Problem 25 - Essential Calculus

Question

Use substitution to find the integral. $$ \int \frac{\sin x}{\cos x(\cos x-1)} d x $$

Step-by-Step Solution

Answer

The simplified form of the integral is \( \ln \left|\frac{\cos x - 1}{\cos x}\right| + C \)

Step 1: Identifying substitution

Observe that the derivative of \( \cos x \) is \( -\sin x \), which appears in the numerator. So let’s take \( u = \cos x \). Therefore, \( du = -\sin x dx\).

Step 2: Simplifying the integral using substitution

Substituting \( u \) and \( du \) into the integral, it transforms as: \(-\int \frac{1}{u(u - 1)} du \).

Step 3: Solving the integral

This integral can be solved using Partial Fraction Decomposition. Let's write \(\frac{1}{u(u - 1)}\) as \(\frac{A}{u} + \frac{B}{u - 1}\). After simplifying, it gives us \(A = -1\) and \(B = 1\). Now, we can rewrite the integral as \( -\int (-\frac{1}{u} + \frac{1}{u - 1}) du \). The integral of the individual terms can be computed directly which results in \(-(-\ln |u| + \ln |u - 1|) + C\).

Step 4: Substituting back

Substitute \( u = \cos x \) back into the solution and simplify, it results in \(-\ln |\cos x| + \ln |\cos x - 1| + C\).

Key Concepts

Partial Fraction Decomposition

When faced with an integral that involves fractions, such as the integral \[ \int \frac{1}{u(u - 1)} du \], we employ the technique known as partial fraction decomposition. This method helps break down a complex rational expression into simpler, easily integrable parts.

To start, decompose the given expression, \( \frac{1}{u(u - 1)} \), into a sum of simpler fractions, generally in the form:

• \( \frac{A}{u} + \frac{B}{u - 1} \)

Determine the coefficients \( A \) and \( B \) by making a common denominator and equating it to the original numerator after solving for variables. For this problem, solving these equations gives:

• \( A = -1 \)
• \( B = 1 \)

By substituting these values back, we rewrite the integral as:
\[ -\int \left(-\frac{1}{u} + \frac{1}{u - 1}\right) du \].

Partial fraction decomposition turns a challenging integral into manageable pieces that can be evaluated individually.

Trigonometric Integration

Integration involving trigonometric functions often requires strategic substitutions or transformations. In the given problem, \( \int \frac{\sin x}{\cos x(\cos x-1)} d x \),
we notice the presence of the trigonometric functions \( \sin x \) and \( \cos x \).

The substitution \( u = \cos x \) and \( du = -\sin x dx \) effectively translates the trigonometric integral into a more approachable algebraic form.
By converting trigonometric expressions using identities or substitution:

• This leverages the relationship between trigonometric derivatives and simplifies the integration process.
• The integral becomes algebraic, allowing for straightforward calculus operations without the complexity of trigonometric manipulation.

Trigonometric integration is thus simplified by identifying appropriate substitutions or identities which align trigonometric expressions into solvable forms.

Calculus Integral Problems

Solving calculus integral problems requires a toolkit of strategies, especially when dealing with integrals of the form:
\( \int \frac{\sin x}{\cos x(\cos x-1)} d x \).

In this scenario, substitution methods streamline what would otherwise be a tricky process.
To tackle such problems effectively:

• Firstly, recognize useful substitutions that simplify the expression, like \( u = \cos x \) to transform the integral into an algebraic context.
• Secondly, employ specialized techniques such as partial fraction decomposition to further simplify and evaluate the integral.
• Finally, ensure the result is expressed in the initial variable, using back-substitution, to match the context of the original problem.

Mastering calculus integral problems involves recognizing the types of expressions involved and applying the appropriate breakdown and solving strategies. This combination transforms complicated integrations into orderly, solvable ones.

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Other exercises in this chapter

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Problem 25

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Problem 26

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty} \frac{e^{x}}{1+e^{x}} d x $$

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