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## anonymous 4 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

The region is between a circle and a line, apparently.

2. anonymous

lol no graphing :p just find the integral

3. blockcolder

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

4. blockcolder

$$y=x\sec{\theta}$$ would be the substitution I would use but the bounds are horrible.

5. anonymous

that's what i did too and got $\ln (\sec \theta + \tan \theta)$

6. blockcolder

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

7. anonymous

?!

8. blockcolder

Huh? Which part is bugging you?

9. anonymous

the substitution of theta thingy...i dont know that

10. blockcolder

The bounds, you mean?

11. anonymous

yeah

12. 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

13. anonymous

Sorry I meant keeping x as constant

14. anonymous

uhh i think i got a weid answer for that @_@ with ln and such

15. .Sam.

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

16. anonymous

Then after you do upper limit - lower limit, integrate with dx with the limits 0 to 3/2

17. anonymous

uhmm yep i got that...then i dont know how to integrate it

18. anonymous

@lgba, make sure you substitute back u = arctan(y/x) before taking upper limit-lower limit

19. anonymous

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

20. anonymous

Now take upper limit - lower limt

21. .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

22. .Sam.

$\tiny \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)$

23. anonymous

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

24. anonymous

wow those are small @.Sam.

25. .Sam.

wolfram doesn't know how to integrate this, lol

26. anonymous

$\small \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{x^2+(\sqrt{2x-x^2})^2}}{x} )-[\ln (\frac{x/\sqrt3}{x})+\frac{\sqrt{x^2+(x/\sqrt3)^2}}{x} )]$

27. anonymous

darn my teacher -_-

28. 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 $\int\limits_{0}^{3/2}\ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )dx-\int\limits_{0}^{3/2}\ln ((\frac{1}{\sqrt{3}})+2 )dx$

29. anonymous

ugh can i still quit learning this?

30. anonymous

you just succeeded i hurting my poor brain :p

31. anonymous

Try wolfram :P The second integral is easy. I have no idea on first part :P

32. anonymous

@.Sam. said wolfram cant do it

33. anonymous

@jazy, what was that ??

34. 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

35. 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

36. anonymous

ugh enough =_=

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