/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Evaluate the integrals. $$\int... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 5

Evaluate the integrals. $$\int \sin ^{3} x d x$$

Short Answer

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

Step by step solution

01

Apply the Reduction Formula

The integral \( \int \sin^3 x \, dx \) is a type of trigonometric integral, which can be simplified using trigonometric identities. First, recall that \( \sin^3 x = (\sin^2 x) \sin x \). To simplify \( \sin^2 x \), we use the identity \( \sin^2 x = 1 - \cos^2 x \). This transforms the integral into \( \int (1 - \cos^2 x) \sin x \, dx \).
02

Use Substitution Method

Let \( u = \cos x \), which gives \( du = -\sin x \, dx \). Therefore, \( \sin x \, dx = -du \). Substitute \( u \) and \( du \) into the integral, changing it to \( \int (1 - u^2) (-du) \). This becomes \( -\int (1 - u^2) \ du = -\left( \int du - \int u^2 du \right) \).
03

Integrate

We integrate each term separately:- \( \int du = u \)- \( \int u^2 du = \frac{u^3}{3} \)Substituting these back, we get:\[-(u - \frac{u^3}{3}) + C\] where \(C\) is the constant of integration.
04

Substitute Back Original Variable

Substitute back \( u = \cos x \) into the integrated expression to return to the original variable:\(- \left( \cos x - \frac{\cos^3 x}{3} \right) + C\), simplifying to:\( - \cos x + \frac{\cos^3 x}{3} + C \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

reduction formula
Reduction formulas are a handy tool in evaluating trigonometric integrals, especially when powers of sine or cosine are involved. The basic idea is to break down a complex integral into simpler, more manageable parts using known identities.
  • For the integral \( \int \sin^n(x) \, dx \), we notice that expressing \( \sin^n(x) \) in terms of \( \sin(x) \) and \( \sin^{n-2}(x) \) helps.
  • We can use trigonometric identities like \( \sin^2(x) = 1 - \cos^2(x) \) to simplify the expression.
  • This method systematically reduces the power of the sine (or cosine), gradually converting the original problem into simpler integrals that are easier to manage.
In our exercise, we expressed \( \sin^3(x) \) as \( (\sin^2(x))(\sin(x)) \) and then used the identity \( \sin^2(x) = 1 - \cos^2(x) \). This expansion allowed us to transform the integral into a form suitable for substitution, leading to a simpler integral to solve.
substitution method
The substitution method, also known as u-substitution, is widely used in calculus to simplify integrals by making them more manageable. By changing variables, complicated integrals can often be transformed into a standard form.
  • First, we identify a part of the integral that can be represented as a single variable \( u \). This is typically a function that, when differentiated, appears elsewhere in the integral.
  • In our problem, we let \( u = \cos(x) \). The derivative \( du = -\sin(x) \, dx \) neatly replaces \( \sin(x) \, dx \) in the integral.
  • This conversion effectively changes the integral from trigonometric form to a simple polynomial form in terms of \( u \).
By rewriting the integral \( \int (1 - \cos^2(x)) \sin(x) \, dx \) in terms of \( u \) as \( \int (1 - u^2)(-du) \), the problem becomes easier to solve with basic polynomial integration techniques.
integration techniques
When dealing with integrals, especially trigonometric ones, a variety of integration techniques come into play. Each technique serves to simplify the integration process.
  • Elementary Techniques: Direct application of anti-differentiation for simple integrals where the antiderivative is obvious.
  • Polynomial Integration: Decomposing expressions into powers and using basic rules like \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
  • Trigonometric Identities: Using fundamental identities to rewrite expressions in a more workable form, such as \( \sin^2(x) = 1 - \cos^2(x) \).
In solving \( \int \sin^3(x) \, dx \), we applied polynomial techniques to integrate terms like \( \int u^2 \, du \) after transforming by substitution, arriving finally at the expression \( -\cos(x) + \frac{\cos^3(x)}{3} + C \). Each of these techniques can greatly simplify problems that initially seem complex, turning them into manageable tasks.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Evaluate the integrals in Exercises \(39-50\). $$\int \frac{\sqrt{x+1}}{x} d x$$ (Hint: Let \(x+1=u^{2}\) )

In Exercises \(9-16,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{2 x+1}{x^{2}-7 x+12} d x$$

Require the use of various trigonometric identities before you evaluate the integrals. $$\int \sin ^{3} \theta \cos 2 \theta d \theta$$

Integration by parts leads to a rule for integrating es that usually gives good results:$$\begin{aligned}\int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \\\&=y f(y)-\int f(y) d y \\ &=x f^{-1}(x)-\int f(y) d y\end{aligned}$$. The idea is to take the most complicated part of the integral, in this case \(f^{-1}(x)\), and simplify it first. For the integral of \(\ln x,\) we get $$\begin{aligned}\int \ln x \, d x &=\int y e^{y} d y \\\&=y e^{y}-e^{y}+C \\\&=x \ln x-x+C \end{aligned}$$ For the integral of \(\cos ^{-1} x\) we get $$\begin{aligned}\int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y \, d y \\\&=x \cos ^{-1} x-\sin y+C \\\&=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C\end{aligned}$$ Use the formula $$\int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y$$ Express your answers in terms of \(x.\) $$\int \tan ^{-1} x d x$$

As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x$$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\)-axis, diverges also. By comparing the two integrals, we see that, for every finite value \(b>1,\) $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x$$ (Check your book to see image) However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. a. Calculate it. b. This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite amount of paint cannot cover an infinite surface. But if we fill the horn with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.