009B Sample Midterm 2, Problem 5 - Revision history

Kayla Murray at 01:03, 13 November 2017

2017-11-13T01:03:55Z

Kayla Murray: Replaced content with " Evaluate the integral: :: $\int \tan^4 x ~dx$
(insert picture of handwritten solution) 009B Sample Midterm 2, Problem 5 Detailed So..."

2017-11-05T01:44:53Z

Replaced content with "<span class="exam"> Evaluate the integral: ::<math>\int \tan^4 x ~dx</math> <hr> (insert picture of handwritten solution) 009B Sample Midterm 2, Problem 5 Detailed So..."

Kayla Murray at 18:02, 27 March 2017

2017-03-27T18:02:12Z

Kayla Murray at 20:53, 14 March 2017

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Kayla Murray at 01:30, 27 February 2017

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Kayla Murray at 01:48, 8 February 2017

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Kayla Murray at 16:57, 7 February 2017

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Kayla Murray at 03:57, 7 February 2017

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Kayla Murray at 16:35, 6 February 2017

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 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 2, Problem 5 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 01:44, 5 November 2017
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	::<math>\int \tan^4 x ~dx</math>		::<math>\int \tan^4 x ~dx</math>

		+	<hr>
		+	(insert picture of handwritten solution)

		+	[[009B Sample Midterm 2, Problem 5 Detailed Solution\|'''<u>Detailed Solution for this Problem</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\|'''1.''' Recall the trig identity~~
−	\|-
−	~~\|       <math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>~~
−	\|-
−	~~\|'''2.''' Recall~~
−	\|-
−	~~\|       <math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>~~
−	\|-
−	~~\|'''3.''' How would you integrate  <math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>~~
−	\|-
−	\|
−	~~        You can 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,~~
−	\|-
−	~~\|        <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:  ~~
−	\|-
−	~~\|First, we write~~
−	\|-
−	~~\|        <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx.</math>~~
−	\|-
−	~~\|Using the trig identity  <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math>~~
−	\|-
−	~~\|we have~~
−	\|-
−	~~\|       <math style="vertical-align: -5px">\tan^2(x)=\sec^2(x)-1.</math>~~
−	\|-
−	~~\|Plugging in the last identity into one of the  <math style="vertical-align: -5px">\tan^2(x),</math>  we get~~
−	\|-
−	\|
−	~~        <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:  ~~
−	\|-
−	~~\|So, we have~~
−	\|-
−	~~\|        <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  <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>~~
−	\|-
−	~~\|So, we have~~
−	\|-
−	~~\|         <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:  ~~
−	\|-
−	~~\|We integrate to get~~
−	\|-
−	\|
−	~~        <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:  ~~
−	\|-
−	~~\|         <math>\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>~~
−	\|}
	[[009B_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]		[[009B_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]

@@ Line 12: / Line 12: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
 |-
-|'''2.''' Also,
+|'''2.''' Recall
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Recall the trig identity &nbsp;<math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
+|'''1.''' Recall the trig identity
 |-
-|'''2.''' Also, &nbsp;<math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">\sec^2x=\tan^2x+1</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 could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
+&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>
@@ Line 22: / Line 26: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Thus, &nbsp;<math style="vertical-align: -15px">\int \sec^2(x)\tan(x)~dx=\int u~du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Thus,
 |}

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Recall the trig identity
+|'''1.''' Recall the trig identity &nbsp;<math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
+|'''2.''' Also, &nbsp;<math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>
 |-
-|'''2.''' Also,
+|'''3.''' How would you integrate &nbsp;<math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
+&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -2px">u=\tan x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=\tan x.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Thus, <math style="vertical-align: -15px">\int \sec^2(x)\tan(x)~dx=\int u~du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Thus, &nbsp;<math style="vertical-align: -15px">\int \sec^2(x)\tan(x)~dx=\int u~du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C.</math>
 |}
@@ Line 38: / Line 34: @@
 |&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 <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math>
+|Using the trig identity &nbsp;<math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math>
 |-
 |we have
@@ Line 44: / Line 40: @@
 |&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 <math style="vertical-align: -5px">\tan^2(x),</math> we get
+|Plugging in the last identity into one of the &nbsp;<math style="vertical-align: -5px">\tan^2(x),</math>&nbsp; we get
 |-
+|
@@ Line 65: / Line 61: @@
 |&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 <math style="vertical-align: 0px">u</math>-substitution.
+|For the first integral, we need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let <math style="vertical-align: -5px">u=\tan(x).</math>
+|Let &nbsp;<math style="vertical-align: -5px">u=\tan(x).</math>
 |-
-|Then, <math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
+|Then, &nbsp;<math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
 |-
 |So, we have

