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

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations:    
 
!Foundations:    
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|-
 
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|Recall:
 
 
|-
 
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|'''1.''' Recall the trig identity
 
|'''1.''' Recall the trig identity
 
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|-
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|&nbsp; &nbsp; <math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
 
|-
 
|-
 
|'''2.''' Also,  
 
|'''2.''' Also,  
 
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|-
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|&nbsp; &nbsp; <math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>
 
|-
 
|-
 
|'''3.''' How would you integrate <math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>
 
|'''3.''' How would you integrate <math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>
 
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&nbsp; &nbsp; 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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&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.  
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -2px">u=\tan x.</math>  
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
 
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|-
 
|
 
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&nbsp; &nbsp; 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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&nbsp; &nbsp; &nbsp; &nbsp; 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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|}
  

Revision as of 16:30, 7 February 2017

Evaluate the integral:


Foundations:  
1. Recall 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^2x=\tan^2x+1}
2. Also,
        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}
3. 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 \begin{array}{rcl} \displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\int \tan^2(x) (\sec^2(x)-1)~dx}\\ &&\\ & = & \displaystyle{\int \tan^2(x)\sec^2(x)~dx-\int \tan^2(x)~dx}\\ &&\\ & = & \displaystyle{\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx} \end{array}}

by 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 \begin{array}{rcl} \displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\frac{u^3}{3}-(\tan(x)-x)+C}\\ &&\\ & = & \displaystyle{\frac{\tan^3(x)}{3}-\tan(x)+x+C.} \end{array}}


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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