Question

For the following exercises, evaluate the integral using the specified method.\(\int \frac{1}{x^{2} \sqrt{x^{2}+16}} d x\) using trigonometric substitution

Short answer

The integral evaluates to \(-\frac{\sqrt{x^2+16}}{16x} + C\).

Step-by-step solution

Identify the substitution

We are dealing with the integral \(\int \frac{1}{x^{2} \sqrt{x^{2}+16}} \ dx\). This suggests using a trigonometric substitution to handle the \(\sqrt{x^2+16}\) term. Since this matches the form \(\sqrt{x^2 + a^2}\), we choose the substitution \(x = 4 \tan(\theta)\). This implies that \(\sqrt{x^2 + 16} = 4\sec(\theta)\). This substitution will simplify the integral.

Differentiate the substitution

From our substitution \(x = 4 \tan(\theta)\), differentiate with respect to \(\theta\) to find \(dx\). We have \(dx = 4 \sec^2(\theta) \ d\theta\). We will use this expression to replace \(dx\) in the integral.

Substitute and Simplify

Substitute \(x = 4 \tan(\theta)\), \(dx = 4 \sec^2(\theta) \ d\theta\), and \(\sqrt{x^2 + 16} = 4\sec(\theta)\) into the integral: \[\int \frac{1}{x^2 \sqrt{x^2 + 16}} \, dx = \int \frac{1}{(4\tan(\theta))^2 \cdot 4\sec(\theta)} \cdot 4\sec^2(\theta) \, d\theta \]. Simplify this to \[\int \frac{\sec(\theta)}{16\tan^2(\theta)} \, d\theta\].

Simplify the Integral Expression

Further simplify the integral to \[\frac{1}{16}\int \frac{\sec(\theta)}{\tan^2(\theta)} \, d\theta\]. Write \(\tan(\theta)\) as \(\frac{\sin(\theta)}{\cos(\theta)}\) and \(\sec(\theta)\) as \(\frac{1}{\cos(\theta)}\), transforming the integral to \[\frac{1}{16}\int \frac{\cos(\theta)}{\sin^2(\theta)} \, d\theta\] or \[\frac{1}{16}\int \csc^2(\theta) \cot(\theta) \, d\theta\].

Integrate Using Known Trigonometric Identities

Utilize the identity: \(\int \csc^2(\theta) \cot(\theta) \, d\theta = - \csc(\theta)\). Hence, \[\frac{1}{16} \left(- \csc(\theta)\right) + C\], where \(C\) is the constant of integration. This leads to the solution: \[-\frac{1}{16} \csc(\theta) + C\].

Back-Substitute for \(x\)

Recall \(x = 4 \tan(\theta)\), which implies \(\tan(\theta) = \frac{x}{4}\). Therefore, \(\sin(\theta) = \frac{x}{\sqrt{x^2 + 16}}\) and \(\csc(\theta) = \frac{\sqrt{x^2 + 16}}{x}\). Substitute back to get \[-\frac{1}{16} \cdot \frac{\sqrt{x^2 + 16}}{x} + C \].

Key concepts

Integration Techniques

When tackling integrals, especially ones that become cumbersome, integration techniques come to the rescue. One approach in simplifying complex integrals is **trigonometric substitution**, particularly when dealing with expressions like \( \sqrt{x^2 + a^2} \). Let's break this down:

• Identify the pattern in the expression. For \( \sqrt{x^2 + 16} \), it's of the form \( \sqrt{x^2 + a^2} \), prompting us to use trigonometric identities.
• Choose an appropriate trigonometric substitution. For this expression, recognizing \( a^2 = 16 \) translates to using \( x = 4 \tan(\theta) \), as it simplifies \( x^2 + 16 \) using trigonometric identities.
• Differentiate your substitution choice. Substitution leads us to find \( dx = 4 \sec^2(\theta) \, d\theta \), simplifying the differential aspect of integration.

Trigonometric substitution simplifies the integral by transforming it into a manageable trigonometric identity, allowing us to integrate more easily.

Calculus

Calculus, at its core, provides tools for analyzing change through differentiation and integration. Here, we focus on integration, which sums infinitely small quantities to determine total size or value. In this exercise:

• ***Integration*** evaluates the area under a curve. The selected method—trigonometric substitution—simplifies complex integrals resembling geometric shapes.
• Through trigonometric identities, it simplifies integration by converting variables into sine, cosine, tangent, etc., which have simpler integrals.
• Our original integral \( \int \frac{1}{x^2 \sqrt{x^2 + 16}} \, dx \), transformed using trigonometric substitutions and identities, results in a simpler form: \( \int \csc^2(\theta) \cot(\theta) \, d\theta \).

This illustrates how calculus not only introduces integration but dynamically reshapes the integral for easier processing through substitution and identities.

Definite Integrals

Definite integrals are a vital part of calculus, fundamentally different from indefinite integrals, as they compute the integral across a specific interval. Despite this exercise's focus on indefinite integration using trigonometric substitution, it sets the foundation for understanding definite integrals.Key ideas to consider:

• **Definite integral** provides the total accumulation of quantity over a specified range \([a, b]\), like area or displacement.
• By knowing the antiderivative, definite integrals evaluate to a numerical value using \( F(b) - F(a) \).
• Though this problem doesn't use definite limits, showing how to handle complex forms via substitution prepares you for evaluating definite integrals later.

Understanding these concepts isn't merely about finding antiderivatives; it's about grasping the full spectrum of calculus' power in measuring and transforming periodic, irregular, or bounded phenomena.