@@ Line 36: / Line 36: @@
 |First, we write
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx.</math>
+|&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 <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math>
 |-
-|we have <math style="vertical-align: -5px">\tan^2(x)=\sec^2(x)-1.</math>
+|we have
 |-
 |Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x),</math> we get
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\int \tan^2(x) (\sec^2(x)-1)~dx}\\
 &&\\
@@ Line 61: / Line 63: @@
 |So, we have
 |-
-|&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>
+|&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 <math style="vertical-align: 0px">u</math>-substitution.
@@ Line 71: / Line 73: @@
 |So, we have
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx.</math>
+| &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>
 |}
@@ Line 80: / Line 82: @@
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\frac{u^3}{3}-(\tan(x)-x)+C}\\
 &&\\
@@ Line 91: / Line 93: @@
 !Final Answer: &nbsp;
 |-
-| &nbsp;&nbsp; <math>\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>
+| &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>''']]

@@ Line 10: / Line 10: @@
 |Recall:
 |-
-|'''1.''' <math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
+|'''1.''' Recall the trig identity
 |-
-|'''2.''' <math style="vertical-align: -13px">\int \sec^2 x~dx=\tan x+C</math>
+|&nbsp; &nbsp; <math style="vertical-align: -1px">\sec^2x=\tan^2x+1</math>
 |-
 |'''3.''' How would you integrate <math style="vertical-align: -12px">\int \sec^2(x)\tan(x)~dx?</math>
 |-
+|
-::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>
+&nbsp; &nbsp; 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, <math style="vertical-align: -15px">\int \sec^2(x)\tan(x)~dx=\int u~du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C.</math>
+&nbsp; &nbsp; Thus, <math style="vertical-align: -15px">\int \sec^2(x)\tan(x)~dx=\int u~du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C.</math>
 |}
@@ Line 28: / Line 32: @@
 !Step 1: &nbsp;
 |-
-|First, we write <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx.</math>
+|First, we write
 |-
-|Using the trig identity <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math> we have <math style="vertical-align: -5px">\tan^2(x)=\sec^2(x)-1.</math>
+|Using the trig identity <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math>
 |-
 |Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x),</math> we get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -13px">\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>
 |-
 |by using the identity again on the last equality.
@@ Line 42: / Line 57: @@
 !Step 2: &nbsp;
 |-
-|So, we have <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx.</math>
+|So, we have
 |-
-|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>
+|Then, <math style="vertical-align: -5px">du=\sec^2(x)dx.</math>
 |-
 |So, we have
@@ Line 56: / Line 77: @@
 |We integrate to get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+x+C.</math>
 |}

@@ Line 22: / Line 22: @@
 ::Thus, <math style="vertical-align: -15px">\int \sec^2(x)\tan(x)~dx=\int u~du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C.</math>
 |}
 '''Solution:'''
@@ Line 27: / Line 28: @@
 !Step 1: &nbsp;
 |-
-|First, we write <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx</math>.
+|First, we write <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx.</math>
 |-
-|Using the trig identity <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1</math>, we have <math style="vertical-align: -5px">\tan^2(x)=\sec^2(x)-1</math>.
+|Using the trig identity <math style="vertical-align: -5px">\sec^2(x)=\tan^2(x)+1,</math> we have <math style="vertical-align: -5px">\tan^2(x)=\sec^2(x)-1.</math>
 |-
-|Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x)</math>, we get
+|Plugging in the last identity into one of the <math style="vertical-align: -5px">\tan^2(x),</math> we get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -13px">\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>,
+| &nbsp;&nbsp; <math style="vertical-align: -13px">\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>
 |-
-|using the identity again on the last equality.
+|by using the identity again on the last equality.
 |}
@@ Line 41: / Line 42: @@
 !Step 2: &nbsp;
 |-
-|So, we have <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx</math>.
+|So, we have <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 <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>.
+|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>
 |-
 |So, we have
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx</math>.
+| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx.</math>
 |}
@@ Line 55: / Line 56: @@
 |We integrate to get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+x+C</math>.
+| &nbsp;&nbsp; <math style="vertical-align: -13px">\int \tan^4(x)~dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+x+C.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"