Question

In the following exercises, compute each integral using appropriate substitutions. \(\int \frac{e^{t} \cos ^{-1}\left(e^{t}\right)}{\sqrt{1-e^{2 t}}} d t\)

Short answer

The integral evaluates to \( (\cos^{-1}(e^t))^2 + C \).

Step-by-step solution

Identify the substitution

The given integral is \( \int \frac{e^{t} \cos^{-1}(e^{t})}{\sqrt{1-e^{2t}}} \, dt \). Notice the \( \cos^{-1}(e^t) \) and the square root \( \sqrt{1-e^{2t}} \). Let us try a substitution: set \( u = e^t \). Then, \( du = e^t \, dt \), or equivalently, \( dt = \frac{du}{u} \). This substitution aims to simplify the integral by changing variables.

Rewrite in terms of \( u \)

Substitute \( u = e^t \) into the integral:\[ \int \frac{u \cos^{-1}(u)}{\sqrt{1-u^2}} \frac{du}{u} \]. This simplifies to \[ \int \frac{\cos^{-1}(u)}{\sqrt{1-u^2}} \, du \]. The integral now represents the standard integral of \( \cos^{-1}(u) \) differentiated with respect to \( u \).

Recognize the standard integral

The integral \( \int \frac{\cos^{-1}(u)}{\sqrt{1-u^2}} \, du \) is known to be the antiderivative form of the inverse trigonometric function. Specifically, it results in \( \cos^{-1}(u)^2 \) up to a constant of integration.

Insert the original variable

Returning to the original variable \( t \) using \( u = e^t \), the expression becomes \( (\cos^{-1}(e^t))^2 + C \), where \( C \) is the constant of integration.

Key concepts

Substitution Method

In integral calculus, the substitution method is a powerful tool used to simplify the integration process by changing the variable of integration. This technique often makes integrals more manageable by converting them to a simpler or standard form. In the original exercise, we applied the substitution method by letting \( u = e^t \). When we made this substitution, the differential \( dt \) was also rewritten in terms of \( u \), using \( du = e^t dt \) or \( dt = \frac{du}{u} \).

• Purpose: It simplifies complex integrals where the expression includes functions of other functions, such as inverse trigonometric functions or nested exponentials.
• How it works: By substituting a part of the integral with a new variable, we often transform it into a recognizable or standard form which can be easily integrated using known formulas.

When you are making a substitution, always ensure to express the entire integral in terms of the new variable, including converting the limits of integration if handling a definite integral.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse functions of the basic trigonometric functions. These functions are essential when working with angles and can often pop up in integration problems. For instance, the integral \( \int \frac{\cos^{-1}(u)}{\sqrt{1-u^2}} \, du \) from the solution is linked directly to the function \( \cos^{-1}(u) \).

• Common Inverse Trigonometric Functions: These include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), among others.
• Properties: These functions return an angle given a ratio of sides, and they have domain and range restrictions based on their corresponding trigonometric functions.

In the step-by-step solution, recognizing the integral as a standard form related to \( \cos^{-1}(u) \) allowed the solving process to move forward more effectively. This identification is crucial as it connects to known antiderivative forms.

Definite and Indefinite Integrals

In calculus, integrals can be classified as definite or indefinite. Understanding the distinction between these two types is vital.

• Definite Integrals: These are evaluated over a specific interval \([a, b]\). They yield a number representing the area under the curve within the given limits.
• Indefinite Integrals: These do not have specified limits of integration and result in a family of functions, plus a constant of integration \( C \).

In the exercise, we were working with an indefinite integral as there were no specified boundaries for the integral. This is why you see the inclusion of a constant \( C \) in the final solution. When converting back to the original variable \( t \), the indefinite nature allows for a general solution.