substitution.

• Trigonometric Substitution: Good for handling integrals involving square roots, as it uses trigonometric identities to simplify expressions.
• Simple Substitution: Often called \(u\)-substitution, it simplifies integrals by changing variables, making it easier to integrate polynomial expressions.

Understanding which method to apply leads to smoother problem-solving and enhances comprehension of integral calculus.

Polynomial Integrals

Polynomial integrals are among the simplest types of integrals. They occur when the integrand is a polynomial, such as \(8 \int u^3 (1 - u^2) \, du\), which simplifies to \(8 \int u^3 - 8 \int u^5 \, du\).

For polynomials, the power rule is used: if \(n\) is not equal to \(-1\), \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration. The power rule makes calculating polynomial integrals straightforward, allowing for quick solutions once the integrand is expressed in a polynomial form.