Problem 56 - Calculus Early Transcendentals

Question

Additional integrals Evaluate the following integrals. $$\int \csc ^{10} x \cot ^{3} x d x$$

Step-by-Step Solution

Answer

Answer: \(-\frac{1}{4}\cot^4 x-\frac{11}{6}\cot^6 x-\frac{23}{4}\cot^8 x-\frac{73}{5}\cot^{10} x-\frac{211}{12}\cot^ {12} x-\frac{47}{14}\cot^ {14} x-\frac{1}{16}\cot^ {16} x + C\)

Step 1: Use the Pythagorean Identity

In order to solve this integral, first observe that: $$\csc^2 x - \cot^2 x = 1$$ From this identity, we can express \(\csc^2 x\) in terms of \(\cot^2 x\) as: $$\csc^2 x = 1+ \cot^2 x$$ We can replace \(\csc ^{10} x\) in the integral: $$\int \csc ^{10} x \cot ^{3} x d x = \int (\csc ^2 x)^5 \cot ^3 x d x = \int (1 + \cot ^2 x)^5 \cot^3 x d x$$

Step 2: Perform U-Substitution

To simplify this expression, use u-substitution. Let: $$u = \cot x$$ Now differentiate \(u\) with respect to \(x\): $$\frac{du}{dx} = -\csc ^2 x$$ Now solve for \(dx\) in terms of \(du\): $$dx = -\frac{1}{\csc^2 x} du$$ Now we can rewrite the integral in terms of \(u\) and \(du\): $$\int (1 + u^2)^5 u^3 d(-\frac{1}{\csc^2 x} du) = -\int (1+u^2)^5 u^3\csc^2 x du$$ Remember that \(\csc^2 x = u^2 + 1\), so we can rewrite the integral as: $$-\int (1+u^2)^5 u^3(u^2+1) du$$

Step 3: Expand and Simplify

Now, expand the expression \((1+u^2)^5\) and multiply it with \(u^3(u^2+1)\). After expanding and simplifying, we have: $$-\int (1+10u^2+45u^4+100u^6+105u^8+46u^{10}+u^{12}) u^3(u^2+1) du$$ Now perform the integration term by term: $$-\int( u^3+11u^5+46u^7+146u^9+211u^{11}+47u^{13}+u^{15}) du = -\frac{1}{4}u^4-\frac{11}{6}u^6-\frac{46}{8}u^8-\frac{146}{10}u^{10}-\frac{211}{12}u^{12}-\frac{47}{14}u^{14}-\frac{1}{16}u^{16}+ C$$

Step 4: Substitute Back and Simplify

Now, substitute back the original expression for \(u\): $$-\frac{1}{4}(\cot^4 x)-\frac{11}{6}(\cot^6 x)-\frac{46}{8}(\cot^8 x)-\frac{146}{10}(\cot^{10} x)-\frac{211}{12}(\cot^ {12} x)-\frac{47}{14}(\cot^ {14} x)-\frac{1}{16}(\cot^ {16} x) + C$$ So, the solution to the given integral is: $$\int \csc ^{10} x \cot ^{3} x d x = -\frac{1}{4}\cot^4 x-\frac{11}{6}\cot^6 x-\frac{23}{4}\cot^8 x-\frac{73}{5}\cot^{10} x-\frac{211}{12}\cot^ {12} x-\frac{47}{14}\cot^ {14} x-\frac{1}{16}\cot^ {16} x + C$$

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a fundamental concept in trigonometry. This identity states that for any angle \(x\), the equation \(\sin^2 x + \cos^2 x = 1\) holds true. However, there are other forms of this identity that involve the reciprocal trigonometric functions: cosecant (csc) and cotangent (cot).
In the context of the original exercise, another variation of the Pythagorean Identity is used: \(\csc^2 x - \cot^2 x = 1\). This can be rearranged to express \(\csc^2 x\) in terms of \(\cot^2 x\):

• \(\csc^2 x = 1 + \cot^2 x\)

This alteration is crucial for simplifying the integral in the problem. By expressing \(\csc^2 x\) in terms of \(\cot^2 x\), the integral is transformed into a simpler form to work with trigonometric functions during the integration process.

U-Substitution

U-Substitution is a technique used in integration to make the integral easier to solve. It's akin to using the inverse of the chain rule from differentiation.
The idea is to substitute a part of the integrand with a new variable \(u\), then rewrite and solve the integral in terms of \(u\). This technique simplifies the problem, particularly when dealing with complex expressions.
For the given integral \(\int \csc^{10} x \cot^3 x \, dx\), we use \(u = \cot x\). This substitution simplifies the integral:

• Differentiating \(u\) with respect to \(x\) gives \(\frac{du}{dx} = -\csc^2 x\).
• Solving for \(dx\), we get \(dx = -\frac{1}{\csc^2 x} \, du\).

Now the integral can be rewritten in terms of \(u\) as \(-\int (1 + u^2)^5 u^3 (u^2 + 1) \, du\), making it a polynomial that can be integrated more simply.

Trigonometric Integrals

Trigonometric integrals involve integrating products or powers of trigonometric functions. These types of integrals require specific identities and substitutions to be simplified and solved.
When dealing with integrals of higher powers, like in the exercise \(\int \csc^{10} x \cot^3 x \, dx\), recognizing trigonometric relationships and identities is crucial. Using replacements such as \(\csc^2 x = 1 + \cot^2 x\) helps maneuver around cumbersome trigonometric expressions.
The exercise demonstrates a method to tackle these integrals by reducing them to polynomial forms using trigonometric identities.

• By expressing functions in terms of \(u\), as shown in the integration process, the trigonometric relationships become clearer and simpler to integrate.
• The strategy usually entails distributing and expanding expressions before integrating.
• Post-integration, the solution is expressed back in terms of the original variable by reversing the substitution process.

Understanding these techniques allows you to solve a variety of trigonometric integrals efficiently.