/*! 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 24 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 24

Evaluate the following integrals. $$\int e^{3 x} \cos 2 x d x$$

Short Answer

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Question: Evaluate the integral \(\int e^{3x}\cos(2x)dx\). Answer: \(\int e^{3x}\cos(2x)dx = -\frac{2}{5}e^{3x}\sin(2x) - \frac{3}{5}e^{3x}\cos(2x) + C\)

Step by step solution

01

Apply integration by parts for the first time.

Integration by parts states that \(\int u dv = uv - \int v du\). We need to choose our two functions, u(x) and dv(x). To make it easier to integrate or differentiate, let: \(u = e^{3x}\), therefore \(du = 3e^{3x} dx\) \(dv = \cos(2x) dx\), therefore \(v = \frac{1}{2} \sin(2x)\) Now, we can apply the integration by parts formula: \(\int e^{3x}\cos(2x)dx = e^{3x}\frac{1}{2}\sin(2x) - \int \frac{1}{2}\sin(2x)(3e^{3x})dx\)
02

Apply integration by parts again.

In the new integral, again, we define u(x) and dv(x) as: \(u = 3e^{3x}\), therefore \(du = 9e^{3x} dx\) \(dv = \frac{1}{2}\sin(2x) dx\), therefore \(v = -\frac{1}{4}\cos(2x)\) Now, apply the integration by parts formula for the second integral: \(- \int \frac{1}{2}\sin(2x)(3e^{3x})dx = -3e^{3x}\left(-\frac{1}{4}\cos(2x)\right) - \int -\frac{1}{4}\cos(2x)(9e^{3x})dx\)
03

Simplify and rearrange the equation.

Simplify the equation by combining the constant terms and rearrange it to isolate the original integral: \(\int e^{3x}\cos(2x)dx = \frac{1}{2}e^{3x}\sin(2x) + \frac{3}{4}e^{3x}\cos(2x) + \frac{9}{4}\int e^{3x}\cos(2x)dx\) Now, we have the same integral on both sides of the equation. \(-\frac{5}{4}\int e^{3x}\cos(2x)dx = \frac{1}{2}e^{3x}\sin(2x) + \frac{3}{4}e^{3x}\cos(2x)\) Now we can solve for the original integral: \(\int e^{3x}\cos(2x)dx = -\frac{2}{5}e^{3x}\sin(2x) - \frac{3}{5}e^{3x}\cos(2x) + C\) So the final answer is: \(\int e^{3x}\cos(2x)dx = -\frac{2}{5}e^{3x}\sin(2x) - \frac{3}{5}e^{3x}\cos(2x) + C\)

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Most popular questions from this chapter

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty).\) a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\) b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\) c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)

Use numerical methods or a calculator to approximate the following integrals as closely as possible. $$\int_{0}^{\pi / 2} \ln (\sin x) d x=\int_{0}^{\pi / 2} \ln (\cos x) d x=-\frac{\pi \ln 2}{2}$$

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}.\) a. Find the work required to launch an object in terms of \(m.\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

Challenge Show that with the change of variables \(u=\sqrt{\tan x}\) the integral \(\int \sqrt{\tan x} d x\) can be converted to an integral amenabl to partial fractions. Evaluate \(\int_{0}^{\pi / 4} \sqrt{\tan x} d x\)

Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 2 and an illustration.
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