## lgbasallote 2 years ago $\large \int_0^{3/2} \int_{x/\sqrt 3}^{\sqrt{2x - x^2}} \frac{dydx}{\sqrt{x^2 + y^2}}$ help?

1. blockcolder

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

2. lgbasallote

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. lgbasallote

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. lgbasallote

?!

8. blockcolder

Huh? Which part is bugging you?

9. lgbasallote

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

10. blockcolder

The bounds, you mean?

11. lgbasallote

yeah

12. shivam_bhalla

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

Sorry I meant keeping x as constant

14. lgbasallote

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. shivam_bhalla

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

17. lgbasallote

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

18. shivam_bhalla

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

19. shivam_bhalla

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

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. shivam_bhalla

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

wow those are small @.Sam.

25. .Sam.

wolfram doesn't know how to integrate this, lol

26. shivam_bhalla

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

27. lgbasallote

darn my teacher -_-

28. shivam_bhalla

$\ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2x}}}{x} )-[\ln ((\frac{1}{\sqrt{3}})+\frac{\sqrt{4x^2}}{x} )]$ $\ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )-[\ln ((\frac{1}{\sqrt{3}})+2 )]$ 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

ugh can i still quit learning this?

30. lgbasallote

you just succeeded i hurting my poor brain :p

31. shivam_bhalla

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

32. lgbasallote

@.Sam. said wolfram cant do it

33. shivam_bhalla

@jazy, what was that ??

34. shivam_bhalla

@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

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

ugh enough =_=