Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int \frac{\sin x}{\cos ^{2} x+\cos x-6} d x\)

Short answer

The integral is \( \frac{1}{5}\ln |\cos x + 3| - \frac{1}{5}\ln |\cos x - 2| + C \).

Step-by-step solution

Substitute a Trig Identity

To simplify the integrand, we use a trigonometric substitution. Set \[ u = \cos x \], so \[ du = -\sin x \, dx \]. This implies \[ \sin x \, dx = -du \]. Substitute these into the integral to get:\[ \int \frac{\sin x}{\cos ^{2} x + \cos x - 6} \, dx = \int \frac{-du}{u^2 + u - 6} \]. This changes the integral into one in terms of the rational function \( -\int \frac{du}{u^2 + u - 6} \).

Factor the Denominator

Now, factor the quadratic in the denominator:\[ u^2 + u - 6 = (u + 3)(u - 2) \].So the integral becomes:\[ -\int \frac{du}{(u+3)(u-2)} \].

Decompose into Partial Fractions

Use partial fraction decomposition to express \[ \frac{1}{(u+3)(u-2)} = \frac{A}{u+3} + \frac{B}{u-2} \]. Multiply through by \((u+3)(u-2)\) to obtain:\[ 1 = A(u-2) + B(u+3) \].

Solve for Coefficients

To find the values of A and B, choose strategic values for u:- Let \( u = 2 \): \[ 1 = A(2-2) + B(2+3) \Rightarrow 1 = 5B \Rightarrow B = \frac{1}{5} \].- Let \( u = -3 \): \[ 1 = A(-3-2) + B(-3+3) \Rightarrow 1 = -5A \Rightarrow A = -\frac{1}{5} \].Thus, the decomposition is: \[ \frac{1}{(u+3)(u-2)} = \frac{-1/5}{u+3} + \frac{1/5}{u-2} \].

Integrate the Partial Fractions

Replace into the integrated expression:\[ -\int \left( \frac{-1/5}{u+3} + \frac{1/5}{u-2} \right) du = \frac{1}{5} \int \frac{1}{u+3}du - \frac{1}{5} \int \frac{1}{u-2}du \].Evaluate each integral:\[ \int \frac{1}{u+3} du = \ln |u+3| + C_1 \]\[ \int \frac{1}{u-2} du = \ln |u-2| + C_2 \].Together:\[ \frac{1}{5}\ln |u+3| - \frac{1}{5}\ln |u-2| + C \].

Substitute Back in Terms of x

Substitute back \( u = \cos x \):\[ \frac{1}{5}\ln |\cos x + 3| - \frac{1}{5}\ln |\cos x - 2| + C \].

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals that contain trigonometric functions. This method takes advantage of trigonometric identities to transform the integrals into a more manageable form. When dealing with integrals involving trigonometric functions like sine or cosine, converting these into an algebraic form can ease the integration process. For example, we can make a substitution such as \( u = \cos x \), which implies \( du = -\sin x \, dx \). This turns the integrand involving sine and cosine into a function of \( u \), often simplifying the problem into a rational function. This substitution helps in eliminating the trigonometric functions and paves the way for easier manipulation and solution of the integral.

Partial Fraction Decomposition

Partial fraction decomposition is a technique that breaks down complex fractions into simpler parts, allowing easier integration. This method is specifically useful when dealing with rational functions, which are fractions containing polynomials in the numerator and denominator. Once the integrand is a rational function, we can express it as a sum of simpler fractions. Each of these fractions can then be integrated individually. For example, if we have a fraction like \( \frac{1}{(u+3)(u-2)} \), we can express it as a sum: \( \frac{A}{u+3} + \frac{B}{u-2} \), where \( A \) and \( B \) are constants that can be determined through strategic placement of values or solving a system of equations. This breakdown into partial fractions is a fundamental step for integration because it makes complex integrations much more straightforward.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. These functions frequently appear in calculus problems, particularly in integration exercises. Converting an integrand into a rational function is often the goal when using techniques like trigonometric substitution. This transformation can transform problematic trigonometric terms into algebraic forms, simplifying the integration process. In the given exercise, after substituting \( u = \cos x \), we obtained the rational function \( \frac{1}{u^2 + u - 6} \), which after factoring, became \( \frac{1}{(u+3)(u-2)} \). Handling rational functions requires know-how of polynomial manipulations and techniques like partial fraction decomposition to solve integrals effectively.

Integration Techniques

Integration is a core concept in calculus, used to find areas under curves and solve differential equations, among other things. Various integration techniques are applied depending on the form of the integrand. These include substitution, integration by parts, and partial fractions. In this exercise, we used substitution to simplify trigonometric terms and then applied partial fraction decomposition to simplify a rational function. Each of these techniques serves a unique purpose and can handle different kinds of functions, making it essential to understand which technique is best suited for the problem at hand. Techniques like these allow multiple problems to be simplified into a sum of logarithms or simple polynomials, making them much easier to integrate and providing a solid framework for solving complex calculus problems.