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Chapter 6: Problem 24

Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int e^{x} \cos 2 x d x $$

Short Answer

Expert verified
\[ \int e^{x} \cos 2x dx = e^{x}/5 (\cos 2x + 2\sin 2x ) + C \]

Step by step solution

01

Identify 'u' and 'dv'

First, assign \(u\) and \(dv\) from the given integral. It is preferential to choose \(e^{x}\) as \(dv\) and \(\cos(2x)\) as \(u\) due to the ease in integrating \(e^{x}\). So you have \(u = \cos(2x)\) and \(dv = e^{x} dx\).
02

Find 'du' and 'v'

Next, find \(du\) by differentiating \(u\) and find \(v\) by integrating \(dv\). This gives \(du = -2\sin(2x) dx\) and \(v = e^{x}\).
03

Apply Integration by Parts

Now apply the formula for Integration by Parts \[ \int u dv = uv -\int v du \] which gives: \[ \int e^{x} \cos 2x dx = e^{x} \cos 2x +2 \int e^{x} \sin 2x dx\]. However, there is another integral to solve here.
04

Use Integration by Parts Again

We need to use Integration by Parts again for the integral now inside of the expression. This time, choose \(\sin(2x)\) as \(u\), then \(dv\) stays the same. This results in \(du = 2\cos(2x) dx\). Substituting into our main equation again, leads to expansion without new integral: \[ \int e^{x} \cos 2x dx = e^{x} \cos 2x +2e^{x} \sin 2x - 4 \int e^{x} \cos 2x dx\].
05

Solve for the Integral

Now you see that the integral we started with has appeared again on the right hand side. This calls for some algebraic manipulation: Add \(4 \int e^{x} \cos 2x dx\) to both sides, you will end up with \(5 \int e^{x} \cos 2x dx = e^{x} \cos 2x +2e^{x} \sin 2x\). Therefore, our original integral is \( \int e^{x} \cos 2x dx = e^{x}/5 (\cos 2x + 2\sin 2x ) + C\), where \(C\) is the constant of integration.

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