Problem 16 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{1}{\theta^{2}} \csc \frac{1}{\theta} d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln|\csc(\frac{1}{\theta}) + \cot(\frac{1}{\theta})| + C \)."

Step 1: Identify a Possible Substitution

We begin by examining the integrand \( \frac{1}{\theta^2} \csc \frac{1}{\theta} \) to identify a substitution that simplifies it. Notice that \( \csc \frac{1}{\theta} = \frac{1}{\sin \frac{1}{\theta}} \) hints that substituting \( u = \frac{1}{\theta} \) could simplify the expression. Thus, let \( u = \frac{1}{\theta} \).

Step 2: Differentiate and Modify Differential

Differentiate \( u = \frac{1}{\theta} \) with respect to \( \theta \) to find \( du \). Then, \( du = -\frac{1}{\theta^2} d\theta \). Solving for \( d\theta \), we find \( d\theta = -\theta^2 \, du \).

Step 3: Substitute and Simplify the Integral

Substitute \( u = \frac{1}{\theta} \) and \( d\theta = -\theta^2 \, du \) into the integral. The integral becomes:\( \int \frac{1}{\theta^2} \csc(u) (-\theta^2 \, du) = -\int \csc(u) \, du \).

Step 4: Evaluate the Standard Integral

Now the integral simplifies to \(-\int \csc(u) \, du \), which has a known antiderivative. The antiderivative of \( \csc(u) \) is \(-\ln|\csc(u) + \cot(u)| + C\). Therefore, the integral evaluates to:\[ \int \csc(u) \, du = \ln|\csc(u) + \cot(u)| + C \].

Step 5: Substitute Back in Terms of \( \theta \)

Finally, substitute \( u = \frac{1}{\theta} \) back into the antiderivative to return to the original variable. Thus, the solution is:\[ \ln|\csc(\frac{1}{\theta}) + \cot(\frac{1}{\theta})| + C \].

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify complex integrals into a more manageable form. The idea is to change variables using a substitution that resembles the standard forms for which antiderivatives are known.

In the context of the given problem, the substitution technique is employed by identifying a component of the integrand that can be replaced by a single variable. Specifically, we saw that substituting \( u = \frac{1}{\theta} \) transforms the integrand \( \frac{1}{\theta^2} \csc \frac{1}{\theta} \) into a simpler form. This substitution takes advantage of the reciprocal relationship of trigonometric functions.

• This method involves differentiating \( u \) with respect to \( \theta \), which helps in expressing \( d\theta \) in terms of \( du \).
• Simplifying the integral using this new substitution often allows the integral to match a standard form.

Thus, the complex problem is unraveled into an easily integrable function, with the final step in the process requiring a simple reverse substitution to return to the original variable.

Antiderivatives

Finding the antiderivative of a function is the core task of integration. It allows you to reverse the process of differentiation to discover the original function. In this particular exercise, once we simplified the integral with the substitution, it turned into \(-\int \csc(u) \, du\).

The key is recognizing that \(\csc(u)\) has a known antiderivative. The antiderivative for \( \csc(u) \) is \( -\ln|\csc(u) + \cot(u)| + C \), where \(C\) is the constant of integration.

• This means you have to remember or derive this antiderivative as a standard result.
• Using this known result, the integration becomes a straightforward task.

Thus, by using substitution, we reduced complex expressions to a standard form that allows us to directly apply known antiderivative results.

Trigonometric Functions

Trigonometric functions play a crucial role in calculus, especially in evaluating integrals. They include functions like sine, cosine, and cosecant, each having unique properties and identities that are often used in integration.

In the problem we tackled, the trigonometric function \(\csc(\theta)\), which is \(1/\sin(\theta)\), is central. By substituting \(u = \frac{1}{\theta}\), the trigonometric function undergoes simplification in the integral as \(\csc(u)\).

• Understanding how trigonometric functions transform under substitution is vital in solving integrals.
• Additionally, recognizing standard integrals involving these functions and their antiderivatives is key to quickly solving problems.

For trigonometric integrals, being familiar with their identities and properties, such as the reciprocal identities, helps simplify the integration process, allowing for more efficient problem-solving.