Problem 10 - Advanced Engineering Mathematics

Question

In Problems 1-10, evaluate the given trigonometric integral. $$ \int_{0}^{2 \pi} \frac{1}{\cos \theta+2 \sin \theta+3} d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{2\pi}{3} \).

Step 1: Simplify the Denominator

Rewrite the denominator as \( rac{1}{3} \left( 1 + \frac{2}{3} \sin \theta + \frac{1}{3} \cos \theta \right) \). This form allows us to use the Weierstrass substitution. Factor the constant 3 out of the entire integral, simplifying the integral to \( \frac{1}{3} \int_{0}^{2 \pi} \frac{3}{\cos \theta + 2 \sin \theta + 3} \, d\theta \).

Step 2: Apply the Weierstrass Substitution

Use the substitution \( t = \tan \frac{\theta}{2} \) (Weierstrass substitution), where \( \sin \theta = \frac{2t}{1+t^2} \) and \( \cos \theta = \frac{1-t^2}{1+t^2} \). This makes the integral \( \int \frac{1}{\frac{1-t^2}{1+t^2} + \frac{4t}{1+t^2} + 3} \, d\theta \).

Step 3: Simplify further using substitution

Substitute the expressions for \( \sin \theta \) and \( \cos \theta \) into the integral. Simplify to get the integral in terms of \( t \) and \( dt \). After simplification, the integral becomes \( \int \frac{1+t^2}{1 - t^2 + 4t + 3} \, dt \).

Step 4: Perform Partial Fraction Decomposition

The denominator now simplifies to \( 1 - t^2 + 4t + 3 = (t+3)^2 - t^2 \). To decompose, let \( A, B \) be constants such that \( \frac{1}{(t+3)^2 - t^2} = \frac{A}{(t+3)} + \frac{B}{t^2 + 1} \). Solve for \( A \) and \( B \).

Step 5: Integrate Using Standard Integral Forms

After solving for \( A \) and \( B \), integrate the terms separately: \( \int \frac{A}{t+3} \dt \) and \( \int \frac{B}{t^2+1} \, dt \). The latter is a standard integral yielding \( B \arctan(t) \).

Step 6: Back-substitute and Evaluate

Back-substitute \( t = \tan \frac{\theta}{2} \) to express your answer in terms of \( \theta \). Evaluate the definite integral from 0 to \( 2\pi \). Simplify to find a concise expression.

Key Concepts

Weierstrass substitution

Weierstrass substitution is a powerful technique used in integral calculus to simplify certain types of trigonometric integrals. This method involves substituting trigonometric functions with expressions involving the tangent of half an angle. The substitution is given by:

• Replace \( \theta \) with \( t = \tan \frac{\theta}{2} \). This substitution transforms periodic trigonometric functions into rational functions, which are typically easier to integrate.
• It converts functions like \( \sin \theta \) and \( \cos \theta \) into rational expressions in terms of \( t \): \( \sin \theta = \frac{2t}{1+t^2} \) and \( \cos \theta = \frac{1-t^2}{1+t^2} \).

This approach not only simplifies the integration process but also allows us to exploit the simplicity of polynomials and rational functions. This method is particularly useful when dealing with expressions that might not factor easily using standard trigonometric identities.

Partial fraction decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. This method is particularly useful when integrating rational functions, as it converts the integrand into a sum of more manageable terms. In the context of integral calculus, once a trigonometric function is expressed in terms of a rational function (like after applying Weierstrass substitution), we can:

• Write the rational function as a sum of simpler fractions. This involves expressing the denominator, typically as a product of linear and sometimes irreducible quadratic factors.
• Assign coefficients (like \(A, B, C\), etc.) to these simpler terms, and solve for these coefficients to match the original numerator of the rational expression.
• Once we know the values of these coefficients, the integral of the original complex expression is transformed into the sum of integrals of these simpler fractions.

This transformation is useful because the simpler fractions often represent elementary integrals, such as \( \ln |x| \) or \( \arctan x \), which can be directly integrated using standard methods.

Integral calculus

Integral calculus is one of the two main branches of calculus, with the other being differential calculus. It focuses on the accumulation of quantities, such as areas under curves and total change, providing solutions for problems involving continuous rates of change.At its core, integral calculus is concerned with:

• Finding the antiderivative, or the integral, of a function. This involves reversing the process of differentiation.
• Evaluating the definite integral, which calculates the area under the curve of a graph of a function, represented as \( \int_{a}^{b} f(x) \, dx \).

Integral calculus is especially crucial when solving problems in physics and engineering, where it helps in determining quantities like mass, volume, and displacement, based on continuous distributions. Furthermore, trigonometric integrals, like the one we're solving, often appear in practical applications requiring precise, comprehensive solutions.

Trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where both sides are defined. These identities are essential tools in solving trigonometric integrals, simplifying expressions, and proving other mathematical statements.Major types of trigonometric identities we often use in integral calculus include:

• Basic identities like \( \sin^2 \theta + \cos^2 \theta = 1 \), which are foundational in simplifying expressions.
• Double angle and half-angle identities, which help transform integrals into a more familiar form.
• Sum-to-product and product-to-sum formulas that allow the conversion of products of sin and cos into sums and differences, easing integration.

In conjunction with techniques like Weierstrass substitution, these identities provide a robust toolkit to tackle intricate trigonometric integrals, reducing them to forms that are more straightforward to integrate. By mastering these identities, students can approach integration problems with greater confidence and efficiency.