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MATTW20 Group Title

Help Trigonometric Substitution \[\int\limits_{}^{}\sqrt{64-x ^{2}}\]

  • 10 months ago
  • 10 months ago

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  1. MATTW20 Group Title
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    @hartnn

    • 10 months ago
  2. SACAPUNTAS Group Title
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    \[\cos^2 \theta = 1 - \sin^2 \theta\]

    • 10 months ago
  3. MATTW20 Group Title
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    ok

    • 10 months ago
  4. SACAPUNTAS Group Title
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    Do you see how that helps us here?

    • 10 months ago
  5. MATTW20 Group Title
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    yes because that will get you \[\sqrt{64\cos ^{2}\theta }\]

    • 10 months ago
  6. MATTW20 Group Title
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    and then dx=8cos t dt t for theta

    • 10 months ago
  7. SACAPUNTAS Group Title
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    Good. Keep going!

    • 10 months ago
  8. MATTW20 Group Title
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    now this is where i'm getting kind of lost

    • 10 months ago
  9. SACAPUNTAS Group Title
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    The expression will be kind of messy, I think. You'll have to integrate it by parts. Do you know how to do that?

    • 10 months ago
  10. MATTW20 Group Title
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    Yeah we passed that section. I just am not super familiar with it. but could you just write the expression I'm supposed to integrate by parts so I make sure everything is correct up to that point

    • 10 months ago
  11. MATTW20 Group Title
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    and then i'll work on integrating that

    • 10 months ago
  12. SACAPUNTAS Group Title
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    Sure, one sec

    • 10 months ago
  13. SACAPUNTAS Group Title
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    Actually I think I spaced out a bit. I'm just getting \(\int d\theta\)

    • 10 months ago
  14. MATTW20 Group Title
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    no the answer is something weird \[\frac{ x }{ 2 }\sqrt{64-x ^{2}}+32\sin^{-1} (\frac{ x }{ 8 })+C\]

    • 10 months ago
  15. SACAPUNTAS Group Title
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    Ok, that looks like the result of integration by parts after all. So how did you get that? :D

    • 10 months ago
  16. MATTW20 Group Title
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    back of the book lol

    • 10 months ago
  17. MATTW20 Group Title
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    that's why I'm having trouble I can't get there

    • 10 months ago
  18. myininaya Group Title
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    So what did you @MATTW20 let x=?

    • 10 months ago
  19. myininaya Group Title
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    Oh I see you let x=8sin(theta) and then you said dx=8cos(theta) d theta Just plug into your expression.

    • 10 months ago
  20. myininaya Group Title
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    \[\int\limits_{}^{}\sqrt{64\cos^2(\theta) } 8 \cos(\theta) d \theta \] That is what you have got so far, right?

    • 10 months ago
  21. MATTW20 Group Title
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    correct

    • 10 months ago
  22. myininaya Group Title
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    \[\int\limits_{}^{}8 \cos(\theta) \cos(\theta) d \theta ,\text{ assuming } \cos(\theta)>0\] so we have that we are trying to evaluate: \[8\int\limits_{}^{}\cos^2(\theta) d \theta \] The double angle identity comes in use for integating cos^2(theta) and even sin^2(theta)

    • 10 months ago
  23. surjithayer Group Title
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    \[put x=8\sin \theta ,dx=8\cos \theta d \theta \] \[I=\int\limits \sqrt{64-64\sin ^{2}\theta} 8\cos \theta d \theta =32\int\limits 2\cos ^{2}\theta d \theta\] \[=32\int\limits \left( 1+\cos 2\theta \right)d \theta=32\left( \theta+\frac{ \sin 2\theta }{2 } \right)+c\] |dw:1381785998999:dw| \[I=32\left( \theta+\frac{ 2\sin \theta \cos \theta }{ 2 } \right)+c\] \[I=32\left( \sin^{-1} \frac{ x }{8 }+\frac{ x }{8 }*\cos \frac{ \sqrt{64-x ^{2}} }{8 } \right)+c\]

    • 10 months ago
  24. myininaya Group Title
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    \[\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)=\cos^2(\theta)-(1-\cos^2(\theta))\] \[\cos(\theta)=\cos^2(\theta)-1+\cos^2(\theta)=2\cos^2(\theta)-1\] Solve for cos^2(theta) This is just the way I remember this formula. I remember other formulas and derive other formulas from the ones I have memorized. \[\frac{1}{2}(\cos(\theta)+1)=\cos^2(\theta) \]

    • 10 months ago
  25. MATTW20 Group Title
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    ok

    • 10 months ago
  26. myininaya Group Title
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    And yes I forgot to write the other 8.. Errr...

    • 10 months ago
  27. zepdrix Group Title
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    \[\Large \color{teal}{\cos^2(\theta)\quad=\quad \frac{1}{2}(\cos2\theta+1)}\]Woops! Careful with your half-angle formula there missy! c:

    • 10 months ago
  28. myininaya Group Title
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    Yeah I missed my 2.

    • 10 months ago
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