lgbasallote
  • lgbasallote
\[\large \int_0^{3/2} \int_{x/\sqrt 3}^{\sqrt{2x - x^2}} \frac{dydx}{\sqrt{x^2 + y^2}}\] help?
Mathematics
chestercat
  • chestercat
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blockcolder
  • blockcolder
The region is between a circle and a line, apparently.
lgbasallote
  • lgbasallote
lol no graphing :p just find the integral
blockcolder
  • blockcolder
What? No graphing? o_O I'd convert to polar coordinates but apparently that's thrown out of the window.

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blockcolder
  • blockcolder
\(y=x\sec{\theta}\) would be the substitution I would use but the bounds are horrible.
lgbasallote
  • lgbasallote
that's what i did too and got \[\ln (\sec \theta + \tan \theta)\]
blockcolder
  • blockcolder
I meant \(y=x \tan\theta\). Trying it anyway... \[y=\sqrt{2x-x^2}\Rightarrow \theta=\tan^{-1}\left ( \frac{\sqrt{2x-x^2}}{x}\right )=\alpha\\ y=\frac{x}{\sqrt3} \Rightarrow \theta=\frac{\pi}{6}\\ dy=x\ \sec^2\theta d\theta\\ \Large \int_0^{3/2}\int_{\pi/6}^\alpha \frac{x \sec^2\theta d\theta dx}{x \sec\theta}\]
lgbasallote
  • lgbasallote
?!
blockcolder
  • blockcolder
Huh? Which part is bugging you?
lgbasallote
  • lgbasallote
the substitution of theta thingy...i dont know that
blockcolder
  • blockcolder
The bounds, you mean?
lgbasallote
  • lgbasallote
yeah
anonymous
  • anonymous
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
anonymous
  • anonymous
Sorry I meant keeping x as constant
lgbasallote
  • lgbasallote
uhh i think i got a weid answer for that @_@ with ln and such
.Sam.
  • .Sam.
\[\int\limits \frac{1}{\sqrt{x^2+y^2}} \, dy\] Let \[y=x \tan (u) ~~and ~~dy=x \sec ^2(u) d u\] \[u=\tan^{-1}(\frac{y}{x})\] \[\text{Then}\sqrt{x^2+y^2}\text{ =}\sqrt{x^2 \tan ^2(u)+x^2}\text{ = }x \sec (u)\] \[\int\limits \sec (u) \, du\] \[[\ln (\tan (u)+\sec (u))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
anonymous
  • anonymous
Then after you do upper limit - lower limit, integrate with dx with the limits 0 to 3/2
lgbasallote
  • lgbasallote
uhmm yep i got that...then i dont know how to integrate it
anonymous
  • anonymous
@lgba, make sure you substitute back u = arctan(y/x) before taking upper limit-lower limit
anonymous
  • anonymous
Contiuing from @.Sam. 's work so basically you get \[[\ln (\tan (\arctan(y/x))+\sec (\arctan(y/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln (y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln (y/x)+(\sqrt{x^2+y^2}/x)]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
anonymous
  • anonymous
Now take upper limit - lower limt
.Sam.
  • .Sam.
\[\small \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\] lol
.Sam.
  • .Sam.
\[\tiny \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\]
anonymous
  • anonymous
Well sorry I messed up the bracket's earlier. It should be \[[\ln ((y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln ((y/x)+\frac{\sqrt{x^2+y^2}}{x} )]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
lgbasallote
  • lgbasallote
wow those are small @.Sam.
.Sam.
  • .Sam.
wolfram doesn't know how to integrate this, lol
anonymous
  • anonymous
\[\small \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{x^2+(\sqrt{2x-x^2})^2}}{x} )-[\ln (\frac{x/\sqrt3}{x})+\frac{\sqrt{x^2+(x/\sqrt3)^2}}{x} )]\]
lgbasallote
  • lgbasallote
darn my teacher -_-
anonymous
  • anonymous
\[ \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2x}}}{x} )-[\ln ((\frac{1}{\sqrt{3}})+\frac{\sqrt{4x^2}}{x} )]\] \[ \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )-[\ln ((\frac{1}{\sqrt{3}})+2 )]\] So now, it's upto you to integrate \[\int\limits_{0}^{3/2}\ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )dx-\int\limits_{0}^{3/2}\ln ((\frac{1}{\sqrt{3}})+2 )dx\]
lgbasallote
  • lgbasallote
ugh can i still quit learning this?
lgbasallote
  • lgbasallote
you just succeeded i hurting my poor brain :p
anonymous
  • anonymous
Try wolfram :P The second integral is easy. I have no idea on first part :P
lgbasallote
  • lgbasallote
@.Sam. said wolfram cant do it
anonymous
  • anonymous
@jazy, what was that ??
anonymous
  • anonymous
@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
anonymous
  • anonymous
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
lgbasallote
  • lgbasallote
ugh enough =_=

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