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Chapter 8: Problem 7

Evaluate the integrals. \(\int \sin ^{5} x d x\)

Short Answer

Expert verified
The integral is \(-\cos(x) + \frac{2}{3}\cos^3(x) - \frac{1}{5}\cos^5(x) + C \).

Step by step solution

01

Rewrite the Integrand Using Trigonometric Identities

We start by rewriting \( \sin^{5}(x) \) in a form that is easier to integrate. Using the identity \( \sin^{2}(x) = 1 - \cos^{2}(x) \), rewrite \( \sin^{5}(x) \) as \( \sin^{4}(x) \sin(x) = (\sin^{2}(x))^2 \sin(x) = (1 - \cos^{2}(x))^2 \sin(x) \).
02

Substitute for Integration

Let \( u = \cos(x) \), making \( du = -\sin(x) \, dx \) or \( -du = \sin(x) \, dx \). The integrand becomes \[ \int (1-u^2)^2 \cdot (-du) \], or \[ -\int (1 - 2u^2 + u^4) \, du \].
03

Integrate the Simplified Expression

Distribute the negative sign and integrate each term separately: \[ -\int 1 \, du + 2\int u^2 \, du - \int u^4 \, du \]. This gives us \(-u + \frac{2}{3}u^3 - \frac{1}{5}u^5 + C \).
04

Substitute Back Using Original Variable

Replace \( u \) with \( \cos(x) \) to return to the original variable: \(-\cos(x) + \frac{2}{3}\cos^3(x) - \frac{1}{5}\cos^5(x) + C \).
05

Write the Final Solution

The integral \( \int \sin^5(x) \, dx \) evaluates to \(-\cos(x) + \frac{2}{3}\cos^3(x) - \frac{1}{5}\cos^5(x) + C \), where \( C \) is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identities
Trigonometric identities are handy tools used to simplify trigonometric expressions. They help us express complex trigonometric functions in simpler forms, making them easier to work with. In this problem, we started by using the identity for sine squared, which is \( \sin^2(x) = 1 - \cos^2(x) \). This identity allows us to break down higher powers of sine into more manageable expressions.
  • Purpose: Simplifying expressions to make integration feasible.
  • Common Identities: Include Pythagorean identities like \( \sin^2(x) + \cos^2(x) = 1 \).
  • Application: Used in transforming \( \sin^5(x) \) into \( (1 - \cos^2(x))^2 \sin(x) \).
Remember, many of these identities stem from the basic trigonometric property that relates sine and cosine. Knowing how to manipulate and apply these identities is crucial for problems involving integration of trigonometric functions.
Substitution Method
The substitution method is a fundamental technique in calculus for simplifying the process of integration. It involves changing the variable of integration to a new variable, simplifying the integrand. In our example, we let \( u = \cos(x) \), turning the integral into something more straightforward.

  • Why Use It? It can convert a difficult integral into an easier one, often turning a trigonometric expression into a polynomial.
  • How It Works: Choose a substitution for a part of the integrand, find the differential (\( du \)), and replace the variable and expression accordingly.
  • In Practice: When \( u = \cos(x) \), we get \( du = -\sin(x) \ dx \), so \( -du = \sin(x) \ dx \). This replacement simplifies the integral considerably.
Key to this method is choosing the right substitution to reveal an easier path to integration, almost like solving a puzzle piece by piece.
Indefinite Integrals
Indefinite integrals involve finding the antiderivative of a function, essentially performing the reverse operation of differentiation. This type of integration does not result in a specific value but rather a function plus a constant, denoted as \( +C \).

  • Definition: The integral of a function without specified limits, represented as \( \int f(x) \, dx \).
  • Constant of Integration: The \( C \) represents an infinite number of possible integration outcomes, accounting for potential vertical shifts of the antiderivative function.
  • Example in Use: In our exercise, we found \( -u + \frac{2}{3}u^3 - \frac{1}{5}u^5 + C \) and later converted it back in terms of \( x \). This entire expression represents the family of functions whose derivatives can give us back the original integrand of \( \sin^5(x) \).
Indefinite integrals are essential for understanding the general solution of integration problems, implicitly defining a broad range of functions rather than a specific point or interval.

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Most popular questions from this chapter

Pollinating flowers A biologist models the time in minutes until a bee arrives at a flowering plant with an exponential distribution having a mean of 4 minutes. If 1000 flowers are in a field, how many can be expected to be pollinated within 5 minutes?

Prove that the sum \(T\) in the Trapezoidal Rule for \(\int_{a}^{b} f(x) d x\) is a Riemann sum for \(f\) continuous on \([a, b] .\) (Hint: Use the Intermediate Value Theorem to show the existence of \(c_{k}\) in the sub interval \(\left[x_{k-1}, x_{k}\right]\) satisfying \(f\left(c_{k}\right)=\left(f\left(x_{k-1}\right)+f\left(x_{k}\right)\right) / 2 . )\)

$$ \begin{array}{c}{\text { a. Use a CAS to evaluate }} \\ {\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x} \\ {\text { where } n \text { is an arbitrary positive integer. Does your CAS find }} \\ {\text { the result? }}\end{array} $$$$ \begin{array}{l}{\text { b. In succession, find the integral when } n=1,2,3,5, \text { and } 7 .} \\ {\text { Comment on the complexity of the results. }}\end{array} $$$$ \begin{array}{l}{\text { c. Now substitute } x=(\pi / 2)-u \text { and add the new and old }} \\ {\text { integrals. What is the value of }} \\\ {\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x ?} \\\ {\text { This exercise illustrates how a little mathematical ingenuity }} \\\ {\text { Solves a problem not immediately amenable to solution by a }} \\\ {\text { CAS. }}\end{array} $$

\begin{equation} \begin{array}{l}{\text { Consider the infinite region in the first quadrant bounded by the }} \\ {\text { graphs of } y=\frac{1}{x^{2}}, y=0, \text { and } x=1 .} \\ {\text { a. Find the area of the region. }} \\ {\text { b. Find the volume of the solid formed by revolving the region }} \\ {\text { (i) about the } x \text { -axis; (ii) about the } y \text { -axis. }}\end{array} \end{equation}

\begin{equation} \begin{array}{l}{\text { Exercises } 71-74 \text { are about the infinite region in the first quadrant }} \\ {\text { between the curve } y=e^{-x} \text { and the } x \text { -axis. }}\end{array} \end{equation} Find the volume of the solid generated by revolving the region about the \(x\) -axis.
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