Problem 38 - Calculus: Early Transcendentals

Question

Evaluate the indefinite integral. \( \displaystyle \int \frac{dt}{\cos^2 t \sqrt{1 + \tan t}} \)

Step-by-Step Solution

Answer

The indefinite integral is \( 2 \sqrt{1 + \tan t} + C. \)

Step 1: Identify the Substitution

To simplify the integral, let's perform a substitution. Notice that the expression involves terms related to both \( \tan t \) and \( \cos^2 t \), which are related to \( \sin t \) through trigonometric identities. Since \( 1 + \tan t \) is under a square root, consider substituting \( u = \tan t \). This substitution simplifies the terms involved.'

Step 2: Differentiate to Find \( dt \)

Differentiate \( u = \tan t \) with respect to \( t \). This gives us \( du = \sec^2 t \, dt \). Since \( \cos^2 t = \frac{1}{\sec^2 t} \), we rearrange to find \( dt \): \( dt = \frac{du}{\sec^2 t} = \cos^2 t \, du \).

Step 3: Substitute into the Integral

Substitute \( u = \tan t \) and \( dt = \cos^2 t \, du \) into the original integral. The integral becomes: \[ \int \frac{\cos^2 t \, du}{\cos^2 t \sqrt{1 + u}} = \int \frac{du}{\sqrt{1 + u}}. \] Notice how the \( \cos^2 t \) terms cancel out.

Step 4: Integrate with respect to \( u \)

We now have \( \int \frac{du}{\sqrt{1 + u}} \). This can be directly integrated by recognizing it as a standard form. Use the formula \( \int \frac{du}{\sqrt{ax+b}} = \frac{2}{a} \sqrt{ax+b} + C \). Here, \( a = 1 \) and \( b = 1 \), so the integral is: \[ 2\sqrt{1 + u} + C. \]

Step 5: Substitute Back for \( t \)

Finally, substitute back \( u = \tan t \) into the expression. This gives us: \[ 2 \sqrt{1 + \tan t} + C. \] This represents the original integral in terms of \( t \).

Key Concepts

Indefinite Integrals

Indefinite integrals represent a family of functions whose derivative corresponds to the function inside the integral. If you see an integral symbol without specified limits, like in our problem, we call it an "indefinite integral."

For example, when computing the indefinite integral \( \int f(x) \, dx \), we are searching for a function \( F(x) \) such that \( F'(x) = f(x) \). We then add a constant \( C \) at the end to represent all possible vertical shifts of \( F(x) \), as indefinite integrals are generally expressed up to an arbitrary constant of integration.

Indefinite integration is fundamental in calculus as it helps us rewind a derivative to understand the original function, much like reversing a process.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving square roots or expressions that become simpler when transitioning into trigonometric functions. This is particularly useful when dealing with problems that include \/em{trigonometric identities}.

Consider the substitution \( u = \tan t \) from our original exercise. This decision stems from recognizing that expressions like \( \tan t \) and \( \cos^2 t \) relate fundamentally through trigonometric identities. By switching to trigonometric functions, we can leverage simpler rules and transcendent properties.

When making a trigonometric substitution, always:

• Identify the relations between existing terms and possible trigonometric identities.
• Perform the differentiation to express \( dt \) in the new variable.
• Substitute back after integrating completely to reach the original variable.

Trigonometric substitution can drastically reduce complexity, resulting in more straightforward integrals.

Integration by Substitution

Often called "u-substitution," this technique is designed to simplify the integral evaluation process. It works by identifying a section of the integral to substitute with a new variable, usually "\( u \)," in order to make the integral more manageable.

From our example, we substituted \( u = \tan t \), recognizing that this substitution neatly transformed the complicated integral into a simpler form. This requires rearranging to find \( dt \) as expressed through \( du \), facilitating easier integration.

When performing integration by substitution, it's crucial to:

• Select a substitution that simplifies the integral expression.
• Differentiate the substitution to find \( du \) in terms of \( dt \).
• Rewrite the integral entirely in terms of \( u \) before attempting integration.

Once the integration is completed in terms of \( u \), you'll substitute back to the original variable to put the solution in terms of the original function. Integration by substitution is a powerful way to transform complex integrals into simpler, solvable forms.