Problem 51 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(37-54\). $$ \int_{\pi / 2}^{\pi} 2 \cot \frac{\theta}{3} d \theta $$

Step-by-Step Solution

Answer

The evaluated integral is \(6 \ln{\sqrt{3}}\).

Step 1: Identify the Integration Method

The function to be integrated is \(2 \cot \frac{\theta}{3}\). Recognize that \(\cot{x} = \frac{\cos{x}}{\sin{x}} \). We will employ substitution to simplify this integral.

Step 2: Perform Substitution

Let \( u = \frac{\theta}{3} \). This implies that \( du = \frac{1}{3} d\theta \), which in turn means \( d\theta = 3 du \). The limits of integration change from \( \theta = \frac{\pi}{2} \to u = \frac{\pi}{6} \) and \( \theta = \pi \to u = \frac{\pi}{3} \). The integral becomes \( 6 \int_{\pi/6}^{\pi/3} \cot u \ du \).

Step 3: Integrate Using Fundamental Formulas

Recall that the integral of \(\cot{u}\) is \(\ln|\sin{u}|\). Thus, the expression becomes \( 6 [\ln|\sin{u}|]_{\pi/6}^{\pi/3} \).

Step 4: Evaluate the Integral

Compute the antiderivative at the bounds: \[ 6 \left[ \ln|\sin{\frac{\pi}{3}}| - \ln|\sin{\frac{\pi}{6}}| \right] \]. Evaluator the sine values: \( \sin{\frac{\pi}{3}} = \frac{\sqrt{3}}{2} \) and \( \sin{\frac{\pi}{6}} = \frac{1}{2} \).

Step 5: Simplify the Result

Substitute the values into the expression: \[ 6 \left[ \ln\left(\frac{\sqrt{3}}{2}\right) - \ln\left(\frac{1}{2}\right) \right] = 6 \ln{\sqrt{3}} \]. Use the property \(\ln{a} - \ln{b} = \ln{\frac{a}{b}}\) to simplify.

Step 6: Final Answer

Thus, the final evaluation of the integral \( \int_{\pi / 2}^{\pi} 2 \cot \frac{\theta}{3} d \theta \) is \( 6 \ln{\sqrt{3}} \).

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique used in calculus to simplify complex integrals. Imagine you have an integral like \( \int f(g(x))g'(x) \, dx \). By substituting \( u = g(x) \), you convert the integral into \( \int f(u) \, du \). This often makes integration much easier to handle.

The given exercise requires integrating \( \int_{\pi / 2}^{\pi} 2 \cot \frac{\theta}{3} \, d\theta \). The substitution method helps by letting \( u = \frac{\theta}{3} \). This converts \( d\theta \) to \( 3 \, du \). Substitution is not only a strategic move to simplify the function but also to handle the variable changes smoothly.

Remember, when substituting, you must also change the limits of integration. In this exercise, as \( \theta \) changes from \( \pi/2 \) to \( \pi \), \( u \) changes from \( \pi/6 \) to \( \pi/3 \). This step ensures that the bounds of integration correspond to the new variable \( u \) instead of the original \( \theta \).

Trigonometric Integrals

Trigonometric integrals involve integrating functions of trigonometric functions, like sine, cosine, and tangent. These integrals often require specific techniques or formulas. One of the most useful formulas for this kind of problem is recognizing that

• \( \int \cot u \, du = \ln|\sin u| + C \)

In this exercise, after substituting and simplifying, we need to integrate \( \int \cot u \, du \).

The cotangent function, \( \cot u \), is the ratio \( \frac{\cos u}{\sin u} \), which often appears in calculus problems. Using the aforementioned formula helps simplify our task significantly. The integral of \( \cot u \) directly gives us \( \ln|\sin u| \), which is key to solving the integral in this problem.

Understanding how to integrate trigonometric functions and using substitution can reduce the difficulty of seemingly complex integrals.

Definite Integrals

Definite integrals provide a number, rather than a function, representing the area under the curve from one point to another. In this exercise, we are dealing with a definite integral with limits \( \pi/2 \) and \( \pi \) for \( \theta \), changing to \( \pi/6 \) and \( \pi/3 \) for \( u \) using substitution.

After computing the integral of \( \cot u \), we apply the Fundamental Theorem of Calculus. This involves evaluating the antiderivative \( \ln|\sin u| \) at the new limits \( u = \pi/3 \) and \( u = \pi/6 \), and finding the difference.

The expression simplifies to \( 6 [\ln(\sin(\pi/3)) - \ln(\sin(\pi/6))] \). By evaluating \( \sin(\pi/3) \) as \( \frac{\sqrt{3}}{2} \) and \( \sin(\pi/6) \) as \( \frac{1}{2} \), we deduce that the integral equals \( 6 \ln(\sqrt{3}) \).

This value represents the area under the curve of the original function from \( \theta = \pi/2 \) to \( \theta = \pi \), calculated through transformation and understanding trigonometric properties.