Evaluate the integral:
| ExpandFoundations: |
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| Review integration by parts |
Solution:
| ExpandStep 1: |
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| We proceed using integration by parts. 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(2x)} and 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 dv=e^{-2x}dx} . 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=2\cos(2x)dx} and 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 v=\frac{e^{-2x}}{-2}} . |
| So, 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 \int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}-\int \frac{e^{-2x}2\cos(2x)~dx}{-2}=\frac{\sin(2x)e^{-2x}}{-2}+\int e^{-2x}\cos(2x)~dx} . |
| ExpandStep 2: |
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| Now, we need to use integration by parts again. 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(2x)} and 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 dv=e^{-2x}dx} . 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=-2\sin(2x)dx} and 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 v=\frac{e^{-2x}}{-2}} . |
| So, 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 \int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}-\int e^{-2x}\sin(2x)~dx} . |
| ExpandStep 3: |
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| Notice that the integral on the right of the last equation in Step 2 is the same integral that we had at the beginning of the problem. |
| So, if we add the integral on the right to the other side of the equation, 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 2\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}} . |
| Now, we divide both sides by 2 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 \int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-4}+\frac{\cos(2x)e^{-2x}}{-4}} . |
| Thus, the final answer is 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 e^{-2x}\sin (2x)~dx=\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C} . |
| ExpandFinal 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{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C} |