## anonymous 5 years ago Compute the integral of 1/(x*sqrt(x^3-25)) Hint: let u=x^(3/2)

1. anonymous

I don't know that your hint gets you there. Who gave the hint?

2. anonymous

it is on the review sheet

3. anonymous

i know the answer too if that helps. its 2/15 arcsec(x^(3/2)/5)+c

4. anonymous

In doing it, they manipulated in a way to use trig method.

5. anonymous

Yeah, you have to brush up on your trig method (some variation of tan^2=sec2-1). Using hint x under the square root turns to sq root of u^2-25 and you can do the trig method.

6. anonymous

put (x^(3/2)-25)=t^2

7. anonymous

use this substitution

8. anonymous

I think he wants us to use the trig method. I just keep getting stuck. I have u=x^(3/2) du=3/2sqrt(x)dx 2/3du=sqrt(x)dx I would be able to solve from there but i dont have a sqrt(x)

9. anonymous

use my method and multiply denominator and numerator by x^0.5 and put x^(3/2) as t^2+25

10. anonymous

He's going to test us on the trig method...

11. anonymous

who?

12. anonymous

the teacher

13. anonymous

oh...but even ur hint uses substitution

14. anonymous

right, but isnt there a difference between trig substitution and regular substitution?

15. anonymous

if u want to use trig then use sec^(2/3)t=x

16. anonymous

got it?

17. anonymous

not really. Can you go through the process step by step?

18. anonymous

use (25*sec (t))^(4/3)=x so x^(3/2)=25*sec^2(t) so dx=25^(4/3)*(4/3)*sec^(1/3)(t)*sec(t)*tan(t)

19. anonymous

dx=c*sec^(4/3)(t)*tan(t) dt c=constant

20. anonymous

now solve further i m very slow in typing

21. anonymous

Jas, give me a minute. I have the method his instructor is expecting.

22. anonymous

OK, using hint $u=x ^{3/2}$Convert this all the way through you get $u ^{2/3}=x$

23. anonymous

if u will solve my expression then it will become simplified

24. anonymous

Yes, but he is in school, he is learning a really helpful method, technique.

25. anonymous

Just hold on one minute, let me finish, so the work don't get all muddled.

26. anonymous

i have already told similar method before but he wants trig method

27. anonymous

28. anonymous

$2/3 du =\sqrt{x}$

29. anonymous

if u will use x^(3/2)-25=t^2 then it will be more easier

30. anonymous

New$\int\limits_{?}^{?}[(2/3) du]/[(u ^{2/3}\sqrt{u ^{2}-25}$

31. anonymous

what happened to 2/3du=sqrt(x)?

32. anonymous

it is 2/3du=sqrt(x)dx

33. anonymous

$\int\limits_{?}^{?}[(2/3) du]/25\sec \theta \tan \theta$

34. anonymous

I guess that should be derivative theta

35. anonymous

i mean, what happens to the sqrt(x)?

36. anonymous

multiply deno and nume by x and cha... made mistake in calculation

37. anonymous

In the substitution there is an extra $\sqrt{5 \sec \theta}$ and that is = to sq rt of x. That is swallowed up in dtheta.

38. anonymous

OK, I just noticed the square root of x is in bottom. How do I get it on top to be swallowed by dtheta, Jas?

39. anonymous

you solve complete expression in u first it will be 1/u*(......)

40. anonymous

(.....)=u^2-25

41. anonymous

now put u=sec(t)

42. anonymous

One second, there is a fix for this, don't change it too much or uswag would get lost in translation.

43. anonymous

otherwise change directly which i already proposed earlier

44. anonymous

OK, that is already wiped out uswag. when I bring sq rt x and dx on top, I have an extra sq rt x in bottom, they cancel each other out.

45. anonymous

there is no sqrt but there is x

46. anonymous

Yes, but that x is converted to u^(2/3). In my conversion I keep u^2=5 sec theta and I wipe out sq root 5 theta or sq rt x.

47. anonymous

Uswag, you seem to be following along. Does this make any sense?

48. anonymous

k...i was doing in other way

49. anonymous

i need to go

50. anonymous

Can we start the explanation from 2/3 du/ u^(2/3)(sqrt(u^2-25)..... i got lost in the explanation after that. so its 2/3 du= sqrt(x)dx so we have sqrt(x)dx on top.

51. anonymous

Thanks Jas, you been a great help.

52. anonymous

Thanks Jas

53. anonymous

thanks 4 being my fan

54. anonymous

${[2/(3\sqrt{x})]du}/[\sqrt{x}u ^{2}\sqrt{u ^{2}-25}]$

55. anonymous

$2/75\int\limits_{?}^{?}d \theta/\sec \theta \tan \theta$

56. anonymous

Now you have a function you can integrate. Once you integrate you have to go back and change to x. These trig techniques are explained nicely in MIT lectures http://www.youtube.com/watch?v=CXKoCMVqM9s&feature=BFa&list=PL590CCC2BC5AF3BC1&index=25

57. anonymous

I'll write down some of the substitutions I made$u^{2}-25=5^{2}\sec ^{2}\theta-5^{2}$

58. anonymous

And that is equal to $5^{2}\tan ^{2}\theta$

59. anonymous

$u =5\sec \theta$

60. anonymous

$\sqrt{u ^{2}-25}=\sqrt{5^{2}\sec ^{2}\theta-5^{2}}=\sqrt{5^{2}\tan ^{2}\theta}=5\tan \theta$

61. anonymous

A lot of info, makes sense?

62. anonymous

yeah, after seeing the info plus that it does. Thanks!

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