Difference between revisions of "009B Sample Midterm 1, Problem 4"
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| <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math> | | <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math> | ||
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| − | |'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math> | + | |'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math> |
|- | |- | ||
| | | | ||
| − | You could use <math style="vertical-align: 0px">u</math>-substitution. | + | You could use <math style="vertical-align: 0px">u</math>-substitution. |
|- | |- | ||
| − | | Let <math style="vertical-align: -2px">u=\sin x.</math> | + | | Let <math style="vertical-align: -2px">u=\sin x.</math> |
|- | |- | ||
| − | | Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> | + | | Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> |
|- | |- | ||
| Thus, | | Thus, | ||
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| <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math> | | <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math> | ||
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| − | |Using the identity <math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math> | + | |Using the identity <math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math> |
|- | |- | ||
| − | |we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math> | + | |we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math> |
|- | |- | ||
|If we use this identity, we have | |If we use this identity, we have | ||
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!Step 2: | !Step 2: | ||
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| − | |Now, we use <math style="vertical-align: 0px">u</math>-substitution. | + | |Now, we use <math style="vertical-align: 0px">u</math>-substitution. |
|- | |- | ||
| − | |Let <math style="vertical-align: -5px">u=\cos(x).</math> | + | |Let <math style="vertical-align: -5px">u=\cos(x).</math> |
|- | |- | ||
| − | |Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> | + | |Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> |
|- | |- | ||
|Therefore, | |Therefore, | ||
Revision as of 16:59, 26 February 2017
Evaluate the integral:
| 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 \sin^2x+\cos^2x=1} |
| 2. 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 \sin^2x\cos 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=\sin 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=\cos 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 \begin{array}{rcl} \displaystyle{\int \sin^2x\cos x~dx} & = & \displaystyle{\int u^2~du}\\ &&\\ & = & \displaystyle{\frac{u^3}{3}+C}\\ &&\\ & = & \displaystyle{\frac{\sin^3x}{3}+C.} \end{array}} |
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\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.} |
| Using the 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 \sin^2x+\cos^2x=1,} |
| 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 \sin^2x=1-\cos^2x.} |
| If we use this identity, 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 \begin{array}{rcl} \displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int (\sin x) (1-\cos^2x)\cos^2x~dx}\\ &&\\ & = & \displaystyle{\int (\cos^2x-\cos^4x)\sin(x)~dx.} \end{array}} |
| Step 2: |
|---|
| Now, we 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=\cos(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=-\sin(x)dx.} |
| Therefore, |
|
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\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\ &&\\ & = & \displaystyle{\frac{-u^3}{3}+\frac{u^5}{5}+C}\\ &&\\ & = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+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{\cos^5x}{5}-\frac{\cos^3x}{3}+C} |