009B Sample Midterm 2, Problem 4 - Revision history

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Evaluate the integral:

::<math>\int e^{-2x}\sin (2x)~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Review integration by parts
|}

'''Solution:'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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 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>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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:  
|-
|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:  
|-
|    <math>\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C</math>
|}
[[009B_Sample_Midterm_2|'''Return to Sample Exam''']]

← Older revision		Revision as of 01:10, 13 November 2017
Line 3:		Line 3:
	::<math>\int e^{-2x}\sin (2x)~dx</math>		::<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:  ~~
−	\|-
−	~~\|'''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 can 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>~~
−	\|-
−	~~\|        Therefore,~~
−	\|-
−	\|
−	~~        <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.~~
−	\|-
−	~~\|        Therefore, 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>~~
−	\|}

		+	[[009B Sample Midterm 2, Problem 4 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|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>~~
−	\|-
−	~~\|Thus, 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:  ~~
−	\|-
−	~~\|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>~~
−	\|-
−	~~\|Therefore, 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:  ~~
−	\|-
−	~~\|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.~~
−	\|-
−	~~\|Thus, 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:  ~~
−	\|-
−	~~\|        <math>\frac{e^{-2x}}{-4}(\sin(2x)+\cos(2x))+C</math>~~
−	\|}
	[[009B_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]		[[009B_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Integration by parts tells us <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
+|'''1.''' Integration by parts tells us
 |-
-|'''2.''' How would you integrate <math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
 |-
+|
-::You could use integration by parts.
+&nbsp; &nbsp; &nbsp; &nbsp; You can 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>
+&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>
 |-
+|
-::Thus, <math style="vertical-align: -15px">\int e^x\sin x~dx\,=\,e^x\sin(x)-\int e^x\cos(x)~dx.</math>
+&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>
 |-
+|
-::Now, we need to use integration by parts a second time.
+&nbsp; &nbsp; &nbsp; &nbsp; 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,
+&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>
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <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}\\
 &&\\
@@ Line 34: / Line 42: @@
 |-
+|
-::Notice, we are back where we started.
+&nbsp; &nbsp; &nbsp; &nbsp; Notice, we are back where we started.
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; Therefore, 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>
+&nbsp; &nbsp; &nbsp; &nbsp; <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>
+&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>
 |}
 '''Solution:'''
@@ Line 50: / Line 58: @@
 !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>
+|We proceed using integration by parts.
 |-
-|So, we get
+|Thus, we get
 |-
-−
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <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.}\\
+& = & \displaystyle{\frac{\sin(2x)e^{-2x}}{-2}+\int e^{-2x}\cos(2x)~dx.}
 \end{array}</math>
 |}
@@ Line 65: / Line 77: @@
 !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>
+|Now, we need to use integration by parts again.
 |-
-|So, we get
+|Therefore, 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>.
+&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>
 |}
@@ Line 76: / Line 92: @@
 !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.
+|Notice that the integral on the right of the last equation in Step 2
 |-
-|So, if we add the integral on the right to the other side of the equation, we get
+|Thus, if we add the integral on the right to the other side of the equation, we get
 |-
+| &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>
 |-
 |Now, we divide both sides by 2 to get
 |-
+| &nbsp; &nbsp; &nbsp; &nbsp; <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
 |-
+|&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>
 |}
 {| 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>
+| &nbsp;&nbsp; &nbsp; &nbsp; <math>\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|Review integration by parts
+|Integration by parts tells us <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
 |}