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

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
|'''1.''' Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
+
|'''1.''' Recall the trig identity
 
|-
 
|-
|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
+
| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>
 +
|-
 +
|'''2.''' How would you integrate &nbsp;<math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
 
|-
 
|-
 
|
 
|
::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,  
+
&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=\sin x.</math>  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=\cos x~dx.</math>  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Thus,  
 
|-
 
|-
 
|
 
|
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int \sin^2x\cos x~dx} & = & \displaystyle{\int u^2~du}\\
 
\displaystyle{\int \sin^2x\cos x~dx} & = & \displaystyle{\int u^2~du}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{\frac{u^3}{3}+C}\\
 
& = & \displaystyle{\frac{u^3}{3}+C}\\
 
&&\\
 
&&\\
& = & \displaystyle{\frac{\sin^3x}{3}+C.} \\
+
& = & \displaystyle{\frac{\sin^3x}{3}+C.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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|First, we write  
 
|First, we write  
 
|-
 
|-
|
+
| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
::<math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
+
|-
 +
|Using the identity &nbsp;<math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math>  
 
|-
 
|-
|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1,</math> we get  
+
|we get  
 
|-
 
|-
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
::<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>  
 
 
|-
 
|-
 
|If we use this identity, we have
 
|If we use this identity, we have
 
|-
 
|-
 
|  
 
|  
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int (\sin x) (1-\cos^2x)\cos^2x~dx}\\
 
\displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int (\sin x) (1-\cos^2x)\cos^2x~dx}\\
 
&&\\
 
&&\\
& = & \displaystyle{\int (\cos^2x-\cos^4x)\sin(x)~dx.}\\
+
& = & \displaystyle{\int (\cos^2x-\cos^4x)\sin(x)~dx.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|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,
+
|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -5px">u=\cos(x).</math>  
 +
|-
 +
|Then, &nbsp;<math style="vertical-align: -5px">du=-\sin(x)dx.</math>  
 
|-
 
|-
|  
+
|Therefore,
::<math>\begin{array}{rcl}
+
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\
 
\displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{\frac{-u^3}{3}+\frac{u^5}{5}+C}\\
 
& = & \displaystyle{\frac{-u^3}{3}+\frac{u^5}{5}+C}\\
 
&&\\
 
&&\\
& = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C.}\\
+
& = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp;&nbsp; <math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
+
| &nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
 
|}
 
|}
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 12:01, 9 April 2017

Evaluate the integral:


Foundations:  
1. Recall the trig identity
       
2. How would you integrate  

        You can 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:  
       

Return to Sample Exam