Problem 29 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x$$

Step-by-Step Solution

Answer

Answer: The result of evaluating the integral is $\frac{64}{3}\sin^3(\arcsin\frac{x}{4}) + C$.

Step 1: Perform substitution

Let us perform a substitution for the expression inside the square root: Let \(x = 4\sin(\theta)\). Therefore, \(x^2 = 16\sin^2(\theta)\) and \(dx = 4\cos(\theta)d\theta\). Now the integral becomes: $$\int \frac{4^2\sin^2(\theta)}{\sqrt{16-16\sin^2(\theta)}} \cdot 4\cos(\theta) d\theta$$

Step 2: Simplify the integral

We can now simplify the integral using the identity: $$\cos^2(\theta) = 1 - \sin^2(\theta)$$ So, the integral becomes: $$\int \frac{16\sin^2(\theta)}{\sqrt{16\cos^2(\theta)}} \cdot 4\cos(\theta) d \theta$$ Now, taking the constants outside the integral and simplifying, we have: $$64 \int \frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)}} \cdot \cos(\theta) d \theta$$ $$64 \int \sin^2(\theta) \cdot \cos(\theta) d\theta$$

Step 3: Integrate

Now, we can integrate the integral: $$64 \int \sin^2(\theta) \cdot \cos(\theta) d\theta$$ Let's perform another substitution: \(u = \sin(\theta)\) so \(du = \cos(\theta) d\theta\). We now have: $$64 \int u^2 du$$ Integrating, we get: $$64\cdot\frac{u^3}{3} + C$$ Now, substitute back for \(u\): $$\frac{64}{3}\sin^3(\theta) + C$$

Step 4: Return to original variable

Now, replace \(\theta\) with the original variable, \(x\), by inverting the substitution: $$\theta = \arcsin\frac{x}{4}$$ So, the final answer is: $$\frac{64}{3}\sin^3(\arcsin\frac{x}{4}) + C$$

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integral calculus to simplify integrals involving square roots. Often, when an integrand contains a square root of a quadratic expression, standard methods of integration are not effective. In such cases, a variable within the integrand is substituted with a trigonometric function that simplifies the square root and makes the integral more manageable.

To implement this technique, one commonly uses three basic trigonometric identities: for a square root of a sum \(a^2 + x^2\), one might use \(x = a\tan(\theta)\); for a square root of a difference \(a^2 - x^2\), \(x = a\sin(\theta)\) or \(x = a\cos(\theta)\) could be used; and for \(x^2 - a^2\), \(x = a\sec(\theta)\) is a typical choice. In the given exercise, the substitution \(x = 4\sin(\theta)\) was used to simplify the square root in the denominator, shifting the problem from an algebraic to a trigonometric form, which we could integrate more directly.

Square Root Simplification

The process of square root simplification is crucial in calculus, especially when dealing with integrals. Simplifying the square root expressions often turns a seemingly complicated problem into one with a straightforward solution. By employing trigonometric identities or substituting certain expressions, we can transform the square root into a form that is easier to integrate.

In the context of the given problem, after substituting \(x^2\) with \(16\sin^2(\theta)\), the expression under the square root became \(16 - 16\sin^2(\theta)\), which was simplified using the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\). This transformation eliminated the square root altogether, reducing the integral to a basic trigonometric function that could be integrated without difficulty.

Integration Techniques

To solve integrals in calculus, various integration techniques are employed, depending on the form and complexity of the function we're integrating. These techniques include substitution (both simple and trigonometric), integration by parts, partial fraction decomposition, and sometimes more specialized methods such as integrating trigonometric integrals or the use of hyperbolic functions.

In the case of our exercise, we've seen two main techniques at play. Initially, trigonometric substitution transformed the algebraic expression into a trigonometric integral. Subsequently, the method of simple substitution was used where \(u = \sin(\theta)\) and \(du = \cos(\theta)d\theta\) to evaluate the integral of \(\sin^2(\theta)\cos(\theta)d\theta\). After substituting \(u\), the problem was reduced to integrating \(u^2 du\), a standard power function integral.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the trigonometric functions and are used to solve for angles given the values of trigonometric ratios. They are particularly important when dealing with trigonometric substitutions in integration as they allow us to revert back to the original variable from the trigonometric variable.

Examples include \(\arcsin(x)\), \(\arccos(x)\), and \(\arctan(x)\). In our exercise, after integrating using the trigonometric substitution, we needed to return to the original variable \(x\). This was done using the inverse function \(\arcsin\), since we originally set \(x = 4\sin(\theta)\). By applying \(\arcsin\) to \(\frac{x}{4}\), we obtained \(\theta\) as \(\arcsin\frac{x}{4}\) and replaced it back into the solution to provide the result in terms of \(x\), completing the integration process.