Difference between revisions of "009B Sample Midterm 2, Problem 5"
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|First, we write | |First, we write | ||
|- | |- | ||
| − | | <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int \tan^2(x) \tan^2(x)~dx.</math> | + | | <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 <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 |
| + | |- | ||
| + | | <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 | ||
|- | |- | ||
| | | | ||
| − | <math>\begin{array}{rcl} | + | <math>\begin{array}{rcl} |
\displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\int \tan^2(x) (\sec^2(x)-1)~dx}\\ | \displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\int \tan^2(x) (\sec^2(x)-1)~dx}\\ | ||
&&\\ | &&\\ | ||
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|So, we have | |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> | + | | <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 <math style="vertical-align: 0px">u</math>-substitution. | ||
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|So, we have | |So, we have | ||
|- | |- | ||
| − | | <math style="vertical-align: -13px">\int \tan^4(x)~dx=\int u^2~du-\int (\sec^2(x)-1)~dx.</math> | + | | <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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|- | |- | ||
| | | | ||
| − | <math>\begin{array}{rcl} | + | <math>\begin{array}{rcl} |
\displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\frac{u^3}{3}-(\tan(x)-x)+C}\\ | \displaystyle{\int \tan^4(x)~dx} & = & \displaystyle{\frac{u^3}{3}-(\tan(x)-x)+C}\\ | ||
&&\\ | &&\\ | ||
| Line 91: | Line 93: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | <math>\frac{\tan^3(x)}{3}-\tan(x)+x+C</math> | + | | <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>''']] | ||
Revision as of 17:48, 7 February 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}
| Foundations: |
|---|
| 1. Recall the trig identity |
| 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 \sec^2x=\tan^2x+1} |
| 2. Also, |
| 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 \sec^2 x~dx=\tan x+C} |
| 3. How would you integrate 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 \sec^2(x)\tan(x)~dx?} |
|
You could use 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 u} -substitution. |
| Let 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 u=\tan x.} |
| Then, 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 du=\sec^2(x)dx.} |
|
Thus, 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 \sec^2(x)\tan(x)~dx=\int u~du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C.} |
Solution:
| Step 1: |
|---|
| First, we write |
| 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=\int \tan^2(x) \tan^2(x)~dx.} |
| Using the trig identity 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 \sec^2(x)=\tan^2(x)+1,} |
| we have |
| 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 \tan^2(x)=\sec^2(x)-1.} |
| Plugging in the last identity into one of the 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 \tan^2(x),} we get |
|
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 \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}} |
| by using the identity again on the last equality. |
| Step 2: |
|---|
| So, we have |
| 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=\int \tan^2(x)\sec^2(x)~dx-\int (\sec^2x-1)~dx.} |
| For the first integral, we need to use 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 u} -substitution. |
| Let 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 u=\tan(x).} |
| Then, 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 du=\sec^2(x)dx.} |
| So, we have |
| 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=\int u^2~du-\int (\sec^2(x)-1)~dx.} |
| Step 3: |
|---|
| We integrate to get |
|
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 \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}} |
| Final Answer: |
|---|
| 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 \frac{\tan^3(x)}{3}-\tan(x)+x+C} |