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anonymous

  • one year ago

Evaluate the integral using integration by parts w/ the indicated choices of u and dv how do i do this? equation inside! thanks!!

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  1. anonymous
    • one year ago
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    |dw:1442193899384:dw|

  2. thomas5267
    • one year ago
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    So the equation is: \[ \int\theta\cos(\theta)\,d\theta\\ u=\theta\\ dv=\cos(\theta)\,d\theta\\ \int u\,dv=uv-\int v\,du \]

  3. anonymous
    • one year ago
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    yes:) i'm a bit confused, so i find du and v, right? would du = ø and v = -cosødø ?

  4. zepdrix
    • one year ago
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    Hmm if \(\large\rm u=\theta\) then \(\large\rm du\ne\theta\) You're taking a derivative with respect to theta :o \(\large\rm u'=1\) yes?

  5. anonymous
    • one year ago
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    yes :) ohh so du = 1 ?

  6. zepdrix
    • one year ago
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    \[\large\rm \color{orangered}{u'}=1\]\[\large\rm \color{orangered}{\frac{du}{d\theta}}=1\qquad\to\qquad du=1\cdot d\theta\]

  7. anonymous
    • one year ago
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    ohh okay :) so which means du = dø ? and so would v be sinø ? :/

  8. zepdrix
    • one year ago
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    To get from \(\large\rm dv\) to \(\large\rm v\) you have to integrate. So going backwards from cosine gives us sine, ya that sounds right! :)\[\large\rm dv=\cos\theta~d\theta\qquad\to\qquad v=\sin\theta\]

  9. anonymous
    • one year ago
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    ooh yay!! :) and so now i do this?|dw:1442194685139:dw|

  10. zepdrix
    • one year ago
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    I wish you wouldn't use the "empty set" for your theta XD lol but ya looks good so far!

  11. anonymous
    • one year ago
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    ohh haha oops :P it's just more convenient using that symbol instead of drawing it :P and so would it get sin2ø - (-cosø)(ø) ?

  12. anonymous
    • one year ago
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    getting sin2ø + cos2ø + c ?

  13. zepdrix
    • one year ago
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    Hmm

  14. zepdrix
    • one year ago
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    In general: \(\large\rm a\cos(x)\ne\cos(ax)\) You can't just bring stuff inside of the trig function willy nilly like that.

  15. anonymous
    • one year ago
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    ohh okay :( how do i do this part then? |dw:1442195029862:dw|

  16. zepdrix
    • one year ago
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    We leave this alone \(\large\rm \theta\sin\theta\) that will be part of our final answer

  17. zepdrix
    • one year ago
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    \[\large\rm -\int\limits\sin\theta~d\theta\]Hmm looks like we made some kind of boo boo here on this integral.

  18. zepdrix
    • one year ago
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    If you ignore the negative out front, what is the integral of sine?

  19. anonymous
    • one year ago
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    cos ?

  20. zepdrix
    • one year ago
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    Hmm, no we're going backwards. Going forwards, derivative of sine is cosine.

  21. anonymous
    • one year ago
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    ohh - cos ?

  22. zepdrix
    • one year ago
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    \[\large\rm -\color{orangered}{\int\limits\limits\sin\theta~d\theta}=\quad -\color{orangered}{(-\cos \theta+c)}\]Mmm ya that sounds better!

  23. zepdrix
    • one year ago
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    So you have:\[\large\rm \int\limits \theta \cos \theta~d \theta=\theta \sin \theta-(-\cos \theta+c)\]

  24. anonymous
    • one year ago
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    ohh okay and so i simply it to øsinø + cos ø - c ?

  25. zepdrix
    • one year ago
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    Or even: \(\large\rm \theta\sin\theta+\cos\theta+C\) absorb the negative into the c, since it can represent any negative or positive value.

  26. anonymous
    • one year ago
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    ahh okay! thank you!!:)

  27. zepdrix
    • one year ago
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    yay team \c:/

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