Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \frac{\csc ^{2} 2 t}{\sqrt{1+\cot 2 t}} d t $$

Step-by-Step Solution

Answer

\(-\sqrt{1+\cot 2t} + C\)

Step 1: Recognize a substitution

Start by recognizing that the integrand involves functions of \(2t\). We can use a trigonometric identity or substitution to simplify this problem. Notice that \(\cot 2t\) is related to \(\csc 2t\), so let's try a substitution. Let \( u = \cot 2t \). Then, the derivative is \( du = -2 \csc^2 2t \, dt \). To solve for \( dt \), we get \( dt = -\frac{1}{2 \csc^2 2t} \, du \).

Step 2: Substitute and Simplify

Substitute \( u = \cot 2t \) and \( dt = -\frac{1}{2\csc^2 2t} \, du \) into the integral. The integral becomes:\[\int \frac{\csc^2 2t}{\sqrt{1+\cot 2t}} \, dt = \int \frac{\csc^2 2t}{\sqrt{1+u}} \left( -\frac{1}{2\csc^2 2t} \right) \, du \]This simplifies to:\[ -\frac{1}{2} \int \frac{1}{\sqrt{1+u}} \, du \]

Step 3: Solve the Integral

The integral \( \int \frac{1}{\sqrt{1+u}} \, du \) is known to be the inverse hyperbolic sine function or can be expressed as:\[ \int \frac{1}{\sqrt{1+u}} \, du = 2\sqrt{1+u} + C \]Apply this result to obtain:\[ -\frac{1}{2} \int \frac{1}{\sqrt{1+u}} \, du = -\sqrt{1+u} + C \]

Step 4: Back-Substitute

Re-substitute \( u = \cot 2t \) back into the expression:\[-\sqrt{1+\cot 2t} + C\]

Step 5: Simplified Final Expression

Re-write the final expression for clarity:\[-\sqrt{1+\cot 2t} + C \] is the simplified result of the given integral. This is our final answer.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a strategic method in calculus used to simplify integrals involving trigonometric expressions. In this exercise, we performed trigonometric substitution to tackle the integration of \( \int \frac{\csc ^{2} 2 t}{\sqrt{1+\cot 2 t}} \, d t \). The key to using trigonometric substitution effectively is identifying the trigonometric forms or identities that allow for a simpler integral.

Here, we recognized that \( \cot 2t \) is a key part of the expression with its derivative \( \csc^2 2t \) also present in the integral. We used the substitution \( u = \cot 2t \). This choice is logical because the derivative \( \frac{d}{dt} [\cot 2t] = -2 \csc^2 2t \) is conveniently related to the integrand. By substituting \( u \) and then modifying \( dt \) appropriately, we transformed a complex trigonometric integral into a simpler algebraic form.

Through this process, we simplified our work and resolved the intricacy of the original integral by leveraging this helpful technique. Always keep in mind that substitutions should simplify the integral into a format you can easily manage or recognize.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, are a core concept of integral calculus. Unlike definite integrals, they do not evaluate to a particular number but rather to a family of functions plus a constant \( C \). The task is to reverse the process of differentiation, identifying what function was originally differentiated to produce a given integrand.

In our exercise, we aimed to find the indefinite integral of a trigonometric expression. After performing trigonometric substitution, the complex integrand \( \int \frac{1}{\sqrt{1+u}} \, du \) simplified significantly, making it manageable. By recognizing the resulting integral \( \int \frac{1}{\sqrt{1+u}} \, du \) aligns with common forms, we quickly arrived at the antiderivative.

The integration yielded the result expressed as \( 2\sqrt{1+u} + C \), which upon further simplification became \( -\sqrt{1+u} + C \). Always remember that indefinite integrals include a constant of integration \( C \), representing all possible antiderivatives. When solving indefinite integrals, testing the derivative of your result can verify accuracy.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables at which the functions are defined. They are incredibly useful for simplifying trigonometric expressions and are pivotal in solving integrals and derivatives linked with trigonometric functions.

In our integration exercise, identifying and utilizing the correct identities simplified the operation dramatically. We specifically recognized relationships between trigonometric functions, like the link between \( \cot \theta \) and \( \csc \theta \). These relationships guided our choice of substitution.

Understanding and applying identities such as \( \cot( heta) = \frac{\cos(\theta)}{\sin(\theta)} \) and \( \csc(\theta) = \frac{1}{\sin(\theta)} \) can transform complex trigonometric expressions into simpler forms. Mastery of these identities allows for versatile manipulation of problems in integral calculus, often opening the path to solve what initially seems unsolvable. They are essential tools in any calculus student's toolkit.