Difference between revisions of "009B Sample Midterm 3, Problem 4"

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<span class="exam">Evaluate the integral:
 
<span class="exam">Evaluate the integral:
  
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::<math>\int \sin (\ln x)~dx</math>
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::<math>\int \sin (\ln x)~dx.</math>
  
  
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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|-
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|Integration by parts tells us that <math>\int u~dv=uv-\int vdu</math>.
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|Integration by parts tells us that <math>\int u~dv=uv-\int vdu.</math>
 
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|-
 
|How could we break up <math>\sin (\ln x)~dx</math> into <math>u</math> and <math>dv?</math>
 
|How could we break up <math>\sin (\ln x)~dx</math> into <math>u</math> and <math>dv?</math>
 
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|-
 
|
 
|
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::Notice that <math>\sin (\ln x)</math> is one term. So, we need to let <math>u=\sin (\ln x)</math> and <math>dv=dx</math>.
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::Notice that <math>\sin (\ln x)</math> is one term. So, we need to let <math>u=\sin (\ln x)</math> and <math>dv=dx.</math>
 
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|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|-
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|We proceed using integration by parts. Let <math>u=\sin(\ln x)</math> and <math>dv=dx</math>. Then, <math>du=\cos(\ln x)\frac{1}{x}dx</math> and <math>v=x</math>.
+
|We proceed using integration by parts. Let <math>u=\sin(\ln x)</math> and <math>dv=dx.</math> Then, <math>du=\cos(\ln x)\frac{1}{x}dx</math> and <math>v=x.</math>
 
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|-
 
|Therefore, we get  
 
|Therefore, we get  
 
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|-
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|<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx</math>.
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|
 +
::<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx.</math>
 
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|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Now, we need to use integration by parts again. Let <math>u=\cos(\ln x)</math> and <math>dv=dx</math>. Then, <math>du=-\sin(\ln x)\frac{1}{x}dx</math> and <math>v=x</math>.
+
|Now, we need to use integration by parts again. Let <math>u=\cos(\ln x)</math> and <math>dv=dx.</math> Then, <math>du=-\sin(\ln x)\frac{1}{x}dx</math> and <math>v=x.</math>  
 
|-
 
|-
 
|Therfore, we get  
 
|Therfore, we get  
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|-
 
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|<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\bigg(x\cos(\ln x)+\int \sin(\ln x)~dx\bigg)=x\sin(\ln x)-x\cos(\ln x)-\int \sin(\ln x)~dx</math>.
 
 
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|-
 
|
 
|
 +
::<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\bigg(x\cos(\ln x)+\int \sin(\ln x)~dx\bigg)=x\sin(\ln x)-x\cos(\ln x)-\int \sin(\ln x)~dx.</math>
 
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|}
  
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|So, if we add the integral on the right to the other side of the equation, we get
 
|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 \sin(\ln x)~dx=x\sin(\ln x)-x\cos(\ln x)</math>.
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|
 +
::<math>2\int \sin(\ln x)~dx=x\sin(\ln x)-x\cos(\ln x)</math>.
 
|-
 
|-
 
|Now, we divide both sides by 2 to get  
 
|Now, we divide both sides by 2 to get  
 
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|-
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|<math>\int \sin(\ln x)~dx=\frac{x\sin(\ln x)}{2}-\frac{x\cos(\ln x)}{2}</math>.
+
|
 +
::<math>\int \sin(\ln x)~dx=\frac{x\sin(\ln x)}{2}-\frac{x\cos(\ln x)}{2}.</math>
 
|-
 
|-
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|Thus, the final answer is <math>\int \sin(\ln x)~dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
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|Thus, the final answer is <math>\int \sin(\ln x)~dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C.</math>
 
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|}
  
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|-
 
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|<math>\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
 
|<math>\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
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[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 15:41, 29 March 2016

Evaluate the integral:


Foundations:  
Integration by parts tells us that
How could we break up 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 \sin (\ln x)~dx} into 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} 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?}
Notice that 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 \sin (\ln x)} is one term. So, we need to 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 (\ln 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=dx.}

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(\ln 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=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(\ln x)\frac{1}{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=x.}
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 \sin (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx.}
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(\ln 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=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(\ln x)\frac{1}{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=x.}
Therfore, 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 \sin (\ln x)~dx=x\sin(\ln x)-\bigg(x\cos(\ln x)+\int \sin(\ln x)~dx\bigg)=x\sin(\ln x)-x\cos(\ln x)-\int \sin(\ln x)~dx.}
Step 3:  
Notice that the integral on the right of the last equation is the same integral that we had at the beginning.
So, 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 \sin(\ln x)~dx=x\sin(\ln x)-x\cos(\ln x)} .
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 \sin(\ln x)~dx=\frac{x\sin(\ln x)}{2}-\frac{x\cos(\ln x)}{2}.}
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 \sin(\ln x)~dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+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{x}{2}(\sin(\ln x)-\cos(\ln x))+C}

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