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Chapter 8: Problem 38

First make a substitution and then use integration by parts to evaluate the integral. $$ \int e^{6 x} \cos e^{3 x} d x $$

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{3} [e^{3x} \sin(e^{3x}) + \cos(e^{3x})] + C \).

Step by step solution

01

Make a substitution

Let's use the substitution \( u = e^{3x} \). Then, \( du = 3e^{3x}dx \) or equivalently \( dx = \frac{du}{3u} \). Also, \( e^{6x} = (e^{3x})^2 = u^2 \). Now we can rewrite the integral: \[ \int e^{6x} \cos(e^{3x}) \, dx = \int u^2 \cos(u) \frac{du}{3u} = \frac{1}{3} \int u \cos(u) \, du \]
02

Set up integration by parts

We will use integration by parts on \( \frac{1}{3} \int u \cos(u) \, du \). Recall the integration by parts formula: \( \int v \, dw = vw - \int w \, dv \). Choose \( v = u \) and \( dw = \cos(u) \, du \). Then, \( dv = du \) and \( w = \int \cos(u) \, du = \sin(u) \).
03

Apply integration by parts formula

Substitute \( v \), \( dv \), \( w \), and \( dw \) into the integration by parts formula: \[ \frac{1}{3} \int u \cos(u) \, du = \frac{1}{3} \left[u \sin(u) - \int \sin(u) \, du \right] \] Simplify: \[ = \frac{1}{3} \left[u \sin(u) + \cos(u) \right] + C \] where \( C \) is the constant of integration.
04

Back-substitute the original variable

Our original substitution was \( u = e^{3x} \). Replace \( u \) in our expression to return to the variable \( x \): \[ \frac{1}{3} [e^{3x} \sin(e^{3x}) + \cos(e^{3x})] + C \] This is the evaluated integral in terms of \( x \).

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Key Concepts

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

Substitution Method
The substitution method is a straightforward technique to simplify an integral by substituting variables to make the computation easier. In this original exercise, we started by setting a substitution to reduce complexity. Let's break down the key points:
  • Choose a substitution that simplifies the integral. Here, we used the substitution \( u = e^{3x} \), which reduces the exponent and simplifies the integral.
  • Calculate the derivative of your substitution to express \( dx \) in terms of \( du \). For our substitution, \( du = 3e^{3x}dx \), which gives \( dx = \frac{du}{3u} \).
  • Rearrange the integral using your substitution. The original integral \( \int e^{6x} \cos(e^{3x}) \, dx \) becomes \( \frac{1}{3} \int u \cos(u) \, du \).
This substitution effectively transforms the integral into a simpler form that is easier to integrate in subsequent steps.
Integration by Parts
Integration by parts is a handy tool, especially when dealing with products of functions in integrals. It is based on the product rule for differentiation and follows the formula:\[ \int v \, dw = vw - \int w \, dv \]To use integration by parts effectively, follow these steps:
  • Identify the parts of the integral that will serve as \( v \) and \( dw \). In our case, \( v = u \) and \( dw = \cos(u) \, du \).
  • Calculate \( dv \) and \( w \). Here, \( dv = du \) and \( w = \sin(u) \).
  • Apply the integration by parts formula to get \( \frac{1}{3} \left[u \sin(u) - \int \sin(u) \, du \right] \).
  • Simplify further by integrating \( \sin(u) \) to get the result \( \frac{1}{3} [u \sin(u) + \cos(u)] + C \).
Integration by parts helps break down complex problems into more manageable pieces, allowing you to tackle the integral in stages.
Definite Integrals
While the exercise provided an indefinite integral solution, understanding definite integrals is crucial for solving many real-world problems. Definite integrals compute the net area under a curve from one point to another. Here are some aspects to consider:
  • Definite integrals are evaluated over a specified interval \([a, b]\).
  • The process involves finding the antiderivative of the function and then using the Fundamental Theorem of Calculus.
  • To compute: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \] where \( F(x) \) is the antiderivative of \( f(x) \).
  • For bounded functions, definite integrals can represent physical quantities such as total distance, area, and accumulated change.
Even though definite integrals were not specifically used in this exercise, mastering them enables students to apply integration techniques to practical applications.

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Most popular questions from this chapter

Find the area \(A\) of the portion of the region between the graphs of \(y=\tan ^{3} x\) and \(y=\frac{1}{4} \tan x \sec ^{4} x\) that lies between the lines \(x=0\) and \(x=\pi / 3\).

This exercise is designed to support the assertion that $$ \int_{-\infty}^{\infty} e^{-x^{2} / 2} / \sqrt{2 \pi} d x=1 $$ a. Use Simpson's Rule with \(n=100\) to approximate $$ \int_{-5}^{5} e^{-x^{2} / 2} / \sqrt{2 \pi} d x $$ b. Find the smallest positive integer \(b\) such that when you use Simpson's Rule with \(n=100\), the value your computer or calculator gives for \(\int_{-b}^{b} e^{-x^{2} / 2} / \sqrt{2 \pi} d x\) is 1 .

Determine whether the improper integral converges. If it does, determine the value of the integral. \(\int_{-2}^{\infty} \frac{1}{t^{2}+4 t+8} d t\)

Find the area \(A\) of the region between the graph of \(f\) and the \(x\) axis on the given interval. \(f(x)=\sec ^{4} x ;[-\pi / 3, \pi / 4]\)

Decide whether the region between the graph of the integrand and the \(x\) axis on the interval of integration has finite area. If it does, calculate the area. \(\int_{-1}^{1} \frac{1}{\sqrt{x+1}} d x\)
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