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

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(Created page with "<span class="exam">Evaluate the integral: ::<math>\int \sin^3x \cos^2x~dx</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations:   |-...")
 
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| Review <math style="vertical-align: 0px">u</math>-substitution, and
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|Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
 
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|Trig identities.
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|How would you integrate <math style="vertical-align: -14px">\int \sin^2x\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>
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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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Revision as of 15:08, 8 April 2016

Evaluate the integral:


Foundations:  
Recall the trig identity:
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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