Problem 48 - Thomas Calculus

Question

Use a substitution to change the integral into one you can find in the table. Then evaluate the integral. \(\int \frac{\cos \theta d \theta}{\sqrt{5+\sin ^{2} \theta}}\)

Step-by-Step Solution

Answer

The integral evaluates to \( \ln |\sin \theta + \sqrt{5 + \sin^2 \theta}| + C \).

Step 1: Identify a Suitable Substitution

To simplify the integral \( \int \frac{\cos \theta d \theta}{\sqrt{5+\sin^{2} \theta}} \), we should look for a substitution that relates to the function inside the square root, \(5 + \sin^2 \theta\). Notice that \( \sin^2 \theta \) suggests a trigonometric identity. Let \( u = \sin \theta \), then \( du = \cos \theta \, d\theta \).

Step 2: Substitute in the Integral

Replace \( \sin \theta \) with \( u \) and \( \cos \theta \, d\theta \) with \( du \). The integral becomes: \( \int \frac{du}{\sqrt{5+u^2}} \).

Step 3: Recognize the Standard Form

The integral \( \int \frac{du}{\sqrt{5+u^2}} \) is in the standard form \( \int \frac{du}{\sqrt{a^2+u^2}} \) where \( a^2 = 5 \). This suggests the form of the integral is a standard inverse trigonometric function.

Step 4: Use Standard Integral Formula

Using the standard formula \( \int \frac{du}{\sqrt{a^2+u^2}} = \ln |u + \sqrt{u^2 + a^2}| + C \), we substitute \( a = \sqrt{5} \) to get \( \ln |u + \sqrt{5 + u^2}| + C \).

Step 5: Substitute Back to Original Variable

Since \( u = \sin \theta \), substitute back to get the expression in terms of \( \theta \). The integral evaluates to \( \ln |\sin \theta + \sqrt{5 + \sin^2 \theta}| + C \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a nifty technique in calculus for simplifying certain integrals. When an integral involves expressions like \( a^2 + x^2 \), \( a^2 - x^2 \), or \( x^2 - a^2 \), substituting with trigonometric identities can make things much easier. This is because trigonometric functions naturally handle squares due to the Pythagorean identity.

In our example, the integral \( \int \frac{\cos \theta d \theta}{\sqrt{5+\sin ^{2} \theta}} \) was simplified by recognizing it involved \( \sin^2 \theta \). By letting \( u = \sin \theta \), we made use of the identity that transforms the variable into a form without the sine. The substitution directly translated into \( du = \cos \theta \, d\theta \), which perfectly matched the differential part of the integral. This eased the transformation into an expression in terms of \( u \) and allowed the integral to adopt a standard form that is more straightforward to solve.

The purpose of trigonometric substitution here is to leverage the inherent properties of sine, cosine, and other trigonometric functions to simplify the integration process, especially when dealing with radical expressions. Once the substitution and simplification are complete, it's crucial to revert to the original variable for the final answer.

Definite and Indefinite Integrals

Integrals are foundational tools in calculus used to find areas, volumes, and more. What's special about them is that they can either be definite or indefinite, depending on whether you need a specific numerical value or a general formula.

• An indefinite integral provides a family of functions and includes a constant of integration, \( C \). It looks like \( \int f(x) \, dx = F(x) + C \). This represents a collection of antiderivatives.
• A definite integral, on the other hand, gives the exact area under a curve between specific bounds \( a \) and \( b \), and it results in a numerical value: \( \int_a^b f(x) \, dx = F(b) - F(a) \).

In our solved problem, we dealt with an indefinite integral as there's no specified upper and lower limit. The presence of \( + C \) indicates that we are finding the antiderivative form. The solution uses an inverse trigonometric form with a natural logarithm to express this indefinite antiderivative. This prepares us for potential definite integral problems, by always considering magnitude and bounds during real-world applications.

Inverse Trigonometric Functions

Inverse trigonometric functions play a key role in integration, especially when dealing with radical expressions. These functions such as \( \arcsin \), \( \arccos \), and \( \arctan \), allow us to express angles in terms of ratios and are useful for solving integrals involving square roots.

In the context of our problem, after making the substitution \( u = \sin \theta \), the integral transformed into \( \int \frac{du}{\sqrt{5+u^2}} \). Recognizing this form is vital, as it resembles the inverse trigonometric integral \( \int \frac{du}{\sqrt{a^2+u^2}} \), linked to a standard result involving natural logarithm.

The standard formula used here is typically given as:

• \( \int \frac{du}{\sqrt{a^2+u^2}} = \ln |u+\sqrt{u^2+a^2}| + C \)

This equation relies on the inverse hyperbolic function form, and in practice, it's used to manage complex integrals involving square roots. Once the integration is complete, substituting back the original trigonometric variables ensures the final result reflects the problem's original context, providing both the solution to the problem and understanding of its relation to trigonometric and inverse trigonometric functions.