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lgbasallote

  • 3 years ago

\[\large \int_0^{3/2} \int_{x/\sqrt 3}^{\sqrt{2x - x^2}} \frac{dydx}{\sqrt{x^2 + y^2}}\] help?

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  1. blockcolder
    • 3 years ago
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    The region is between a circle and a line, apparently.

  2. lgbasallote
    • 3 years ago
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    lol no graphing :p just find the integral

  3. blockcolder
    • 3 years ago
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    What? No graphing? o_O I'd convert to polar coordinates but apparently that's thrown out of the window.

  4. blockcolder
    • 3 years ago
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    \(y=x\sec{\theta}\) would be the substitution I would use but the bounds are horrible.

  5. lgbasallote
    • 3 years ago
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    that's what i did too and got \[\ln (\sec \theta + \tan \theta)\]

  6. blockcolder
    • 3 years ago
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    I meant \(y=x \tan\theta\). Trying it anyway... \[y=\sqrt{2x-x^2}\Rightarrow \theta=\tan^{-1}\left ( \frac{\sqrt{2x-x^2}}{x}\right )=\alpha\\ y=\frac{x}{\sqrt3} \Rightarrow \theta=\frac{\pi}{6}\\ dy=x\ \sec^2\theta d\theta\\ \Large \int_0^{3/2}\int_{\pi/6}^\alpha \frac{x \sec^2\theta d\theta dx}{x \sec\theta}\]

  7. lgbasallote
    • 3 years ago
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    ?!

  8. blockcolder
    • 3 years ago
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    Huh? Which part is bugging you?

  9. lgbasallote
    • 3 years ago
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    the substitution of theta thingy...i dont know that

  10. blockcolder
    • 3 years ago
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    The bounds, you mean?

  11. lgbasallote
    • 3 years ago
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    yeah

  12. shivam_bhalla
    • 3 years ago
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    First integrate the inner part and then the outer part \[\large \int\limits_0^{3/2} [{ \int\limits_{x/\sqrt 3}^{\sqrt{2x - x^2}} \frac{dy}{\sqrt{x^2 + y^2}}}]dx\] First integrate the part in the brackets keeping y as constant

  13. shivam_bhalla
    • 3 years ago
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    Sorry I meant keeping x as constant

  14. lgbasallote
    • 3 years ago
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    uhh i think i got a weid answer for that @_@ with ln and such

  15. .Sam.
    • 3 years ago
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    \[\int\limits \frac{1}{\sqrt{x^2+y^2}} \, dy\] Let \[y=x \tan (u) ~~and ~~dy=x \sec ^2(u) d u\] \[u=\tan^{-1}(\frac{y}{x})\] \[\text{Then}\sqrt{x^2+y^2}\text{ =}\sqrt{x^2 \tan ^2(u)+x^2}\text{ = }x \sec (u)\] \[\int\limits \sec (u) \, du\] \[[\ln (\tan (u)+\sec (u))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]

  16. shivam_bhalla
    • 3 years ago
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    Then after you do upper limit - lower limit, integrate with dx with the limits 0 to 3/2

  17. lgbasallote
    • 3 years ago
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    uhmm yep i got that...then i dont know how to integrate it

  18. shivam_bhalla
    • 3 years ago
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    @lgba, make sure you substitute back u = arctan(y/x) before taking upper limit-lower limit

  19. shivam_bhalla
    • 3 years ago
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    Contiuing from @.Sam. 's work so basically you get \[[\ln (\tan (\arctan(y/x))+\sec (\arctan(y/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln (y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln (y/x)+(\sqrt{x^2+y^2}/x)]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]

  20. shivam_bhalla
    • 3 years ago
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    Now take upper limit - lower limt

  21. .Sam.
    • 3 years ago
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    \[\small \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\] lol

  22. .Sam.
    • 3 years ago
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    \[\tiny \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\]

  23. shivam_bhalla
    • 3 years ago
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    Well sorry I messed up the bracket's earlier. It should be \[[\ln ((y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln ((y/x)+\frac{\sqrt{x^2+y^2}}{x} )]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]

  24. lgbasallote
    • 3 years ago
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    wow those are small @.Sam.

  25. .Sam.
    • 3 years ago
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    wolfram doesn't know how to integrate this, lol

  26. shivam_bhalla
    • 3 years ago
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    \[\small \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{x^2+(\sqrt{2x-x^2})^2}}{x} )-[\ln (\frac{x/\sqrt3}{x})+\frac{\sqrt{x^2+(x/\sqrt3)^2}}{x} )]\]

  27. lgbasallote
    • 3 years ago
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    darn my teacher -_-

  28. shivam_bhalla
    • 3 years ago
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    \[ \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2x}}}{x} )-[\ln ((\frac{1}{\sqrt{3}})+\frac{\sqrt{4x^2}}{x} )]\] \[ \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )-[\ln ((\frac{1}{\sqrt{3}})+2 )]\] So now, it's upto you to integrate \[\int\limits_{0}^{3/2}\ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )dx-\int\limits_{0}^{3/2}\ln ((\frac{1}{\sqrt{3}})+2 )dx\]

  29. lgbasallote
    • 3 years ago
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    ugh can i still quit learning this?

  30. lgbasallote
    • 3 years ago
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    you just succeeded i hurting my poor brain :p

  31. shivam_bhalla
    • 3 years ago
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    Try wolfram :P The second integral is easy. I have no idea on first part :P

  32. lgbasallote
    • 3 years ago
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    @.Sam. said wolfram cant do it

  33. shivam_bhalla
    • 3 years ago
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    @jazy, what was that ??

  34. shivam_bhalla
    • 3 years ago
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    @lgbasallote , wolfram works :p http://www.wolframalpha.com/input/?i=integral+of+ln%28%28sqrt%282x-x^2%29%29%2Fx+%2B+sqrt%282%2Fx%29%29dx

  35. shivam_bhalla
    • 3 years ago
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    First part is solved by wolfram alpha. Take upper limit - lower limit for the first. The second part of integral is a cake walk. But I would still be interested for the technique of solving http://www.wolframalpha.com/input/?i=integral+of+ln%28%28sqrt%282x-x%5E2%29%29%2Fx+%2B+sqrt%282%2Fx%29%29dx

  36. lgbasallote
    • 3 years ago
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    ugh enough =_=

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