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

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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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::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> Thus,
 
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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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::<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>
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|First, we write <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx</math>.
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|First, we write  
 
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|Using the trig identity <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1</math>, we have <math style="vertical-align: -5px">\tan^2(x)=\sec^2(x)-1</math>.
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::<math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx.</math>
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|Using the trig identity <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math> we have  
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::<math style="vertical-align: -5px">\tan^2(x)=\sec^2(x)-1.</math>
 
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|Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x)</math>, we get
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|Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x),</math> we get
 
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| &nbsp;&nbsp; <math style="vertical-align: -13px">\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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 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\int \tan^2(x) (\sec^2(x)-1)~dx}\\
 +
&&\\
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& = & \displaystyle{\int \tan^2(x)\sec^2(x)~dx-\int \tan^2(x)~dx}\\
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&&\\
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& = & \displaystyle{\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx.}\\
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\end{array}</math>
 
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|using the identity again on the last equality.
 
|using the identity again on the last equality.
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|So, we have <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx</math>.
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|So, we have  
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|-
 +
|
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::<math style="vertical-align: -13px">\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 <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\tan(x)</math>. Then, <math style="vertical-align: -5px">du=\sec^2(x)dx</math>.
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|For the first integral, we need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\tan(x).</math> Then, <math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
 
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|-
 
|So, we have
 
|So, we have
 
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| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx</math>.
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|
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::<math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx.</math>
 
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|We integrate to get  
 
|We integrate to get  
 
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| &nbsp;&nbsp; <math style="vertical-align: -13px">\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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|  
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::<math>\begin{array}{rcl}
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\displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\frac{u^3}{3}-(\tan(x)-x)+C}\\
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&&\\
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& = & \displaystyle{\frac{\tan^3(x)}{3}-\tan(x)+x+C.}\\
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\end{array}</math>
 
|}
 
|}
  

Revision as of 14:03, 18 April 2016

Evaluate the integral:


Foundations:  
Recall:
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 \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}}
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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