Difference between revisions of "009B Sample Midterm 1, Problem 4"

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!Foundations:    
 
!Foundations:    
 
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|Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
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|'''1.''' Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
 
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|How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
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|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
 
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|-
 
|
 
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::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\sin x.</math> Then, <math style="vertical-align: -1px">du=\cos 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=\sin x.</math> Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> Thus,
 
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::Thus, <math style="vertical-align: -14px">\int \sin^2x\cos x~dx\,=\,\int u^2~du\,=\,\frac{u^3}{3}+C\,=\,\frac{\sin^3x}{3}+C.</math>
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::<math>\begin{array}{rcl}
 +
\displaystyle{\int \sin^2x\cos x~dx} & = & \displaystyle{\int u^2~du}\\
 +
&&\\
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& = & \displaystyle{\frac{u^3}{3}+C}\\
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&&\\
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& = & \displaystyle{\frac{\sin^3x}{3}+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\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx</math>.
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|First, we write  
 
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|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>. If we use this identity, we have
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|
 +
::<math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
 
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| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) (1-\cos^2x)\cos^2x~dx=\int (\cos^2x-\cos^4x)\sin(x)~dx</math>.
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|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1,</math> we get
 
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|-
 
|
 
|
 +
::<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
 +
|-
 +
|If we use this identity, we have
 +
|-
 +
|
 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int (\sin x) (1-\cos^2x)\cos^2x~dx}\\
 +
&&\\
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& = & \displaystyle{\int (\cos^2x-\cos^4x)\sin(x)~dx.}\\
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\end{array}</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\cos(x)</math>. Then, <math style="vertical-align: -5px">du=-\sin(x)dx</math>. Therefore,  
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|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> Therefore,  
 
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| &nbsp;&nbsp; <math style="vertical-align: -14px">\int\sin^3x\cos^2x~dx=\int -(u^2-u^4)~du=\frac{-u^3}{3}+\frac{u^5}{5}+C=\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>.
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|  
 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\
 +
&&\\
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& = & \displaystyle{\frac{-u^3}{3}+\frac{u^5}{5}+C}\\
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&&\\
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& = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C.}\\
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\end{array}</math>
 
|}
 
|}
  

Revision as of 14:15, 18 April 2016

Evaluate the integral:


Foundations:  
1. Recall the trig identity:
2. How would you integrate
You could use -substitution. Let Then, Thus,

Solution:

Step 1:  
First, we write
Using the identity we get
If we use this identity, we have
Step 2:  
Now, we use -substitution. Let Then, Therefore,
Final Answer:  
  

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