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

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!Foundations:    
 
!Foundations:    
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|Trig identity
 
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|1)
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|U substitution
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'''Solution:'''
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!Step 1:  
 
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|2)  
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|First, we write <math>\int \tan^4(x)dx=\int \tan^2(x) \tan^2(x)dx</math>.
 
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|Using the trig identity <math>\sec^2(x)=\tan^2(x)+1</math>, we have <math>\tan^2(x)=\sec^2(x)-1</math>.
 
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|Answers:
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|Plugging in the last identity into one of the <math>\tan^2(x)</math>, we get
 
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|1)
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|<math>\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</math>
 
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|2)
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|using the identity again on the last equality
 
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'''Solution:'''
 
 
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!Step 1: &nbsp;  
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!Step 2: &nbsp;
 
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|So, we have <math>\int \tan^4(x)dx=\int \tan^2(x)\sec^2(x)dx-\int (\sec^2x-1)dx</math>.
 
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|For the first integral, we need to use substitution. Let <math>u=\tan(x)</math>. Then, <math>du=\sec^2(x)dx</math>.
 
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|
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|So, we have
 
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|<math>\int \tan^4(x)dx=\int u^2du-\int (\sec^2(x)-1)dx</math>.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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!Step 3: &nbsp;
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|We integrate to get
 
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| <math>\int \tan^4(x)dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|<math>\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>
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[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 11:33, 27 January 2016

Evaluate the integral:


Foundations:  
Trig identity
U substitution

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 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^2du-\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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