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:
  
−
::<math>\int \sin (\ln x)dx</math>
+
::<math>\int \sin (\ln x)~dx</math>
  
  
Line 18: Line 18:
 
|Therefore, we get  
 
|Therefore, we get  
 
|-
 
|-
−
|<math>\int \sin (\ln x)dx=x\sin(\ln x)-\int \cos(\ln x)dx</math>.
+
|<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx</math>.
 
|}
 
|}
  
Line 28: Line 28:
 
|Therfore, we get  
 
|Therfore, we get  
 
|-
 
|-
−
|<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>.
+
|<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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|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
 
|-
 
|-
−
|<math>2\int \sin(\ln x)dx=x\sin(\ln x)-x\cos(\ln x)</math>.
+
|<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  
 
|-
 
|-
−
|<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>.
 
|-
 
|-
−
|Thus, the final answer is <math>\int \sin(\ln x)dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
+
|Thus, the final answer is <math>\int \sin(\ln x)~dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
 
|}
 
|}
  

Revision as of 14:32, 31 January 2016

Evaluate the integral:


Foundations:  
Review integration by parts

Solution:

Step 1:  
We proceed using integration by parts. Let 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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