Difference between revisions of "009B Sample Midterm 2, Problem 5"

Latest revision as of 17:12, 12 November 2017

Evaluate the integral:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \tan^4 x ~dx}

Solution

Detailed Solution

Return to Sample Exam

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

Difference between revisions of "009B Sample Midterm 2, Problem 5"

Latest revision as of 17:12, 12 November 2017

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