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

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|'''1.''' Integration by parts tells us  
 
|'''1.''' Integration by parts tells us  
 
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|-
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|&nbsp; &nbsp; <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
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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 <math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
 
|'''2.''' How would you integrate <math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
 
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|
 
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&nbsp; &nbsp; You could use integration by parts.
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&nbsp; &nbsp; &nbsp; &nbsp; You could use integration by parts.
 
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|-
 
|
 
|
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&nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\sin(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>  
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&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\sin(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>  
 
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|-
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|&nbsp; &nbsp; 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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|&nbsp; &nbsp; &nbsp; &nbsp; 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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&nbsp; &nbsp; 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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&nbsp; &nbsp; &nbsp; &nbsp; 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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|-
 
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&nbsp; &nbsp; Now, we need to use integration by parts a second time.
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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; 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 <math style="vertical-align: -5px">u=\cos(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -5px">du=-\sin(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Therefore,
 
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|-
 
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&nbsp; &nbsp; <math>\begin{array}{rcl}
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&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}\\
 
\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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&nbsp; &nbsp; 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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&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
 
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|-
 
|
 
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&nbsp; &nbsp; 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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&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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|-
 
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&nbsp; &nbsp; 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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&nbsp; &nbsp; &nbsp; &nbsp; 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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|Then, <math style="vertical-align: -5px">du=2\cos(2x)dx</math> and <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</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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|-
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|So, we get  
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|Thus, we get  
 
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|
 
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|-
 
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&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>
+
&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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|-
 
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|Thus, 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
 
|-
 
|-
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| &nbsp;&nbsp; <math>2\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}.</math>
+
| &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  
 
|Now, we divide both sides by 2 to get  
 
|-
 
|-
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| &nbsp;&nbsp; <math>\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-4}+\frac{\cos(2x)e^{-2x}}{-4}.</math>
+
| &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  
 
|Thus, the final answer is  
 
|-
 
|-
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|&nbsp; &nbsp; <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C.</math>
+
|&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>
 
|}
 
|}
  

Revision as of 17:54, 7 February 2017

Evaluate the integral:


Foundations:  
1. Integration by parts tells us
       
2. How would you integrate

        You could use integration by parts.

        Let 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 u=\sin(x)} and 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 dv=e^x~dx.}

        Then, 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 du=\cos(x)~dx} and 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 v=e^x.}

        Thus, 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^x\sin x~dx=e^x\sin(x)-\int e^x\cos(x)~dx.}

        Now, we need to use integration by parts a second time.

        Let 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 u=\cos(x)} and 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 dv=e^x~dx.}

        Then, 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 du=-\sin(x)~dx} and 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 v=e^x.}
        Therefore,

        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 \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}}

        Notice, we are back where we started.

        Therefore, adding the last term on the right hand side to the opposite side, we get

        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 2\int e^x\sin (x)~dx=e^x(\sin(x)-\cos(x)).}

        Hence, 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^x\sin (x)~dx=\frac{e^x}{2}(\sin(x)-\cos(x))+C.}


Solution:

Step 1:  
We proceed using integration by parts.
Let 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 u=\sin(2x)} and 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 dv=e^{-2x}dx.}
Then, 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 du=2\cos(2x)dx} and 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 v=\frac{e^{-2x}}{-2}.}
Thus, we get

        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 \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}}

Step 2:  
Now, we need to use integration by parts again.
Let 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 u=\cos(2x)} and 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 dv=e^{-2x}dx.}
Then, 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 du=-2\sin(2x)dx} and 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 v=\frac{e^{-2x}}{-2}.}
Therefore, we get

        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=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}-\int e^{-2x}\sin(2x)~dx.}

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
        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 2\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}.}
Now, we divide both sides by 2 to get
        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=\frac{\sin(2x)e^{-2x}}{-4}+\frac{\cos(2x)e^{-2x}}{-4}.}
Thus, the final answer is
        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=\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C.}


Final Answer:  
       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 \frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C}

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