Difference between revisions of "009B Sample Midterm 3, Problem 5"
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!Foundations: | !Foundations: | ||
|- | |- | ||
| − | | | + | |Review u substitution |
|- | |- | ||
| − | | | + | |Trig identities |
| + | |} | ||
| + | |||
| + | '''Solution:''' | ||
| + | |||
| + | '''(a)''' | ||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 1: | ||
|- | |- | ||
| + | |We start by writing <math>\int \tan^3xdx=\int \tan^2x\tan x dx</math>. | ||
|- | |- | ||
| − | | | + | |Since <math>\tan^2x=\sec^2x-1</math>, we have <math>\int \tan^3xdx=\int (\sec^2x-1)\tan x dx=\int \sec^2\tan xdx-\int \tan xdx</math>. |
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 2: | ||
|- | |- | ||
| − | | | + | |Now, we need to use u substitution for the first integral. Let <math>u=\tan(x)</math>. Then, <math>du=\sec^2xdx</math>. So, we have |
|- | |- | ||
| − | |2 | + | |<math>\int \tan^3xdx=\int udu-\int \tan xdx=\frac{u^2}{2}-\int \tan xdx=\frac{\tan^2x}{2}-\int \tan xdx</math>. |
|} | |} | ||
| − | ''' | + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" |
| + | !Step 3: | ||
| + | |- | ||
| + | |For the remaining integral, we need to use u substitution. First, we write <math>\int \tan^3xdx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}dx</math>. | ||
| + | |- | ||
| + | |Now, we let <math>u=\cos x</math>. Then, <math>du=-\sin xdx</math>. So, we get | ||
| + | |- | ||
| + | |<math>\int \tan^3xdx=\frac{\tan^2x}{2}+\int \frac{1}{u}dx=\frac{\tan^2x}{2}+\ln |u|+C=\frac{\tan^2x}{2}+\ln |\cos x|+C</math>. | ||
| + | |} | ||
| + | '''(b)''' | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
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!Step 2: | !Step 2: | ||
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!Final Answer: | !Final Answer: | ||
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| − | | | + | |'''(a)''' <math>\frac{\tan^2x}{2}+\ln |\cos x|+C</math> |
|- | |- | ||
| − | | | + | |'''(b)''' |
| − | + | ||
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 12:56, 31 January 2016
Evaluate the indefinite and definite integrals.
- a)
- b)
| Foundations: |
|---|
| Review u substitution |
| Trig identities |
Solution:
(a)
| Step 1: |
|---|
| We start by writing . |
| Since , we have . |
| Step 2: |
|---|
| Now, we need to use u substitution for the first integral. Let . Then, . So, we have |
| . |
| Step 3: |
|---|
| For the remaining integral, we need to use u substitution. First, we write . |
| Now, we let . Then, . So, we get |
| . |
(b)
| Step 1: |
|---|
| Step 2: |
|---|
| Final Answer: |
|---|
| (a) |
| (b) |