Difference between revisions of "009B Sample Midterm 2, Problem 4"

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(Replaced content with "<span class="exam"> Evaluate the integral: ::<math>\int e^{-2x}\sin (2x)~dx</math> <hr> (insert picture of handwritten solution) 009B Sample Midterm 2, Problem 4 Deta...")
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::<math>\int e^{-2x}\sin (2x)~dx</math>
 
::<math>\int e^{-2x}\sin (2x)~dx</math>
  
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[[009B Sample Midterm 2, Problem 4 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
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!Foundations: &nbsp;
 
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|'''1.''' Integration by parts tells us
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
 
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|'''2.''' How would you integrate &nbsp;<math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; You can use integration by parts.
 
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&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\sin(x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^x~dx.</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=\cos(x)~dx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; Thus, &nbsp;<math style="vertical-align: -15px">\int e^x\sin x~dx=e^x\sin(x)-\int e^x\cos(x)~dx.</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; Now, we need to use integration by parts a second time.
 
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&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\cos(x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^x~dx.</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=-\sin(x)~dx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Therefore,
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int e^x\sin x~dx} & = & \displaystyle{e^x\sin(x)-(e^x\cos(x)-\int -e^x\sin(x)~dx}\\
 
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&&\\
 
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& = & \displaystyle{e^x(\sin(x)-\cos(x))-\int e^x\sin(x)~dx.}\\
 
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\end{array}</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; Notice, we are back where we started.
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Therefore, adding the last term on the right hand side to the opposite side, we get
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">2\int e^x\sin (x)~dx=e^x(\sin(x)-\cos(x)).</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; Hence, &nbsp;<math style="vertical-align: -15px">\int e^x\sin (x)~dx=\frac{e^x}{2}(\sin(x)-\cos(x))+C.</math>
 
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'''Solution:'''
 
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!Step 1: &nbsp;
 
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|We proceed using integration by parts.
 
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|Let &nbsp;<math style="vertical-align: -5px">u=\sin(2x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-2x}dx.</math>
 
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|Then, &nbsp;<math style="vertical-align: -5px">du=2\cos(2x)dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
 
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|Thus, we get
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int e^{-2x}\sin (2x)~dx} & = & \displaystyle{\frac{\sin(2x)e^{-2x}}{-2}-\int \frac{e^{-2x}2\cos(2x)~dx}{-2}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{\sin(2x)e^{-2x}}{-2}+\int e^{-2x}\cos(2x)~dx.}
 
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\end{array}</math>
 
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!Step 2: &nbsp;
 
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|Now, we need to use integration by parts again.
 
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|Let &nbsp;<math style="vertical-align: -5px">u=\cos(2x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-2x}dx.</math>
 
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|Then, &nbsp;<math style="vertical-align: -5px">du=-2\sin(2x)dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
 
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|Therefore, we get
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}-\int e^{-2x}\sin(2x)~dx.</math>
 
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!Step 3: &nbsp;
 
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|Notice that the integral on the right of the last equation in Step 2
 
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|is the same integral that we had at the beginning of the problem.
 
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|Thus, if we add the integral on the right to the other side of the equation, we get
 
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| &nbsp; &nbsp; &nbsp; &nbsp; <math>2\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}.</math>
 
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|Now, we divide both sides by 2 to get
 
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| &nbsp; &nbsp; &nbsp; &nbsp; <math>\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-4}+\frac{\cos(2x)e^{-2x}}{-4}.</math>
 
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|Thus, the final answer is
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C.</math>
 
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!Final Answer: &nbsp;
 
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| &nbsp;&nbsp; &nbsp; &nbsp; <math>\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C</math>
 
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[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:43, 4 November 2017

Evaluate the integral:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^{-2x}\sin (2x)~dx}

(insert picture of handwritten solution)

Detailed Solution for this Problem

Return to Sample Exam

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