Difference between revisions of "009B Sample Midterm 2, Problem 5"

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(Created page with "<span class="exam"> Evaluate the integral: ::<math>\int \tan^4 x ~dx</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations:   |- |Re...")
 
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!Foundations: &nbsp;  
 
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|Review <math>u</math>-substitution and
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|Recall:
 
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|trig identities
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|'''1.''' <math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
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|'''2.''' <math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>
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|How would you integrate <math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>
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::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\tan x.</math> Then, <math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
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::Thus, <math style="vertical-align: -15px">\int \sec^2(x)\tan(x)~dx\,=\,\int u~du\,=\,\frac{u^2}{2}+C\,=\,\frac{\tan^2x}{2}+C.</math>
 
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'''Solution:'''
 
'''Solution:'''

Revision as of 14:20, 8 April 2016

Evaluate the integral:


Foundations:  
Recall:
1. 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 \sec^2x=\tan^2x+1}
2. 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 \sec^2 x~dx=\tan x+C}
How would you integrate 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 \sec^2(x)\tan(x)~dx?}
You could use 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} -substitution. 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=\tan x.} 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=\sec^2(x)dx.}
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 \sec^2(x)\tan(x)~dx\,=\,\int u~du\,=\,\frac{u^2}{2}+C\,=\,\frac{\tan^2x}{2}+C.}


Solution:

Step 1:  
First, we write 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 \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx} .
Using the trig identity 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 \sec^2(x)=\tan^2(x)+1} , we have 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 \tan^2(x)=\sec^2(x)-1} .
Plugging in the last identity into one of the 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 \tan^2(x)} , 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 \tan^4(x)~dx=\int \tan^2(x) (\sec^2(x)-1)~dx=\int \tan^2(x)\sec^2(x)~dx-\int \tan^2(x)~dx=\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx} ,
using the identity again on the last equality.
Step 2:  
So, we have 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 \tan^4(x)~dx=\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx} .
For the first integral, we need to use 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} -substitution. 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=\tan(x)} . 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=\sec^2(x)dx} .
So, we have
   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 \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx} .
Step 3:  
We integrate 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 \tan^4(x)~dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+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{\tan^3(x)}{3}-\tan(x)+x+C}

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