Difference between revisions of "009B Sample Midterm 1, Problem 4"
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!Step 2: | !Step 2: | ||
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| − | |Now, we use <math style="vertical-align: 0px">u</math> substitution. Let <math style="vertical-align: -5px">u=\cos(x)</math>. Then, <math style="vertical-align: -5px">du=-\sin(x)dx</math>. Therefore, | + | |Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\cos(x)</math>. Then, <math style="vertical-align: -5px">du=-\sin(x)dx</math>. Therefore, |
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| <math style="vertical-align: -14px">\int\sin^3x\cos^2x~dx=\int -(u^2-u^4)~du=\frac{-u^3}{3}+\frac{u^5}{5}+C=\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>. | | <math style="vertical-align: -14px">\int\sin^3x\cos^2x~dx=\int -(u^2-u^4)~du=\frac{-u^3}{3}+\frac{u^5}{5}+C=\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>. | ||
Revision as of 13:19, 1 February 2016
Evaluate the integral:
| Foundations: |
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| Review 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, and |
| Trig identities. |
Solution:
| Step 1: |
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| 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 \int\sin^3x\cos^2x~dx=\int (\sin x) (1-\cos^2x)\cos^2x~dx=\int (\cos^2x-\cos^4x)\sin(x)~dx} . |
| Step 2: |
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| 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 \int\sin^3x\cos^2x~dx=\int -(u^2-u^4)~du=\frac{-u^3}{3}+\frac{u^5}{5}+C=\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C} . |
| Final Answer: |
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| 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} |