Problem 81 - Calculus Single Variable

Question

Suppose that \(A\) and \(B\) are constants, not both of which are 0\. Notice that $$ A \cos (x)+B \sin (x)=\sqrt{A^{2}+B^{2}}(\cos (x) \cdot \xi+\sin (x) \cdot \eta) $$ where \(\xi=A / \sqrt{A^{2}+B^{2}}\) and \(\eta=B / \sqrt{A^{2}+B^{2}}\). By observing that \((\xi, \eta)\) is a point on the unit circle, deduce that \(A \cos (x)+\) \(B \sin (x)=\sqrt{A^{2}+B^{2}} \cos (x+\phi)\) for some \(\phi .\) Evaluate $$ \int \frac{1}{(A \cos (x)+B \sin (x))^{k}} d x $$ for \(k=1\) and \(k=2\).

Step-by-Step Solution

Answer

For \(k=1\), the integral is \(\frac{1}{\sqrt{A^2 + B^2}} \ln |\sec(x + \phi) + \tan(x + \phi)| + C\). For \(k=2\), it is \(\frac{1}{A^2 + B^2} \tan(x + \phi) + C\).

Step 1: Simplify the Expression

Given the expression \(A \cos (x) + B \sin (x)\), we can rewrite it as \(\sqrt{A^2 + B^2} (\cos(x) \cdot \xi + \sin(x) \cdot \eta)\), where \(\xi = \frac{A}{\sqrt{A^2 + B^2}}\) and \(\eta = \frac{B}{\sqrt{A^2 + B^2}}\). Since \((\xi, \eta)\) lies on the unit circle \(\xi^2 + \eta^2 = 1\), we can transform \(\cos(x) \cdot \xi + \sin(x) \cdot \eta\) into \(\cos(x + \phi)\), leading to \(A \cos(x) + B \sin(x) = \sqrt{A^2 + B^2} \cos(x + \phi)\).

Step 2: Evaluation of the First Integral when k=1

The integral for \(k=1\) becomes: \(\int \frac{1}{A \cos(x) + B \sin(x)}\, dx = \int \frac{1}{\sqrt{A^2 + B^2} \cos(x + \phi)}\, dx\). This simplifies to \(\frac{1}{\sqrt{A^2 + B^2}} \int \frac{1}{\cos(x + \phi)}\, dx\), which is equivalent to \(\frac{1}{\sqrt{A^2 + B^2}} \int \sec(x + \phi)\, dx\). The integral of \(\sec(x + \phi)\) is \(\ln |\sec(x + \phi) + \tan(x + \phi)| + C\). Thus, \(\int \frac{1}{A \cos(x) + B \sin(x)}\, dx = \frac{1}{\sqrt{A^2 + B^2}} \ln |\sec(x + \phi) + \tan(x + \phi)| + C\).

Step 3: Evaluation of the Second Integral when k=2

For \(k=2\), the integral is \(\int \frac{1}{(A \cos(x) + B \sin(x))^2}\, dx\). This is equivalent to \(\int \frac{1}{(\sqrt{A^2 + B^2} \cos(x + \phi))^2}\, dx = \frac{1}{A^2 + B^2} \int \sec^2(x + \phi)\, dx\). The integral of \(\sec^2(x + \phi)\) is \(\tan(x + \phi) + C\). Therefore, \(\int \frac{1}{(A \cos(x) + B \sin(x))^2}\, dx = \frac{1}{A^2 + B^2} \tan(x + \phi) + C\).

Key Concepts

Trigonometric Integrals

Trigonometric integrals are an essential topic in calculus that deals with integrating functions that involve trigonometric expressions. When faced with these types of integrals, it's crucial to simplify the expression using trigonometric identities. This not only makes the calculation easier but also reveals underlying patterns that aid in finding the solution.

In this exercise, the integral \( \int \frac{1}{(A \cos(x) + B \sin(x))^{k}} \, dx \) needs to be evaluated for two different values of \( k \). To tackle such integrals, rewriting the trigonometric expression into a simpler form is the first step. By using a combination of sine and cosine, we simplify it to \( \sqrt{A^2 + B^2} \cos(x + \phi) \). This transformation significantly reduces the complexity of the integral, making use of trigonometric identities like \( \cos(x + \phi) = \cos(x) \cos(\phi) - \sin(x) \sin(\phi) \).

The approach for integrating trigonometric functions often involves using these identities to transform the problem into one that is easier to solve, such as integrating basic trigonometric functions like \( \sec(x) \) and \( \sec^2(x) \). Once simplified, the integrals can be computed using basic integration techniques, resulting in logarithmic or tangent functions.

Integral Calculus

Integral calculus, the study of integrals and their properties, is a fundamental branch of calculus focused on accumulation and areas under curves. Calculating integrals requires techniques such as substitution, partial fractions, and recognizing patterns within the function.

In this case, integral calculus is used to evaluate expressions involving trigonometric components - specifically, evaluating integrals when different values of \( k \) are applied to the trigonometric function \( A \cos(x) + B \sin(x) \). By simplifying the expression to \( \sqrt{A^2 + B^2} \cos(x + \phi) \), the integrals turn into \( \sec \) and \( \sec^2 \) integrals, which are known to have straightforward solutions. This highlights the power of integral calculus in breaking down complex expressions into manageable calculations, allowing for evaluation using fundamental integration rules.

Integral calculus often requires an understanding of different functions and their derivatives, as it relates to finding areas, solving differential equations, or understanding motion and change. The solution to the exercise thus demonstrates an application of these concepts, showing how integration transforms an otherwise challenging expression.

Trigonometric Identities

Trigonometric identities are fundamental equations involving trigonometric functions that are true for all values of the involved variables. They are essential tools for simplifying and solving equations and integrals involving trigonometric expressions.

In the exercise provided, the identities allow us to convert the complicated expression \( A \cos(x) + B \sin(x) \) into a more manageable form \( \sqrt{A^2 + B^2} \cos(x + \phi) \). This transformation relies on understanding and applying identities like the angle addition formulas, where:

• \( \cos(x + \phi) = \cos(x)\cos(\phi) - \sin(x)\sin(\phi) \)
• \( \sin(x + \phi) = \sin(x)\cos(\phi) + \cos(x)\sin(\phi) \)

These identities are crucial as they break down one expression into the sum or difference of products of sine and cosine.

Understanding trigonometric identities allows students to simplify complex integrals and equations, making the problem-solving process more straightforward and intuitive. Recognizing patterns and substitutions frequently requires knowledge of these identities, which are powerful tools in both algebra and calculus. This exercise showcases how an understanding of trigonometric identities leads to discovering simpler pathways in solving seemingly intricate problems.