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

Latest revision as of 18:10, 12 November 2017

Evaluate the integral:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{-2x}\sin(2x)~dx}

Solution

Detailed Solution

Return to Sample Exam

@@ Line 3: / Line 3: @@
 ::<math>\int e^{-2x}\sin (2x)~dx</math>
+<hr>
+[[009B Sample Midterm 2, Problem 4 Solution|'''<u>Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
-|-
-|'''1.''' Integration by parts tells us <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
-|-
-|'''2.''' How would you integrate <math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
-|-
-|
-::You could use integration by parts.
-|-
-|
-::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>
-|-
-|
-::Thus, <math style="vertical-align: -15px">\int e^x\sin x~dx\,=\,e^x\sin(x)-\int e^x\cos(x)~dx.</math>
-|-
-|
-::Now, we need to use integration by parts a second time.
-|-
-|
-::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,
-|-
-|
-::<math>\begin{array}{rcl}
-\displaystyle{\int e^x\sin x~dx} & = & \displaystyle{e^x\sin(x)-(e^x\cos(x)-\int -e^x\sin(x)~dx}\\
-&&\\
-& = & \displaystyle{e^x(\sin(x)-\cos(x))-\int e^x\sin(x)~dx.}\\
-\end{array}</math>
-|-
-|
-::Notice, we are back where we started.
-|-
-|
-::So, adding the last term on the right hand side to the opposite side, we get
-|-
-|
-::<math style="vertical-align: -13px">2\int e^x\sin (x)~dx\,=\,e^x(\sin(x)-\cos(x)).</math>
-|-
-|
-::Hence, <math style="vertical-align: -15px">\int e^x\sin (x)~dx\,=\,\frac{e^x}{2}(\sin(x)-\cos(x))+C.</math>
-|}
-'''Solution:'''
+[[009B Sample Midterm 2, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|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>
-|-
-|So, we get
-|-
-|
-::<math>\begin{array}{rcl}
-\displaystyle{\int e^{-2x}\sin (2x)~dx} & = & \displaystyle{\frac{\sin(2x)e^{-2x}}{-2}-\int \frac{e^{-2x}2\cos(2x)~dx}{-2}}\\
-&&\\
-& = & \displaystyle{\frac{\sin(2x)e^{-2x}}{-2}+\int e^{-2x}\cos(2x)~dx.}\\
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|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>
-|-
-|So, we get
-|-
-|
-::<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>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|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.
-|-
-|So, if we add the integral on the right to the other side of the equation, we get
-|-
-|
-::<math>2\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}.</math>
-|-
-|Now, we divide both sides by 2 to get
-|-
-|
-::<math>\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-4}+\frac{\cos(2x)e^{-2x}}{-4}.</math>
-|-
-|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>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-| &nbsp;&nbsp; <math>\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C</math>
-|}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

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

Latest revision as of 18:10, 12 November 2017

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