Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int \frac{t^{2} \cos ^{2}\left(t^{3}-2\right)}{\sin ^{2}\left(t^{3}-2\right)} d t\)

Step-by-Step Solution

Answer

\(\frac{1}{3} (-\cot(t^3 - 2) - (t^3 - 2)) + C\)

Step 1: Rewrite the Integrand

First, notice that we can rewrite the integral by utilizing the trigonometric identity \(\frac{\cos^2(x)}{\sin^2(x)} = \cot^2(x)\). Therefore, the integral becomes \(\int t^2 \cot^2(t^3 - 2) \ dt\).

Step 2: Substitution

Let \(u = t^3 - 2\). Then, differentiate \(u\) with respect to \(t\), giving \(\frac{du}{dt} = 3t^2\), which can be rearranged to \(dt = \frac{du}{3t^2}\).

Step 3: Simplify the Integral

Substitute \(u\) and \(dt\) into the integral: \(\int t^2 \cot^2(u) \cdot \frac{du}{3t^2}\). Simplify this to \(\frac{1}{3} \int \cot^2(u) \, du\).

Step 4: Integrate Using Trigonometric Identity

Recognize \(\cot^2(u) = \csc^2(u) - 1\). Thus, \(\frac{1}{3} \int \cot^2(u) \, du = \frac{1}{3} \int (\csc^2(u) - 1) \, du\).

Step 5: Perform the Integration

The integral becomes \(\frac{1}{3} \left( \int \csc^2(u) \, du - \int 1 \, du \right)\). The integral of \(\csc^2(u)\) is \(-\cot(u)\) and the integral of 1 is \(u\). Therefore, this evaluates to \(\frac{1}{3} (-\cot(u) - u) + C\), where \(C\) is the constant of integration.

Step 6: Substitute Back the Original Variable

Replace \(u\) with \(t^3 - 2\) to express the solution in terms of \(t\). Thus, the final answer is \(\frac{1}{3} (-\cot(t^3 - 2) - (t^3 - 2)) + C\).

Key Concepts

Substitution Method

The substitution method is a powerful tool in integral calculus that simplifies seemingly complex integrals. Imagine it like solving a puzzle where you substitute one part of the problem with a simpler piece. Here's how it works:

• Identify a part of the integrand (the expression inside the integral) that can be replaced with a single variable, often labeled as 'u'.
• For instance, in our example, we chose to let \( u = t^3 - 2 \). This variable substitution allows us to manage the integral more easily.
• After substituting, differentiate 'u', providing a new relationship with the original variable. We found \( \frac{du}{dt} = 3t^2 \).
• This enables us to express the differential \( dt \) in terms of \( du \) and other variables, finishing the preparation for integration.

Substitution can significantly streamline an integral, changing it into a form that is easier to integrate compared to the original complexity.

Trigonometric Identities

Trigonometric identities are like hidden keys that can unlock and simplify complex trigonometric expressions. In calculus, they're particularly helpful.
For example, the identity \( \frac{\cos^2(x)}{\sin^2(x)} = \cot^2(x) \) was used to transform the original integrand. By recognizing and applying such identities:

• You can rewrite an integral to a more digestible form.
• The identity helped us transform the fraction containing trigonometric functions into \( \cot^2(t^3 - 2) \), streamlining our problem.

Knowing these identities and when to apply them allows you to reduce complex integrals into manageable tasks. This understanding is vital in calculus, where integrating by hand often involves trigonometric simplifications.

Definite Integrals

While the exercise given is an indefinite integral, understanding definite integrals is equally important.
Definite integrals calculate the area under a curve within a specific interval, providing a number as the result. Indefinite integrals handle functions, thus the constant of integration \( C \) always appears in solutions.

• Definite integrals are evaluated using the Fundamental Theorem of Calculus, linking derivatives and integrals.
• They require limits of integration (typically at the top and bottom of the integral symbol), representing the span over which to find the area.

In calculus, understanding when and how to switch between definite and indefinite integrals is crucial for solving varied problems. Though not directly involved in our exercise, knowing both applications of integrals enriches your problem-solving toolkit in calculus.