/*! 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 29 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 29

Evaluate the following integrals using integration by parts. $$\int e^{-x} \sin 4 x d x$$

Short Answer

Expert verified
Based on the given step-by-step solution, provide a short answer for the integral of \(e^{-x}\sin 4x\) with respect to x: The integral of \(e^{-x}\sin 4x\) with respect to x is given by: \(\int e^{-x}\sin 4x dx = -\frac{3}{4}e^{-x}\cos 4x - \frac{3}{16}e^{-x}\sin 4x + C\)

Step by step solution

01

Find \(\int v' dx\)

To find \(\int v' dx\), let's integrate \(v' = \sin 4x\) with respect to \(x\). $$v = \int \sin 4x dx = -\frac{1}{4}\cos 4x + C_1$$
02

Apply integration by parts formula

Now, we will use the integration by parts formula: $$\int e^{-x} \sin 4x dx = e^{-x}(-\frac{1}{4}\cos 4x) - \int -e^{-x}(-\frac{1}{4}\cos 4x) dx$$
03

Simplify and integrate the remaining integral

Rearranging and simplifying the last equation, we get: $$\int e^{-x} \sin 4x dx = -\frac{1}{4}e^{-x}\cos 4x - \frac{1}{4} \int e^{-x} \cos 4x dx$$ Now, we will apply the integration by parts again. Let \(a = e^{-x}\), and \(b' = \cos 4x\). Find \(a'\) and \(b\). $$a' = -e^{-x}$$ $$\int b' dx = \int \cos 4x dx = \frac{1}{4}\sin 4x + C_2$$ Apply the integration by parts formula again: $$\int e^{-x} \cos 4x dx = e^{-x}\frac{1}{4}\sin 4x - \int -e^{-x}(\frac{1}{4}\sin 4x) dx$$ Simplify and rearrange: $$\int e^{-x} \cos 4x dx = \frac{1}{4}e^{-x}\sin 4x + \frac{1}{4}\int e^{-x}\sin 4x dx$$
04

Solve for the desired integral

Combining our results from Step 3 and Step 4, we have: $$\frac{1}{4}\int e^{-x}\sin 4x dx = -\frac{1}{4}e^{-x}\cos 4x - \frac{1}{4}\left(\frac{1}{4}e^{-x}\sin 4x + \frac{1}{4}\int e^{-x}\sin 4x dx\right)$$ Now, we will rearrange to isolate the integral term we want to find: $$\frac{1}{3}\int e^{-x}\sin 4x dx = -\frac{1}{4}e^{-x}\cos 4x - \frac{1}{16}e^{-x}\sin 4x$$ Finally, we multiply both sides by 3 to obtain our solution: $$\int e^{-x}\sin 4x dx = -\frac{3}{4}e^{-x}\cos 4x - \frac{3}{16}e^{-x}\sin 4x + C$$

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Key Concepts

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

Indefinite Integrals
Indefinite integrals, also known as antiderivatives, are functions that represent the collection of all possible functions whose derivative is the given function. Unlike definite integrals, which find the area under a curve, indefinite integrals yield a family of functions because they include a constant of integration, denoted as \(C\). In the context of integration by parts, indefinite integrals are often used to solve more complex problems by breaking them into simpler parts. Integration by parts is especially useful when the integrand is a product of two functions, as it allows us to integrate each part separately and then combine the results to find the overall integral.
Integration Techniques
Integration by parts is an essential technique used to solve integrals involving products of functions, like the one in our original exercise. This method is based on the product rule for differentiation and is expressed in the formula: \[ \int u \ dv = uv - \int v \ du \]One chooses parts \(u\) and \(dv\) strategically. Here, we've selected \(u = e^{-x}\) and \(dv = \sin 4x \, dx\). The integration by parts technique allows us to transform the original integral into two smaller problems:- The first term \(uv\), which is straightforward to compute.- A new integral \(\int v \, du\) that may be easier to evaluate or requires further application of integration by parts.This cascading approach helps break down complex integrals step by step. Practicing integration by parts solidifies understanding of the interplay between differentiation and integration, key components of calculus.
Trigonometric Functions
Trigonometric functions such as \(\sin\) and \(\cos\) play a pivotal role in many calculus problems. These functions have periodic properties that often lead to repetitive patterns in integration, making them interesting yet challenging.When dealing with integrals involving trigonometric functions, it's important to remember useful results of their derivatives and antiderivatives:
  • Derivative of \(\sin x\) is \(\cos x\).
  • Integral of \(\sin x\) is \(-\cos x + C\).
  • Derivative of \(\cos x\) is \(-\sin x\).
  • Integral of \(\cos x\) is \(\sin x + C\).
In our exercise, we encountered \(\sin 4x\) and \(\cos 4x\). The coefficient of the angle inside the sine and cosine functions affects the results of differentiation and integration by a factor. It is critical to account for these when computing integrals, as seen where integrals of \(\sin 4x\) and \(\cos 4x\) lead to results multiplied by \(\frac{1}{4}\). Understanding these basics facilitates more complex calculations, allowing mastery of deeper calculus concepts.

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Most popular questions from this chapter

Evaluate the following integrals. $$\int_{1}^{\sqrt[3]{2}} y^{8} e^{y^{3}} d y$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational function using the substitution \(u=\tan (x / 2)\) or, equivalently, \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u .\) Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \quad \text { and } \quad \cos x=2 \cos ^{2} \frac{x}{2}-1$$.

For a real number \(a\), suppose \(\lim _{x \rightarrow a^{+}} f(x)=-\infty\) or \(\lim _{x \rightarrow a^{+}} f(x)=\infty .\) In these cases, the integral \(\int_{a}^{\infty} f(x) d x\) is improper for two reasons: \(\infty\) appears in the upper limit and \(f\) is unbounded at \(x=a .\) It can be shown that \(\int_{a}^{\infty} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{\infty} f(x) d x\) for any \(c>a .\) Use this result to evaluate the following improper integrals. $$\int_{1}^{\infty} \frac{d x}{x \sqrt{x-1}}$$

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is a positive integer. \(\int \frac{x}{\sqrt{a x+b}} d x \,\left(\text { Hint: } u^{2}=a x+b .\right)\)

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}(\) coulombs ) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}\).
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