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| | ::<math>\int \tan^4 x ~dx</math> | | ::<math>\int \tan^4 x ~dx</math> |
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| | + | [[009B Sample Midterm 2, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009B Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Foundations:
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| − | |-
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| − | |Recall:
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| − | |-
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| − | |'''1.''' <math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
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| − | |-
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| − | |'''2.''' <math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>
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| − | |-
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| − | |How would you integrate <math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>
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| − | |-
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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=\tan x.</math> Then, <math style="vertical-align: -5px">du=\sec^2(x)dx.</math> Thus,
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int \sec^2(x)\tan(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{\tan^2x}{2}+C.}\\
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| − | \end{array}</math>
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| − | |}
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| − | '''Solution:'''
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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 write
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| − | |
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| − | ::<math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx.</math>
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| − | |-
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| − | |Using the trig identity <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math> we have
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| − | |-
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| − | ::<math style="vertical-align: -5px">\tan^2(x)=\sec^2(x)-1.</math>
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| − | |-
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| − | |Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x),</math> we get
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| − | |-
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\int \tan^2(x) (\sec^2(x)-1)~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int \tan^2(x)\sec^2(x)~dx-\int \tan^2(x)~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx.}\\
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| − | \end{array}</math>
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| − | |-
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| − | |using the identity again on the last equality.
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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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| − | |So, we have
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| − | |-
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| − | |
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| − | ::<math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx.</math>
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| − | |-
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| − | |For the first integral, we need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\tan(x).</math> Then, <math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
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| − | |-
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| − | |So, we have
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| − | |-
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| − | |
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| − | ::<math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx.</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 3:
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| − | |-
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| − | |We integrate to get
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| − | |-
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\frac{u^3}{3}-(\tan(x)-x)+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\tan^3(x)}{3}-\tan(x)+x+C.}\\
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| − | \end{array}</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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| − | !Final Answer:
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| − | |-
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| − | | <math>\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>
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| − | |}
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| | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
