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

From Grad Wiki
Jump to navigation Jump to search
Line 22: Line 22:
 
::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>
 
::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>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
Line 27: Line 28:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we write <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx</math>.
+
|First, we write <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx.</math>  
 
|-
 
|-
−
|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>.
+
|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>
 
|-
 
|-
−
|Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x)</math>, we get
+
|Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x),</math> we get
 
|-
 
|-
−
| &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>,
+
| &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>
 
|-
 
|-
−
|using the identity again on the last equality.
+
|by using the identity again on the last equality.
 
|}
 
|}
  
Line 41: Line 42:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|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>.
+
|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>
 
|-
 
|-
−
|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>.
+
|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>
 
|-
 
|-
 
|So, we have
 
|So, we have
 
|-
 
|-
−
| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx</math>.
+
| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx.</math>
 
|}
 
|}
  
Line 55: Line 56:
 
|We integrate to get  
 
|We integrate to get  
 
|-
 
|-
−
| &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>.
+
| &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>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

Revision as of 08:35, 6 February 2017

Evaluate the integral:


Foundations:  
Recall:
1.
2.
How would you integrate
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}
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 \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}

Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=009B_Sample_Midterm_2,_Problem_5&oldid=2005"