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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 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |Integration by parts tells us <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
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| − | |How would you integrate <math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
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| − | ::You could use integration by parts.
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| − | ::Let <math style="vertical-align: -5px">u=\sin(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math> Then, <math style="vertical-align: -5px">du=\cos(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
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| − | ::Thus, <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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| − | ::Now, we need to use integration by parts a second time.
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| − | ::Let <math style="vertical-align: -5px">u=\cos(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math> Then, <math style="vertical-align: -5px">du=-\sin(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math> So,
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| − | ::<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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| − | |
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| − | ::Notice, we are back where we started. So, adding the last term on the right hand side to the opposite side,
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| − | ::we get <math style="vertical-align: -13px">2\int e^x\sin (x)~dx\,=\,e^x(\sin(x)-\cos(x)).</math>
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| − | ::Hence, <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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| − | |}
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| − | '''Solution:'''
| + | [[009B Sample Midterm 2, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We proceed using integration by parts. Let <math style="vertical-align: -5px">u=\sin(2x)</math> and <math style="vertical-align: 0px">dv=e^{-2x}dx</math>. Then, <math style="vertical-align: -5px">du=2\cos(2x)dx</math> and <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}</math>.
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| − | |So, we get
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| − | | <math style="vertical-align: -14px">\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}-\int \frac{e^{-2x}2\cos(2x)~dx}{-2}=\frac{\sin(2x)e^{-2x}}{-2}+\int e^{-2x}\cos(2x)~dx</math>.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we need to use integration by parts again. Let <math style="vertical-align: -5px">u=\cos(2x)</math> and <math style="vertical-align: 0px">dv=e^{-2x}dx</math>. Then, <math style="vertical-align: -5px">du=-2\sin(2x)dx</math> and <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}</math>.
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| − | |So, we get
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| − | | <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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |Notice that the integral on the right of the last equation in Step 2 is the same integral that we had at the beginning of the problem.
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| − | |So, if we add the integral on the right to the other side of the equation, we get
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| − | | <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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| − | | <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 <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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C</math>
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| − | |}
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| | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
