Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int \frac{\cos x}{\sin x(1-\sin x)} d x\)

Short answer

\( \ln|\sin x| - \ln|1 - \sin x| + C \)

Step-by-step solution

Identify Substitution

Look at the integral \[\int \frac{\cos x}{\sin x(1-\sin x)} d x\]Identify that using substitution simplifies this integral. Let \( u = \sin x \), then \( du = \cos x\, dx \). This substitution will simplify the trigonometric components.

Rewrite the Integral in Terms of u

Use the substitution \( u = \sin x \) and \( du = \cos x\, dx \). Substituting into the integral gives:\[\int \frac{\cos x}{\sin x(1-\sin x)} d x = \int \frac{du}{u(1-u)}\]This simplifies the integral to one involving only rational functions.

Use Partial Fraction Decomposition

The next step is to decompose \( \frac{1}{u(1-u)} \) using partial fractions. Write it as:\[\frac{1}{u(1-u)} = \frac{A}{u} + \frac{B}{1-u}\]Multiply out the right side to form a common denominator and solve for \( A \) and \( B \) by equating coefficients.

Solve for Coefficients A and B

Equating coefficients, we have: \[1 = A(1-u) + Bu\]Substitute convenient values for \( u \) to solve for \( A \) and \( B \). Let \( u = 0 \), so \( A = 1 \). Let \( u = 1 \), so \( B = 1 \). Thus:\[ \frac{1}{u(1-u)} = \frac{1}{u} + \frac{1}{1-u} \]

Integrate Each Term Separately

Now, integrate the expression term by term:\[\int \left( \frac{1}{u} + \frac{1}{1-u} \right) du = \int \frac{1}{u} du + \int \frac{1}{1-u} du\]This results in:\[\ln|u| - \ln|1-u| + C\]

Back-Substitute u = sin x

Replace \( u \) back with \( \sin x \):\[\ln|\sin x| - \ln|1 - \sin x| + C\]This gives the final solution to the integral, expressed in terms of \( x \).

Key concepts

substitution method

The substitution method is a powerful technique in solving integrals, especially when dealing with trigonometric functions. In essence, substitution involves replacing a variable or expression in the integral with another variable to simplify the integration process.
For our given integral \( \int \frac{\cos x}{\sin x(1-\sin x)} dx \), we noticed that trigonometric identities make direct integration cumbersome. By letting \( u = \sin x \), we effectively change the variable from \( x \) to \( u \), which simplifies the integral into rational functions.

• Substitution helps in managing complex expressions.
• It transforms the integral into a new form that is easier to integrate.

The initial derivative \( du = \cos x \, dx \) ensures that the differential part matches the original equation. Once substitution is applied, our focus shifts entirely to the newly defined variable, \( u \). This step transforms our original trigonometric integral into \( \int \frac{du}{u(1-u)} \), allowing us to proceed with integration more easily.

partial fraction decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions, which are easier to integrate. This technique is particularly useful for integrals of the form \( \int \frac{du}{u(1-u)} \).
To utilize partial fraction decomposition, we write \( \frac{1}{u(1-u)} \) as:

• \( \frac{A}{u} + \frac{B}{1-u} \)

We then determine the constants \( A \) and \( B \) by multiplying both sides by \( u(1-u) \) to remove the denominators, leading to the equation \( 1 = A(1-u) + Bu \). By substituting strategic values such as \( u = 0 \) and \( u = 1 \), we find that \( A = 1 \) and \( B = 1 \). This decomposition simplifies the task of integrating into two straightforward integrals, \( \int \frac{1}{u} du \) and \( \int \frac{1}{1-u} du \). After solving for these values, partial fraction decomposition becomes a straightforward yet essential tool for tackling otherwise complex rational functions.

trigonometric substitution

Trigonometric substitution is another integration method used to simplify integrals by substituting trigonometric functions for expressions involving square roots or other identities. Even though this problem primarily uses simple substitution to connect trigonometric functions, it is important to understand how and when trigonometric substitution is applied.
For example, in cases where the integral involves forms such as \( \sqrt{a^2 - x^2} \), trigonometric substitution allows us to convert expressions into familiar trigonometric identities such as \( x = a \sin \theta \).

• This substitution often simplifies the integral into basic trigonometric integrals or identities.
• It is typically used when dealing with the classical derivatives of inverse trigonometric functions.

In our case with \( \sin x \) and \( \cos x \), the original substitution from \( x \) to \( u \) simplifies the trigonometric terms into rational ones without delving into full trigonometric substitutions. However, it's crucial to identify when such substitutions are necessary for solving more complex integrals efficiently.

integration techniques

Integration techniques are a set of strategies used to evaluate complicated integrals. The combination of these techniques allows one to tackle a variety of integrals effectively. Our example \( \int \frac{\cos x}{\sin x(1-\sin x)} dx \) perfectly illustrates the synergy of several methods.

Initially, the substitution method was employed to simplify the trigonometric integral into a rational one, allowing us to apply algebraic techniques easily. Once transformed to \( \int \frac{du}{u(1-u)} \), we used partial fraction decomposition to express the integrand in a simpler form.

• Integration by substitution is often the first step for simplifying expressions involving composite functions.
• Partial fraction decomposition aids in converting complex rational expressions into easily integrable components.
• Recognizing when to utilize specific integration methods is key to efficiently solving integrals with various functions.

These techniques not only simplify the process but also make integration practically feasible for more intricate problems. By employing methods like substitution and fraction decomposition, complex integrals become a collection of simpler and more manageable problems.