Question

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{\cos x}{\left(\sin ^{3} x-4 \sin x\right)} d x$$

Short answer

Answer: The evaluated integral is $$-\frac{1}{4}(\ln|\sin\theta|) + \frac{1}{8}(\ln|\sin\theta - 2|) + \frac{1}{8}(\ln|\sin\theta + 2|) + C$$.

Step-by-step solution

Substitute the sine function with a new variable

Let's substitute $$u = \sin\theta$$. Then, we have $$du = \cos\theta\ d\theta$$. Now the original integral can be rewritten as: $$\int \frac{1}{\left(u^3 - 4u\right)} du$$

Factor out the denominator

Let's factor out the $$u$$ term in the denominator: $$\int \frac{1}{u\left(u^2 - 4\right)} du$$

Use Partial Fraction Decomposition

Now, we can perform partial fraction decomposition on the expression inside the integral: $$ \frac{1}{u(u^2 - 4)} = \frac{A}{u} + \frac{B}{u-2} + \frac{C}{u+2} $$ Multiplying by the common denominator gives us: $$ 1=A(u^2-4) + B(u(u+2)) + C(u(u-2)) $$ Expanding and simplifying, we get: $$ 1=A(u^2-4) + Bu^2 + 2Bu + Cu^2 - 2Cu $$ Combining the like terms, we get: $$ 1=(A+B+C)u^2 + (2B - 2C)u - 4A $$ Now, we can find the values of A, B, and C by comparing the coefficients of the terms: 1. A+B+C=0 2. 2B - 2C=0 => B=C 3. -4A=1 => A=-\frac{1}{4} Now, using B=C from equation (2) and A+B+C=0 and substituting A: $$ -\frac{1}{4} + 2B = 0 $$ Solving for B and C: $$ B = \frac{1}{8}, \quad C=\frac{1}{8} $$ Now, the integral can be rewritten using the partial fractions: $$ \int \left(\frac{-\frac{1}{4}}{u} + \frac{\frac{1}{8}}{u-2} + \frac{\frac{1}{8}}{u+2}\right)du $$

Integrate each term separately

Now, we can integrate each term separately: $$ -\frac{1}{4}\int\frac{1}{u} du + \frac{1}{8}\int\frac{1}{u-2} du + \frac{1}{8}\int\frac{1}{u+2} du $$ $$ -\frac{1}{4}(\ln|u|) + \frac{1}{8}(\ln|u-2|) + \frac{1}{8}(\ln|u+2|) $$

Replace u with original variable

Now that we have our solution in terms of $$u$$, we need to replace $$u$$ with $$\sin\theta$$: $$ -\frac{1}{4}(\ln|\sin\theta|) + \frac{1}{8}(\ln|\sin\theta - 2|) + \frac{1}{8}(\ln|\sin\theta + 2|) $$ Finally, we add the integration constant, $$C$$, to obtain our final solution: $$ -\frac{1}{4}(\ln|\sin\theta|) + \frac{1}{8}(\ln|\sin\theta - 2|) + \frac{1}{8}(\ln|\sin\theta + 2|) + C $$

Key concepts

Integration Techniques

Integration techniques are a set of methods used to solve integrals that are not readily solvable by basic integration. One of the main techniques used in the given exercise is partial fraction decomposition. This approach simplifies a complex rational function into simpler fractions that are easier to integrate.

Here's the process summarized:

• First, if the degree of the numerator is equal to or larger than the degree of the denominator, long division is used to simplify the expression.
• Then, we express the function as a sum of simpler fractions using partial fraction decomposition.
• Finally, we perform integration on each of these simpler fractions individually.

This method is crucial for integrating rational functions and is widely used in calculus problems.

Change of Variables

The change of variables technique is a powerful tool in calculus for simplifying integrals. It involves substituting a part of the integrand with a new variable, making the integral easier to handle. In the context of this exercise, it's utilized by substituting \( u = \sin\theta \), simplifying the integral into a more manageable form.

Steps for using the change of variables are:

• Identify a substitution that simplifies the integrand (in this case, \( u = \sin\theta \)).
• Differentiate the substitution to express \( du \) in terms of the original variable (here, \( du = \cos\theta\ d\theta \)).
• Rewrite the integral using the new variable. This step often transforms a complex function into a simpler one.
• Once integrated, don't forget to replace the variable back into terms of the original variable.

This technique is often a first step in problems involving trigonometric functions or when simplifying the given expressions is necessary.

Long Division in Calculus

Long division in calculus is similar to arithmetic long division, but it is applied to functions, specifically polynomials. It's applied when the degree of the numerator in a fraction is greater or equal to the degree of its denominator. This operation reduces the integrand into a standard form for easier integration.

The steps are as follows:

• Divide the numerator by the denominator keeping track of coefficients of similar terms.
• Subtract the result from the original expression to obtain the remainder.
• If necessary, the remainder becomes a new fraction to continue the process until a simpler expression is obtained.
• This simplified expression often allows for direct integration or other techniques like partial fraction decomposition.

This approach is crucial when dealing with rational functions in calculus, ensuring that integral expressions are of a manageable degree for integration with standard methods.