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| | <span class="exam">(c) <math>\int_1^e \frac{\cos(\ln(x))}{x}~dx</math> | | <span class="exam">(c) <math>\int_1^e \frac{\cos(\ln(x))}{x}~dx</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 3, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' Recall
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| − | | <math>\int \frac{1}{1+x^2}~dx=\arctan(x)+C</math>
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| − | |-
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| − | |'''2.''' How would you integrate <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx?</math> | |
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| − | You could use <math style="vertical-align: 0px">u</math>-substitution.
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| − | | Let <math style="vertical-align: -5px">u=\ln(x).</math>
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| − | | Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
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| − | Thus,
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{u^2}{2}+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{(\ln x)^2}{2}+C.}
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| − | \end{array}</math>
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| − | |-
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| − | |}
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| − | '''Solution:''' | + | [[009B Sample Final 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we notice
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| − | | <math>\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx=\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+(4x)^2}~dx.</math>
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| − | |Now, we use <math>u</math>-substitution.
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| − | |Let <math style="vertical-align: -2px">u=4x.</math>
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| − | |Then, <math style="vertical-align: -1px">du=4dx</math> and <math>\frac{du}{4}=dx.</math>
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| − | |-
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| − | |Also, we need to change the bounds of integration.
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| − | |-
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| − | |Plugging in our values into the equation <math style="vertical-align: -5px">u=4x,</math> we get
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| − | | <math style="vertical-align: -6px">u_1=4(0)=0</math> and <math style="vertical-align: -16px">u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.</math>
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| − | |Therefore, the integral becomes
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| − | |-
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| − | | <math style="vertical-align: -19px">\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du.</math>
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| − | |
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |We now have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx} & = & \displaystyle{\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{4}\arctan(u)\bigg|_0^{\sqrt{3}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{4}\arctan(\sqrt{3})-\frac{1}{4}\arctan(0)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{4}\bigg(\frac{\pi}{3}\bigg)-0}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{12}.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We use <math style="vertical-align: 0px">u</math>-substitution.
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| − | |-
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| − | |Let <math style="vertical-align: -2px">u=1+x^3.</math>
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| − | |Then, <math style="vertical-align: 0px">du=3x^2dx</math> and <math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
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| − | |-
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| − | |Therefore, the integral becomes
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| − | |-
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| − | | <math style="vertical-align: -13px">\frac{1}{3}\int \frac{1}{u^2}~du.</math>
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| − | |-
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| − | |
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |We now have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{x^2}{(1+x^3)^2}~dx} & = & \displaystyle{\frac{1}{3}\int \frac{1}{u^2}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{1}{3u}+C}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{1}{3(1+x^3)}+C.}
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| − | \end{array}</math>
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| − | |-
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| − | |
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| − | |}
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| − |
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| − | '''(c)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We use <math>u</math>-substitution.
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| − | |-
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| − | |Let <math style="vertical-align: -5px">u=\ln(x).</math>
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| − | |-
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| − | |Then, <math style="vertical-align: -15px">du=\frac{1}{x}dx.</math>
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| − | |-
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| − | |Also, we need to change the bounds of integration.
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| − | |-
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| − | |Plugging in our values into the equation <math style="vertical-align: -5px">u=\ln(x),</math>
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| − | |-
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| − | |we get
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| − | |-
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| − | | <math style="vertical-align: -6px">u_1=\ln(1)=0</math> and <math style="vertical-align: -6px">u_2=\ln(e)=1.</math>
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| − | |-
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| − | |Therefore, the integral becomes
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| − | | <math style="vertical-align: -19px">\int_0^1 \cos(u)~du.</math>
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| − | |
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |We now have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_1^e \frac{\cos(\ln(x))}{x}~dx} & = & \displaystyle{\int_0^1 \cos(u)~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\sin(u)\bigg|_0^1}\\
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| − | &&\\
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| − | & = & \displaystyle{\sin(1)-\sin(0)}\\
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| − | &&\\
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| − | & = & \displaystyle{\sin(1).}
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| − | \end{array}</math>
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| − | |-
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| − | |
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| − | |}
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| − |
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' <math>\frac{\pi}{12}</math>
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| − | | '''(b)''' <math>-\frac{1}{3(1+x^3)}+C</math>
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| − | |-
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| − | | '''(c)''' <math>\sin(1)</math>
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| − | |}
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| | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
