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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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| − | |Review <math>u</math>-substitution and
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| − | |-
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| − | |trig identities
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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 <math>\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>\sec^2(x)=\tan^2(x)+1</math>, we have <math>\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>\tan^2(x)</math>, we get
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| − | |-
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| − | |<math>\int \tan^4(x)~dx=\int \tan^2(x) (\sec^2(x)-1)~dx=\int \tan^2(x)\sec^2(x)~dx-\int \tan^2(x)~dx=\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx</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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| − | {| 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 <math>\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>u</math>-substitution. Let <math>u=\tan(x)</math>. Then, <math>du=\sec^2(x)dx</math>.
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| − | |So, we have
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| − | |<math>\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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| − | | <math>\int \tan^4(x)~dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>
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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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| − | |<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>''']] |
