Question

In the following exercises, evaluate each integral in terms of an inverse trigonometric function. \(\int_{1}^{\sqrt{2}} \frac{d x}{|x| \sqrt{x^{2}-1}}\)

Short answer

The integral evaluates to \( \frac{\pi}{4} \).

Step-by-step solution

Determine the Form of the Integral

The integral has the form \[ \int \frac{dx}{|x| \sqrt{x^2 - a^2}} \]which corresponds to the derivative form of an inverse trigonometric function, specifically \( \sec^{-1}(x) \). The function \( \sec^{-1}(x) \) is defined for \( x \geq 1 \) or \( x \leq -1 \). In our integral, we need to evaluate the absolute value part by considering the range of \( x \).

Evaluate the Absolute Value Term

Since the limits of integration are from \( 1 \) to \( \sqrt{2} \), both values fall under the condition \( x \geq 1 \). Therefore, \( |x| = x \) for this range. This simplifies our integral to:\[\int_{1}^{\sqrt{2}} \frac{dx}{x \sqrt{x^2 - 1}}\]

Use Substitution for Inverse Trigonometric Function

Recognize that the form \( \int \frac{1}{x \sqrt{x^2 - 1}} \, dx \) is related to the derivative of the inverse secant function. Specifically, the derivative \( \frac{d}{dx}(\sec^{-1}(x)) = \frac{1}{x \sqrt{x^2 - 1}} \). Thus, the integral can be evaluated as:\[\int \frac{1}{x \sqrt{x^2 - 1}} \, dx = \sec^{-1}(x) + C\]

Apply Limits of Integration

Use the fundamental theorem of calculus to evaluate the definite integral:\[\left[ \sec^{-1}(x) \right]_{1}^{\sqrt{2}} = \sec^{-1}(\sqrt{2}) - \sec^{-1}(1)\]

Evaluate Inverse Secant Values

We need to compute \( \sec^{-1}(\sqrt{2}) \) and \( \sec^{-1}(1) \):- \( \sec^{-1}(1) = 0 \) because \( \sec(0) = 1 \).- \( \sec^{-1}(\sqrt{2}) = \frac{\pi}{4} \) because \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \).Thus, the result of the integral is:\[\frac{\pi}{4} - 0 = \frac{\pi}{4}\]

Conclusion

The evaluated integral is:\[\int_{1}^{\sqrt{2}} \frac{d x}{|x| \sqrt{x^{2}-1}} = \frac{\pi}{4}\]

Key concepts

Definite Integrals

Definite integrals represent the area under a curve between two specific points on the x-axis. Unlike indefinite integrals, which represent a family of functions, definite integrals give a specific numerical value.
This value corresponds to the net area between the function curve and the x-axis within certain limits. The exercise we discussed required evaluating a definite integral between 1 and \( \sqrt{2} \).
To solve a definite integral, we follow these steps:

• Find the antiderivative (or integral) of the function.
• Apply the limits of integration using the Fundamental Theorem of Calculus.
• Subtract the value of the antiderivative at the lower limit from its value at the upper limit.

Secant Inverse

The inverse secant function, denoted as \( \sec^{-1}(x) \), is the inverse of the secant function. The secant function itself is defined as \( \sec(\theta) = \frac{1}{\cos(\theta)} \).
Therefore, \( \sec^{-1}(x) \) is the angle whose secant is \( x \). The range for the inverse secant function is typically between \( 0 \) and \( \pi \), excluding \( \frac{\pi}{2} \), ensuring it is always real and defined for \( x \geq 1 \) or \( x \leq -1 \).
In this exercise, the integral form matched the derivative of \( \sec^{-1}(x) \), allowing us to directly evaluate it using the inverse secant.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving quadratic expressions. It often involves using an identity to transform the integrand into a recognizable form.
In our example, the substitution did not involve changing variables, but recognizing a familiar derivative form.
To deal with integrals like \( \int \frac{dx}{|x| \sqrt{x^2 - 1}} \), we matched the expression to the derivative of \( \sec^{-1}(x) \).

• This allowed us to treat integration as an "undoing" of differentiation.
• The technique simplifies evaluation by directly leading to known inverse trigonometric functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links derivatives and integrals, two major operations in calculus. It states that if a function is continuous over an interval, the integral of a function's derivative gives back the function.
In two parts, the theorem establishes:

• The indefinite integral (antiderivative) of a function is related to the derivative of the function.
• The definite integral of a function over an interval \([a, b]\) is the difference between the values of an antiderivative at \(b\) and \(a\).

Using this theorem, we evaluated \( \int \frac{dx}{x \sqrt{x^2 - 1}} \) by applying the antiderivative \( \sec^{-1}(x) \) and simplifying it with the given limits, \( 1 \) to \( \sqrt{2} \).