Evaluating The Definite Integral ∫₀^π Dθ/(13 + 5cosθ) A Comprehensive Guide

In the realm of calculus, definite integrals play a pivotal role in determining the area under a curve, among other applications. This article delves into the evaluation of a specific definite integral: ∫₀^π dθ/(13 + 5cosθ). This particular integral presents an interesting challenge due to the presence of the cosine function in the denominator. To tackle this, we will employ a clever substitution technique that transforms the trigonometric integral into an algebraic one, which is far easier to solve. We'll walk through each step meticulously, ensuring a clear understanding of the process. The journey involves trigonometric identities, algebraic manipulations, and a touch of complex analysis. By the end of this exploration, you'll not only know the solution to this integral but also gain insights into a powerful method applicable to a broader range of similar problems. So, let's embark on this mathematical adventure and unravel the intricacies of this integral. We will explore the underlying concepts and techniques that make this evaluation possible, emphasizing the beauty and power of calculus in solving real-world problems. The definite integral ∫₀^π dθ/(13 + 5cosθ) is a classic example that showcases the elegance of mathematical transformations and the interconnectedness of different branches of mathematics.

The problem at hand is to evaluate the definite integral:

∫₀^π dθ/(13 + 5cosθ)

This integral represents the area under the curve of the function f(θ) = 1/(13 + 5cosθ) from θ = 0 to θ = π. The presence of the cosine function in the denominator makes direct integration challenging. Therefore, we need to employ a suitable technique to simplify the integral. The most effective approach for this type of integral is to use the Weierstrass substitution, also known as the tangent half-angle substitution. This method allows us to transform the trigonometric integral into an algebraic integral, which is much easier to solve using standard integration techniques. Before diving into the substitution, it's crucial to understand why this method works and what the underlying principles are. The Weierstrass substitution leverages trigonometric identities to express sine and cosine functions in terms of a new variable, typically denoted as t, which is related to the tangent of half the angle. This transformation effectively eliminates the trigonometric functions from the denominator, replacing them with algebraic expressions. This substitution not only simplifies the integral but also provides a systematic way to handle a wide range of trigonometric integrals. The definite integral ∫₀^π dθ/(13 + 5cosθ) is a specific case that highlights the power and versatility of the Weierstrass substitution.

The Weierstrass substitution, also known as the tangent half-angle substitution, is a powerful technique for evaluating integrals involving trigonometric functions. It relies on the following substitutions:

t = tan(θ/2)

From this, we can derive the following relationships:

sin(θ) = (2t) / (1 + t²)

cos(θ) = (1 - t²) / (1 + t²)

dθ = (2 dt) / (1 + t²)

These substitutions are derived from trigonometric identities and the definition of the tangent function. The key idea behind the Weierstrass substitution is to express trigonometric functions in terms of a new variable, t, which is related to the tangent of half the angle. This transformation effectively converts trigonometric integrals into algebraic integrals, which are often easier to evaluate. The derivation of these substitutions involves clever manipulation of trigonometric identities and the use of the chain rule. For example, the expression for sin(θ) can be derived using the double-angle formula for sine and the definition of tangent. Similarly, the expression for cos(θ) can be derived using the double-angle formula for cosine and the definition of tangent. The differential dθ is also expressed in terms of dt using the chain rule and the derivative of the tangent function. The Weierstrass substitution is a versatile technique that can be applied to a wide range of trigonometric integrals. It is particularly useful when the integral involves rational functions of sine and cosine. However, it's important to note that the substitution may not always be the most efficient method for every trigonometric integral. In some cases, other techniques, such as trigonometric identities or integration by parts, may be more suitable. The definite integral ∫₀^π dθ/(13 + 5cosθ) is a prime example of an integral that can be effectively evaluated using the Weierstrass substitution.

Now, let's apply the Weierstrass substitution to our integral. We have:

∫₀^π dθ/(13 + 5cosθ)

Substituting cos(θ) = (1 - t²) / (1 + t²) and dθ = (2 dt) / (1 + t²), we get:

∫ (2 dt) / ((1 + t²) * (13 + 5((1 - t²) / (1 + t²))))

To simplify the integral further, we need to consider the limits of integration. When θ = 0, t = tan(0/2) = 0. When θ = π, t = tan(π/2), which approaches infinity. Therefore, our new limits of integration are from 0 to ∞. The substitution process involves replacing the trigonometric functions with their algebraic equivalents in terms of t. This transformation is crucial because it converts the trigonometric integral into an algebraic integral, which is often easier to solve. The substitution also affects the differential dθ, which needs to be expressed in terms of dt. The limits of integration also need to be adjusted to reflect the change of variable. In this case, the original limits of integration were from 0 to π, which correspond to the angles in the original integral. However, after the substitution, the variable of integration changes from θ to t, and the limits of integration need to be expressed in terms of t. This requires evaluating the tangent function at the original limits of integration. The definite integral ∫₀^π dθ/(13 + 5cosθ) is transformed into an algebraic integral with new limits of integration.

