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

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blockcolderBest ResponseYou've already chosen the best response.0
The region is between a circle and a line, apparently.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
lol no graphing :p just find the integral
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
What? No graphing? o_O I'd convert to polar coordinates but apparently that's thrown out of the window.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
\(y=x\sec{\theta}\) would be the substitution I would use but the bounds are horrible.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
that's what i did too and got \[\ln (\sec \theta + \tan \theta)\]
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
I meant \(y=x \tan\theta\). Trying it anyway... \[y=\sqrt{2xx^2}\Rightarrow \theta=\tan^{1}\left ( \frac{\sqrt{2xx^2}}{x}\right )=\alpha\\ y=\frac{x}{\sqrt3} \Rightarrow \theta=\frac{\pi}{6}\\ dy=x\ \sec^2\theta d\theta\\ \Large \int_0^{3/2}\int_{\pi/6}^\alpha \frac{x \sec^2\theta d\theta dx}{x \sec\theta}\]
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
Huh? Which part is bugging you?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
the substitution of theta thingy...i dont know that
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
The bounds, you mean?
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
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
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
Sorry I meant keeping x as constant
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
uhh i think i got a weid answer for that @_@ with ln and such
 one year ago

.Sam.Best ResponseYou've already chosen the best response.1
\[\int\limits \frac{1}{\sqrt{x^2+y^2}} \, dy\] Let \[y=x \tan (u) ~~and ~~dy=x \sec ^2(u) d u\] \[u=\tan^{1}(\frac{y}{x})\] \[\text{Then}\sqrt{x^2+y^2}\text{ =}\sqrt{x^2 \tan ^2(u)+x^2}\text{ = }x \sec (u)\] \[\int\limits \sec (u) \, du\] \[[\ln (\tan (u)+\sec (u))]_{x/\sqrt3}^{\sqrt{2xx^2}}\]
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
Then after you do upper limit  lower limit, integrate with dx with the limits 0 to 3/2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
uhmm yep i got that...then i dont know how to integrate it
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
@lgba, make sure you substitute back u = arctan(y/x) before taking upper limitlower limit
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
Contiuing from @.Sam. 's work so basically you get \[[\ln (\tan (\arctan(y/x))+\sec (\arctan(y/x)))]_{x/\sqrt3}^{\sqrt{2xx^2}}\] \[[\ln (y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2xx^2}}\] \[[\ln (y/x)+(\sqrt{x^2+y^2}/x)]_{x/\sqrt3}^{\sqrt{2xx^2}}\]
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
Now take upper limit  lower limt
 one year ago

.Sam.Best ResponseYou've already chosen the best response.1
\[\small \log \left(\tan \left(\tan ^{1}\left(\frac{\sqrt{2 xx^2}}{x}\right)+\sec \left(\tan ^{1}\left(\frac{\sqrt{2 xx^2}}{x}\right)\right)\log \left(\tan \left(\tan ^{1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\] lol
 one year ago

.Sam.Best ResponseYou've already chosen the best response.1
\[\tiny \log \left(\tan \left(\tan ^{1}\left(\frac{\sqrt{2 xx^2}}{x}\right)+\sec \left(\tan ^{1}\left(\frac{\sqrt{2 xx^2}}{x}\right)\right)\log \left(\tan \left(\tan ^{1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\]
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
Well sorry I messed up the bracket's earlier. It should be \[[\ln ((y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2xx^2}}\] \[[\ln ((y/x)+\frac{\sqrt{x^2+y^2}}{x} )]_{x/\sqrt3}^{\sqrt{2xx^2}}\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
wow those are small @.Sam.
 one year ago

.Sam.Best ResponseYou've already chosen the best response.1
wolfram doesn't know how to integrate this, lol
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
\[\small \ln ((\sqrt{2xx^2}/x)+\frac{\sqrt{x^2+(\sqrt{2xx^2})^2}}{x} )[\ln (\frac{x/\sqrt3}{x})+\frac{\sqrt{x^2+(x/\sqrt3)^2}}{x} )]\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
darn my teacher _
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
\[ \ln ((\sqrt{2xx^2}/x)+\frac{\sqrt{{2x}}}{x} )[\ln ((\frac{1}{\sqrt{3}})+\frac{\sqrt{4x^2}}{x} )]\] \[ \ln ((\sqrt{2xx^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )[\ln ((\frac{1}{\sqrt{3}})+2 )]\] So now, it's upto you to integrate \[\int\limits_{0}^{3/2}\ln ((\sqrt{2xx^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )dx\int\limits_{0}^{3/2}\ln ((\frac{1}{\sqrt{3}})+2 )dx\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
ugh can i still quit learning this?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
you just succeeded i hurting my poor brain :p
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
Try wolfram :P The second integral is easy. I have no idea on first part :P
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
@.Sam. said wolfram cant do it
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
@jazy, what was that ??
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
@lgbasallote , wolfram works :p http://www.wolframalpha.com/input/?i=integral+of+ln%28%28sqrt%282xx^2%29%29%2Fx+%2B+sqrt%282%2Fx%29%29dx
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
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%282xx%5E2%29%29%2Fx+%2B+sqrt%282%2Fx%29%29dx
 one year ago
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