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jennychan12

  • 2 years ago

integral from 0 to 1 of [e^x(cos(e^x))]dx ?

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  1. jennychan12
    • 2 years ago
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    \[\int\limits_{0}^{1} [e^x \cos (e^x)]dx\]

  2. KingGeorge
    • 2 years ago
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    Do a u-substitution with \(u=\sin(e^x)\), and your solution should pop right out.

  3. hartnn
    • 2 years ago
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    did you mean u=e^x ?

  4. jennychan12
    • 2 years ago
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    but if you do that, then du = e^xcos(e^x) and there's no cos (e^x) in the question.

  5. hartnn
    • 2 years ago
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    try u=e^x

  6. jennychan12
    • 2 years ago
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    that'd just be ucosu

  7. KingGeorge
    • 2 years ago
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    \(u=\sin(e^x)\implies du=e^x\cos(e^x)dx\). So your integral becomes\[\int du=u\]

  8. KingGeorge
    • 2 years ago
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    Substitute back for \(u\), and you get \[\int_0^1 e^x\cos(e^x)dx=\sin(e^x)|_0^1\]

  9. hartnn
    • 2 years ago
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    u =e^x du = e^x dx \(\int cos udu\) ohh..now i could say that u =sin e^x is a better substitution..

  10. jennychan12
    • 2 years ago
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    oh wait. my bad. i thought u said u = sin u ok i see now

  11. KingGeorge
    • 2 years ago
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    Yup, I like to call my method "guessing the solution and proving you're right"

  12. KingGeorge
    • 2 years ago
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    Although if you really had no clue of the solution, \(u=e^x\) would be a fine substitution. You would just have to integrate by parts.

  13. hartnn
    • 2 years ago
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    whenever i see a non-standard angle with sin/cos/.. i'll call that as 'bad angle' and put u= bad angle.... example : sin x^2 , cos log x ....

  14. hartnn
    • 2 years ago
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    how do we need integration by parts ?? O.o u= e^x du=e^xdx

  15. KingGeorge
    • 2 years ago
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    Oh. Right. You don't. Ignore my ramblings.

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