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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
−
|First, we write <math>\int \tan^4(x)dx=\int \tan^2(x) \tan^2(x)dx</math>.  
+
|First, we write <math>\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx</math>.  
 
|-
 
|-
 
|Using the trig identity <math>\sec^2(x)=\tan^2(x)+1</math>, we have <math>\tan^2(x)=\sec^2(x)-1</math>.  
 
|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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|Plugging in the last identity into one of the <math>\tan^2(x)</math>, we get
 
|Plugging in the last identity into one of the <math>\tan^2(x)</math>, we get
 
|-
 
|-
−
|<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>
+
|<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>
 
|-
 
|-
−
|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;
 
|-
 
|-
−
|So, we have <math>\int \tan^4(x)dx=\int \tan^2(x)\sec^2(x)dx-\int (\sec^2x-1)dx</math>.
+
|So, we have <math>\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 substitution. Let <math>u=\tan(x)</math>. Then, <math>du=\sec^2(x)dx</math>.
 
|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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|So, we have
 
|So, we have
 
|-
 
|-
−
|<math>\int \tan^4(x)dx=\int u^2du-\int (\sec^2(x)-1)dx</math>.
+
|<math>\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  
 
|-
 
|-
−
| <math>\int \tan^4(x)dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>
+
| <math>\int \tan^4(x)~dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>
 
|}
 
|}
  

Revision as of 14:26, 31 January 2016

Evaluate the integral:

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}


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^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}

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