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anonymous

  • 5 years ago

∫[0,π,e^(3t)sin(3t),]dt=

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  1. anonymous
    • 5 years ago
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    have u try using substitution?

  2. anonymous
    • 5 years ago
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    You should use integration by parts:\[\int\limits_{}{}u.dv=uv-\int\limits_{}{}v.du\]and you'll have to use it a couple of times. Set the following:\[u=e^{3t}\]and\[dv=\sin(3t)dt\]then \[du=3e^{3t}dt\]and\[v=-\frac{1}{3}\cos(3t)\]Then your integral becomes\[\int\limits_{}{}e^{3t}\sin(3t)dt=-\frac{1}{3}e^{3t}\cos(3t)+\int\limits_{}{}\cos(3t)e^{3t}dt\]

  3. anonymous
    • 5 years ago
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    Now you have to repeat the process on the final integral. Set u=e^(3t) and dv=cos(3t)dt and repeat. You should get\[\int\limits_{}{}e^{3t}\cos(3t)dt=\frac{1}{3}e^{3t}\sin(3t)-\int\limits_{}{}e^{3t}\sin(3t)dt\]

  4. anonymous
    • 5 years ago
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    Substitute this into the equation before it, so that\[\int\limits_{}{}e^{3t}\sin(3t)dt=-\frac{1}{3}e^{3t}\cos(3t)\]\[+\left[ \frac{1}{3}e^{3t}\sin(3t)-\int\limits_{}{}e^{3t}\sin(3t)dt \right]\]

  5. anonymous
    • 5 years ago
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    Add \[\int\limits_{}{}e^{3t}\sin(3t)dt\]to both sides. Then\[2\int\limits_{}{}e^{3t}\sin(3t)dt=\frac{1}{3}e^{3t}\left( \sin(3t)-\cos(3t) \right)\]

  6. anonymous
    • 5 years ago
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    Divide both sides by 2 to get your integral:\[\int\limits_{}{}e^{3t}\sin(3t)=\frac{1}{6}e^{3t}\left( \sin(3t)-\cos(3t) \right)+C\]

  7. anonymous
    • 5 years ago
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    There is a method by which you can pump out IBP problems extremely quickly, but it's hard to explain online...so I stuck to the 'traditional' route.

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