Substituting the limits and simplifying the expression, we get:

∫₀^∞ (2 dt) / ((1 + t²) * (13 + 5((1 - t²) / (1 + t²))))

= ∫₀^∞ (2 dt) / (13(1 + t²) + 5(1 - t²))

= ∫₀^∞ (2 dt) / (13 + 13t² + 5 - 5t²)

= ∫₀^∞ (2 dt) / (18 + 8t²)

= ∫₀^∞ dt / (9 + 4t²)

The simplification process involves algebraic manipulations to combine terms and reduce the complexity of the integrand. This is a crucial step in evaluating the integral because it transforms the expression into a form that can be readily integrated. The simplification typically involves finding common denominators, combining like terms, and factoring out constants. In this case, the expression is simplified by multiplying the denominator by (1 + t²) and then combining the terms. The resulting expression is a rational function of t, which can be integrated using standard techniques. The constant factor of 2 in the numerator is also simplified by dividing both the numerator and denominator by 2. The definite integral ∫₀^∞ dt / (9 + 4t²) is a simplified form of the original integral after applying the Weierstrass substitution and performing algebraic manipulations.

Now we have a standard integral form. We can rewrite the integral as:

∫₀^∞ dt / (9 + 4t²) = (1/4) ∫₀^∞ dt / ((9/4) + t²)

This integral is of the form ∫ dx / (a² + x²), which has the antiderivative (1/a)arctan(x/a). In our case, a² = 9/4, so a = 3/2. Therefore, the antiderivative of our integral is:

(1/4) * (2/3)arctan(t / (3/2)) = (1/6)arctan((2t) / 3)

Now, we evaluate the antiderivative at the limits of integration:

[(1/6)arctan((2t) / 3)] from 0 to ∞

= (1/6) [arctan(∞) - arctan(0)]

= (1/6) [(π/2) - 0]

= π/12

The evaluation of the simplified integral involves recognizing the standard integral form and applying the corresponding antiderivative formula. This is a key step in the integration process because it allows us to find the indefinite integral of the expression. The standard integral form ∫ dx / (a² + x²) is a common integral that arises in many calculus problems. The antiderivative of this integral is (1/a)arctan(x/a), which can be derived using trigonometric substitution or other integration techniques. In this case, the integral is rewritten to match the standard form, and the antiderivative is applied. The antiderivative is then evaluated at the limits of integration to find the definite integral. The definite integral ∫₀^∞ dt / (9 + 4t²) is evaluated using the antiderivative formula and the limits of integration.

Therefore, the value of the definite integral is:

∫₀^π dθ/(13 + 5cosθ) = π/12

This result represents the exact value of the definite integral. The process of evaluating this integral involved a series of steps, including the Weierstrass substitution, algebraic simplification, and the application of standard integration techniques. Each step played a crucial role in arriving at the final result. The Weierstrass substitution transformed the trigonometric integral into an algebraic integral, which was easier to handle. The algebraic simplification reduced the complexity of the integrand, making it amenable to integration. The application of standard integration techniques allowed us to find the antiderivative and evaluate the definite integral. The definite integral ∫₀^π dθ/(13 + 5cosθ) is a specific example that demonstrates the power and versatility of calculus in solving mathematical problems.

In conclusion, we have successfully evaluated the definite integral ∫₀^π dθ/(13 + 5cosθ) using the Weierstrass substitution. This method provides a systematic approach to solving integrals involving trigonometric functions, particularly those with rational expressions of sine and cosine. The final result, π/12, demonstrates the power of calculus in providing exact solutions to complex problems. The process of evaluating this integral involved several key steps, each of which contributed to the final result. The Weierstrass substitution was crucial in transforming the trigonometric integral into an algebraic integral. The algebraic simplification reduced the complexity of the integrand, making it easier to integrate. The application of standard integration techniques allowed us to find the antiderivative and evaluate the definite integral. The definite integral ∫₀^π dθ/(13 + 5cosθ) is a specific case that highlights the elegance and effectiveness of calculus in solving mathematical problems. The Weierstrass substitution is a versatile technique that can be applied to a wide range of trigonometric integrals. It is particularly useful when the integral involves rational functions of sine and cosine. However, it's important to note that the substitution may not always be the most efficient method for every trigonometric integral. In some cases, other techniques, such as trigonometric identities or integration by parts, may be more suitable. The choice of method depends on the specific characteristics of the integral and the desired level of accuracy. The definite integral ∫₀^π dθ/(13 + 5cosθ) is a valuable example that showcases the importance of choosing the right technique for the problem at hand. Understanding the underlying concepts and techniques allows us to tackle a wide range of mathematical challenges. This exploration underscores the beauty and practicality of calculus in solving real-world problems.The journey through this integral highlights the elegance and interconnectedness of mathematical concepts. From trigonometric identities to algebraic manipulations and the application of calculus techniques, each step builds upon the previous one, leading to the final solution. The value of π/12 is not just a numerical result; it represents the area under the curve of a specific function over a defined interval, showcasing the power of calculus in quantifying geometric concepts. This exercise not only reinforces our understanding of integration techniques but also provides a glimpse into the broader applications of calculus in various fields of science and engineering.