Question

Evaluate the integral \(\int \frac{d x}{x \sqrt{x^{2}-1}}\) using the substitution \(x=\sec \theta\). Next, evaluate the same integral using the substitution \(x=\csc \theta .\) Show that the results are equivalent.

Short answer

The integrals are equivalent; both evaluate to an arctrig function up to a constant.

Step-by-step solution

Substitution and Differential for \(x = \sec \theta\)

We start with the substitution \(x = \sec \theta\). The differential is \(dx = \sec \theta \tan \theta \, d\theta\). Substitute these into the integral.

Simplifying the Integral

Replace \(x\) and \(dx\) in the integral and simplify:\[\int \frac{dx}{x\sqrt{x^2 - 1}} = \int \frac{\sec \theta \tan \theta \, d\theta}{\sec \theta \sqrt{(\sec^2 \theta) - 1}}\]The expression \(\sqrt{\sec^2 \theta - 1}\) simplifies to \(\sqrt{\tan^2 \theta} = |\tan \theta|\), but since \(\theta\) is in the range where \(\tan \theta\) is positive, it simplifies to \(\tan \theta\). This simplifies the integral to:\[\int \frac{\sec \theta \tan \theta \, d\theta}{\sec \theta \tan \theta} = \int d\theta\]

Evaluate the Integral

The integral \(\int d\theta\) is simply \(\theta + C_1\), where \(C_1\) is the integration constant.

Back-Substitute \(\theta\)

Back-substitute \(\theta\) using \(x = \sec \theta\), thus \(\theta = \sec^{-1}(x)\). Therefore, the integral evaluates to:\[\theta = \sec^{-1}(x) + C_1\]This implies \(\int \frac{dx}{x \sqrt{x^2 - 1}} = \sec^{-1}(x) + C_1\).

Substitution and Differential for \(x = \csc \theta\)

Next, use the substitution \(x = \csc \theta\), and then \(dx = -\csc \theta \cot \theta \, d\theta\). Substitute these into the integral.

Simplifying the Integral for Second Substitution

Replace \(x\) and \(dx\) in the integral:\[\int \frac{dx}{x\sqrt{x^2-1}} = \int \frac{-\csc \theta \cot \theta \, d\theta}{\csc \theta \sqrt{\csc^2 \theta - 1}}\]The expression \(\sqrt{\csc^2 \theta - 1}\) simplifies to \(\sqrt{\cot^2 \theta} = |\cot \theta|\). Since \(\theta\) is in the range where \(\cot \theta\) is positive, it simplifies to \(\cot \theta\). This gives:\[-\int \frac{\csc \theta \cot \theta \, d\theta}{\csc \theta \cot \theta} = -\int d\theta\]

Evaluate the Integral for Second Substitution

The integral \(-\int d\theta\) is \(-\theta + C_2\), where \(C_2\) is another integration constant.

Back-Substitute \(\theta\) for Second Substitution

Back-substitute \(\theta\) using \(x = \csc \theta\), thus \(\theta = \csc^{-1}(x)\). Therefore, the integral evaluates to:\[-\theta = -\csc^{-1}(x) + C_2\]This implies \(\int \frac{dx}{x \sqrt{x^2 - 1}} = -\csc^{-1}(x) + C_2\).

Equivalency of the Results

The expressions \(\sec^{-1}(x) + C_1\) and \(-\csc^{-1}(x) + C_2\) are equivalent up to a constant because:\[\sec^{-1}(x) = \csc^{-1}(x) - \frac{\pi}{2}\]So, the integration constants can be adjusted to show equivalence between the two solutions.

Key concepts

Integral Calculus

Integral calculus is the branch of calculus that deals with finding the integrals of functions. Integrals can be understood as the opposite of derivatives, and they play a crucial role in calculating areas under curves, volumes, and solving differential equations. Important aspects of integral calculus include indefinite integrals, definite integrals, and techniques of integration.

In this exercise, we are focused on evaluating an indefinite integral, specifically using a method known as trigonometric substitution. This technique is particularly handy when dealing with integrals involving square roots and quadratic expressions. By making an appropriate substitution, you simplify the integral into a more manageable form, often resulting in a basic trigonometric function that is easier to integrate.

Here are some general steps when approaching an integral using trigonometric substitution:

• Identify parts of the integrand that match trigonometric identities.
• Choose a substitution that simplifies the square root (e.g., using substitutions like \(x = \sec \theta\) for expressions containing \(x^2 - 1\)).
• Replace all occurrences of \(x\) and \(dx\) in the integral.
• Simplify the resulting trigonometric integral.
• Evaluate the integral.
• Replace the trigonometric variable back with the original variable using the reverse substitution.

These steps facilitate the simplification needed to perform integration, which is the heart of integral calculus.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for every value of the occurring variables. These identities are fundamental tools in both calculus and algebra, often used to simplify complex expressions and solve integrals, such as in trigonometric substitution.

In solving the integral using substitutions \(x = \sec \theta\) and \(x = \csc \theta\), the following trigonometric identities are particularly important:

• \(\sec^2 \theta - 1 = \tan^2 \theta\)
• \(\csc^2 \theta - 1 = \cot^2 \theta\)

By replacing expressions like \(\sqrt{x^2 - 1}\) with their corresponding trigonometric identity, you can continue simplifying the integral to an easier form like \(\theta\).

These identities not only help in reducing complex terms but also change the problem into one involving simpler trigonometric integrals, which are straightforward to evaluate.

Understanding and using these identities effectively is essential when working with trigonometric substitution, allowing for successful integration of otherwise difficult expressions.

Inverse Trigonometric Functions

Inverse trigonometric functions undo the trigonometric functions such as sine, cosine, and tangent. They return an angle from a given trigonometric ratio. In calculations involving integration, inverse trigonometric functions become crucial during the back substitution step when expressing the final integral result in terms of the original variable.

For the given exercise, after finding the integral in terms of \(\theta\), you reverse back to the original variable \(x\) using:

• \(\theta = \sec^{-1}(x)\) for the substitution \(x = \sec \theta\)
• \(\theta = \csc^{-1}(x)\) for the substitution \(x = \csc \theta\)

This step is crucial because it provides the solution in terms that are meaningful to the problem's context.

Moreover, understanding the relationships between different inverse trigonometric functions, such as \(\sec^{-1}(x) = \csc^{-1}(x) - \frac{\pi}{2}\), is important when checking the equivalency of solutions obtained using different methods of substitution. These relationships ensure that even though the solutions might look different, they actually represent the same quantity up to a constant.