Problem 55 - Calculus Single Variable

Question

Make a substitution before applying the method of partial fractions to calculate the given integral. $$ \int \frac{\cos (x)}{\sin ^{2}(x)-5 \sin (x)+6} d x $$

Step-by-Step Solution

Answer

\( \int \frac{\cos(x)}{\sin^2(x) - 5\sin(x) + 6} dx = \ln \left| \frac{\sin(x) - 3}{\sin(x) - 2} \right| + C \)

Step 1: Identify the Substitution

In this problem, we can use the substitution \( u = \sin(x) \). This simplifies the trigonometric functions and prepares the expression for partial fractions. This also gives us \( du = \cos(x) \, dx \), which we will substitute into the integral.

Step 2: Substitute and Simplify

Substitute \( u = \sin(x) \) and \( du = \cos(x) \, dx \) into the integral: \[\int \frac{\cos(x)}{\sin^2(x) - 5\sin(x) + 6} dx = \int \frac{1}{u^2 - 5u + 6} \, du\]Now, the problem is an integral of a rational function, which can be tackled using partial fraction decomposition.

Step 3: Factor the Denominator

Factor the quadratic expression in the denominator of the new integral:\[ u^2 - 5u + 6 = (u - 2)(u - 3) \]

Step 4: Apply Partial Fraction Decomposition

Express the fraction \( \frac{1}{(u - 2)(u - 3)} \) as a sum of partial fractions:\[ \frac{1}{(u - 2)(u - 3)} = \frac{A}{u - 2} + \frac{B}{u - 3} \]Solve for \( A \) and \( B \) by multiplying both sides by \( (u - 2)(u - 3) \), then equating coefficients or plugging in convenient values of \( u \).

Step 5: Solve for Constants A and B

Set up the equation: \[ 1 = A(u - 3) + B(u - 2) \]Choose convenient values for \( u \):- For \( u = 2 \): \( 1 = A(2 - 3) \Rightarrow A = -1 \) - For \( u = 3 \): \( 1 = B(3 - 2) \Rightarrow B = 1 \) Therefore, \( A = -1 \) and \( B = 1 \).

Step 6: Integrate Each Term

The integral becomes:\[ \int \left( \frac{-1}{u - 2} + \frac{1}{u - 3} \right) du \]Integrate each term separately:- \( \int \frac{-1}{u - 2} \, du = -\ln|u - 2| + C_1 \) - \( \int \frac{1}{u - 3} \, du = \ln|u - 3| + C_2 \) The constants \( C_1 \) and \( C_2 \) can be combined into a single constant \( C \).

Step 7: Substitute Back and Simplify the Result

Replace \( u \) with \( \sin(x) \) to revert to the original variable:\[ \int \frac{\cos(x)}{\sin^2(x) - 5\sin(x) + 6} \, dx = -\ln|\sin(x) - 2| + \ln|\sin(x) - 3| + C \]Combine the logarithms:\[ \ln \left| \frac{\sin(x) - 3}{\sin(x) - 2} \right| + C \]

Key Concepts

Substitution Method

The substitution method is a powerful tool in integral calculus, offering a simpler path to solving integrals by changing variables. In our example, we use this method to convert a complex trigonometric integral into a simpler form. By setting

• \( u = \sin(x) \)
• we transform the integral into terms of \( u \) instead of \( x \).

This substitution simplifies the trigonometric part and makes the subsequent steps more manageable. Along with it, the differential

• \( du = \cos(x) \, dx \)

replaces \( dx \) in the integral, maintaining the equation's integrity. The aim of substitution is to convert the integral into a form which is easier to integrate, often leading to methods like partial fractions.

Partial Fraction Decomposition

Partial Fraction Decomposition is an integration technique used when dealing with rational functions. Once you have performed a substitution leading to a rational expression, this technique breaks that expression into simpler fractions. For our integral,

• The denominator, \( u^2 - 5u + 6 \), factors down to \( (u - 2)(u - 3) \).

By expressing the single fraction as a sum of partial fractions:

• \( \frac{1}{(u - 2)(u - 3)} = \frac{A}{u - 2} + \frac{B}{u - 3} \)

We can solve for constants \( A \) and \( B \) by considering suitable values for \( u \), finding

• \( A = -1 \) and
• \( B = 1 \).

Once determined, these values are used to form simpler integrals that can be integrated independently.

Trigonometric Substitution

While this method wasn't directly used in the original problem, it's worth explaining how it could assist in integral calculus involving trigonometric functions. Trigonometric substitution comes into play when an expression involves square roots combined with constants and square terms.
This kind of substitution relies on trigonometric identities to simplify integrals by setting a variable equal to a trigonometric function. It is especially effective when encountering integrals involving:

• Sqrt terms like \( \sqrt{a^2 - x^2} \), where \( x = a \sin(\theta) \).

This substitution converts complex algebraic expressions into its trigonometric counterparts, easing the integration process by utilizing familiar trigonometric identities.

Integration Techniques

Understanding various integration techniques is crucial for solving diverse calculus problems. Each method has its unique applications:

• The substitution method simplifies expressions by changing variables.
• Partial fraction decomposition is ideal for breaking down rational expressions for easier integration.
• Trigonometric substitution is useful for integrating expressions involving square roots and trigonometric identities.

These techniques often work in conjunction with one another, as seen in the original exercise involving substitution and partial fractions. Mastery of various methods not only helps with solving complex integrals but also builds a foundation for more advanced calculus problems, enhancing mathematical problem-solving skills.