## precal Group Title Integration problem 2 years ago 2 years ago

1. precal

|dw:1349015981615:dw|

2. bahrom7893

This looks like one of the trig substitutions.

3. bahrom7893

or maybe integration by parts would work.

4. precal

|dw:1349016003653:dw|

5. precal

so I will rewrite it as|dw:1349016056313:dw|

6. precal

Am I doing it correct so far?

7. bahrom7893

I think it's best if we leave it inside. Something tells me that will have to stay as u^(3/2)

8. precal

then I don't see with trig sub I am suppose to use

9. bahrom7893

oh nvm

10. justine027

Nope . Use Inverse sec :) or Sine? I forgot what to use but simple as that :D

11. bahrom7893

It is trig, u-sub won't work. Yeah keep it out, it's inverse sin of 6x i believe

12. precal

|dw:1349016200316:dw|

13. justine027

Its nerp it is sin inverse x/6 +C :)

14. precal

|dw:1349016274224:dw| is what my solution is suppose to be. Of course I am more interested in how to get to the solution

15. Callisto

I think we should use sec instead of sin?! Let x = 6sec (theta)

16. bahrom7893

36Sec^2(theta) - 36 = 36 tan^2(theta)

17. bahrom7893

so the bottom is: sqrt (36 tan^2(theta)) = 6tan^2(theta), yea callisto's right

18. bahrom7893

and top is: dx = 6Sec(theta)Tan(theta)dtheta

19. precal

so is my given solution incorrect?

20. bahrom7893

yea precal

21. precal

I have some integrals at the backof my book "Integral involving x^2-a^2; a>0"

22. siddhantsharan

$\int\limits \frac{ 6secutanudu }{ (36\sec^2u - 36)^{3/2} } = \int\limits \frac{ 6secu tanu du }{ 6^3 \tan^3u }$ $1/36 \int\limits secdu/\tan^2u$ $1/36 \int\limits \frac{ cosudu }{ \sin^2u? }$

23. Callisto

cos u/ sin^2 u = cscu cotu $\int \csc u \cot udu = -cscu +C$ Time to draw a triangle...

24. precal

time to draw a triangle? how is that going to help....I wonder if we are taking the correct approach. I was given problems to study for an upcoming exam. Are we on the correct path?

25. precal

I don't see how this is going to get me the solution I was given at all????

26. siddhantsharan

Take sinu as t. => -dt = cosudu $1/36 \int\limits \frac{ -dt }{ t^2 } = \frac{ 1 }{ 36t } + C = cosecu/36 + C = \frac{ \sqrt6 }{ 36\sqrt{x^2 - 6} }$

27. precal

|dw:1349017386008:dw|

28. alanli123

LOL ITS FUNNY CUZ UR NAME IS PRECAL AND UR DOING CAL LOL

29. Callisto

|dw:1349017310113:dw| since we let x = 6 secu secu = x/6 cosu = 6/x You can find the opposite side, hence sin u, and cscu in terms of x

30. precal

gotta study all types of math. I am just more comfortable at the precal and below level

31. siddhantsharan

Yeaah my bad. Thats $\sqrt x$ NOT $\sqrt 6$

32. precal

|dw:1349017504869:dw|

33. Callisto

Yes. And what is sin u and cscu?

34. precal

|dw:1349017612360:dw|

35. Callisto

And cscu?

36. precal

|dw:1349017652671:dw|

37. Callisto

And remember we got $-\frac{1}{36}cscu+C$For the integral?

38. precal

|dw:1349017768354:dw|

39. siddhantsharan

ALTERNATE: $x^2 - 36 = t^2 =>tdt=xdx$ $I = \int\limits \frac{ tdt }{ t^3\sqrt{t^2 + 36}} = \int\limits \frac{ dt }{t^3 \sqrt{ 1 + \frac{ 36 }{ t^2 }} }$

40. precal

Thanks Callisto, you were correct (back to the basics)

41. Callisto

You're welcome :)

42. precal

that is in reference to drawing the triangle. Yes, I got my solution but now I have to go back and study the process

43. siddhantsharan

Take: $1 + \frac{ 36 }{ t^2 } = u = > du = -36dt/t^3$

44. siddhantsharan

You would get the same answer again after the sub. :)

45. precal

@siddhantsharan I am always interested in the alternate approach, especially if it gets me to the answer quicker

46. precal

thanks everyone, gotta take a break and look at this later........

47. Callisto

Perhaps it's easier for you to read. (PS: I'm not giving direct answer!!) First Method: By Trigo Sub.$\int \frac{dx}{(x^2 -36)^\frac{3}{2}}$Let x=6sec u; dx = 6secu tanu du. Then, the integral becomes$=\int \frac{6secu \ tanu\ du}{((6secu)^2 -36)^\frac{3}{2}}$$=\int \frac{6secu \ tanu\ du}{[36(sec^2u -1)]^\frac{3}{2}}$$=\int \frac{6secu \ tanu\ du}{[36(tan^2u)]^\frac{3}{2}}$$=\frac{1}{36}\int \frac{secu \ tanu\ du}{tan^3u}$$=\frac{1}{36}\int cscutanudu$$=-\frac{1}{36}\csc u+C$ Since x=6secu , secu = x/6 , cosu = 6/x By Pyth. Thm., opposite side is $$\sqrt{x^2-6^2}=\sqrt{x^2-36}$$ So, cscu = $$\frac{x}{\sqrt{x^2-36}}$$. Substituting this into the last step, you can get your answer.

48. Callisto

Second method: by substitution, suggested by @siddhantsharan $\int \frac{dx}{(x^2 -36)^\frac{3}{2}}$Let x^2-36 = t^2 ; 2x dx = 2tdt => x dx = t dt => $$dx = \frac{t dt}{ x} = \frac{t dt}{ \sqrt{t^2+36}}$$ PS: we already know x $$\ge$$ 6, so it should be +ve sqrt The integral becomes $\int \frac{tdt}{\sqrt{(t^2 +36)}(t^3)}$$=\int \frac{\frac{tdt}{t}}{\sqrt{\frac{(t^2 +36)}{t^2}}(t^3)}$ $=\int \frac{dt}{\sqrt{1+\frac{36}{t^2}}(t^3)}$Let u = 1+ (36/t^3) ; du = $$\frac{-72}{t^3}$$dt The integral becomes $=-\frac{1}{72}\int \frac{du}{\sqrt{u}}$$=-\frac{1}{72}(2\sqrt{u}) +C$$=-\frac{1}{36}(\sqrt{u}) +C$ u= 1+ (36/t^2) And t^2 = x^2-36, So,$$u = 1 +\frac{36}{x^2-36} = \frac{x^2}{x^2-36}$$ Sub. $$u = \frac{x^2}{x^2-36}$$ into the last step, and simplify it, you should be able to get the answer. It's so amazing :D

49. siddhantsharan

@Callisto Clarified succinctly. Really well done :D

50. precal

Thanks you are so awesome :) Both of you, wish I could give more than 1 medal