Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int \frac{\sin x d x}{1-\cos ^{2} x}\)

Short answer

The integral evaluates to \(\frac{1}{2} \ln \left| \frac{1+\cos x}{1-\cos x} \right| + C.\)

Step-by-step solution

Recognize Trigonometric Identity

First, notice the identity \(1 - \cos^2 x = \sin^2 x\). This is derived from the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\).

Simplify the Integral

Rewrite the integral using the identity: \(\int \frac{\sin x}{\sin^2 x} \, dx = \int \frac{1}{\sin x} \, dx = \int \csc x \, dx.\)

Use Substitution

Let \( u = \cos x \). Then \( du = -\sin x \, dx \), or \( \sin x \, dx = -du \). Substitute into the integral: \( \int \frac{\sin x}{\sin^2 x} \, dx = \int \frac{-du}{1-u^2}. \)

Express in Partial Fractions

The integral can be expressed as a rational function: \( \int \frac{1}{1-u^2} \, du = \int \left( \frac{1}{2(1-u)} + \frac{1}{2(1+u)} \right) \, du. \)

Integrate the Partial Fractions

Integrate each term separately: \(\frac{1}{2} \int \frac{1}{1-u} \, du + \frac{1}{2} \int \frac{1}{1+u} \, du \)The integrals are: \(-\frac{1}{2} \ln |1-u| + \frac{1}{2} \ln |1+u| + C.\)

Substitute Back for Original Variable

Recall \( u = \cos x \). Substitute back to obtain: \(-\frac{1}{2} \ln |1-\cos x| + \frac{1}{2} \ln |1+\cos x| + C = \frac{1}{2} \ln \left| \frac{1+\cos x}{1-\cos x} \right| + C. \)

Key concepts

Substitution Method in Integration

The substitution method is a fundamental technique in calculus integration, used to simplify complicated integrals. This approach involves changing the variable of integration to make the integral easier to solve.
In the given problem, we start by identifying a suitable substitution. We set the new variable, often called "u", to an expression within the integrand that simplifies the integral. In this case, by setting \( u = \cos x \), we facilitate the transformation. This choice is strategic, considering the derivative \( du = -\sin x \, dx \), which matches the differential component, \( \sin x \, dx \), in the integrand.
Substitution works best when it translates the integral into a simpler form. Here, the integral transforms from trigonometric functions of \( x \) to rational functions of \( u \), allowing us to apply further integration techniques like partial fraction decomposition.

Partial Fractions Decomposition

Partial fractions decomposition is a method used to integrate rational functions, which are polynomials divided by polynomials. This technique is particularly effective when dealing with terms that are not easily integrable in their original form.
In the exercise, after substituting \( u = \cos x \), the integral reduces to \( \int \frac{-du}{1-u^2} \). To integrate, we decompose this into simpler fractions. The expression \( 1-u^2 \) is rewritten using its simpler equivalent form in partial fractions: \( \frac{1}{1-u^2} = \frac{1}{2(1-u)} + \frac{1}{2(1+u)} \).
This decomposition leverages the fact that rational expressions can often be broken into a sum of simpler fractions, with linear denominators that are easier to integrate. By breaking it down, each fraction corresponds to a basic integral form directly lead to a logarithmic result, efficiently solving the integral.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the involved variables. They are a crucial tool in simplifying expressions in calculus.
The identity \( 1 - \cos^2 x = \sin^2 x \) stems from the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \). In the initial problem, this identity allows us to rewrite the denominator \( 1 - \cos^2 x \) as \( \sin^2 x \), directly simplifying the integrand to \( \frac{\sin x}{\sin^2 x} = \frac{1}{\sin x} = \csc x \).
Understanding and applying these identities simplifies integration tasks and aids in converting integrals into forms that are easier to tackle with substitution or other methods.

Integral of csc x

The integral of \( \csc x \) is a classic problem often tackled in calculus courses. It involves finding the antiderivative of the cosecant function, which is the reciprocal of the sine function.
Given that \( \int \csc x \, dx \) is a transformed result from earlier simplifications, it refers back to rewriting trigonometric expressions into simpler forms. For students, this integral is memorized as a standard result:

• \( \int \csc x \, dx = -\ln |\csc x + \cot x| + C \).

However, in our context, by substitution and further simplification, the result eventually takes the form of a logarithmic expression involving cosine, \( \ln \left| \frac{1+\cos x}{1-\cos x} \right| \).
Comprehending integrals like \( \csc x \) illustrates how trigonometric integrals often resolve into logarithmic solutions, reflecting deeper connections in calculus.