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

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!Foundations:    
 
!Foundations:    
 
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|'''1.''' Recall the trig identity &nbsp;<math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
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|'''1.''' Recall the trig identity  
 
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|'''2.''' Also, &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: -1px">\sec^2x=\tan^2x+1</math>
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|'''2.''' Also,  
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>
 
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|'''3.''' How would you integrate &nbsp;<math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>
 
|'''3.''' How would you integrate &nbsp;<math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
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&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 
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|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=\tan x.</math>  
 
|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=\tan x.</math>  
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&nbsp; &nbsp; &nbsp; &nbsp; Thus, &nbsp;<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,  
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\int \sec^2(x)\tan(x)~dx} & = & \displaystyle{\int u~du}\\
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&&\\
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& = & \displaystyle{\frac{u^2}{2}+C}\\
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&&\\
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& = & \displaystyle{\frac{\tan^2x}{2}+C.}
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\end{array}</math>
 
|}
 
|}
  

Revision as of 12:53, 14 March 2017

Evaluate the integral:


Foundations:  
1. Recall the trig identity
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sec ^{2}x=\tan ^{2}x+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 can 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 \begin{array}{rcl} \displaystyle{\int \sec^2(x)\tan(x)~dx} & = & \displaystyle{\int u~du}\\ &&\\ & = & \displaystyle{\frac{u^2}{2}+C}\\ &&\\ & = & \displaystyle{\frac{\tan^2x}{2}+C.} \end{array}}


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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