## anonymous 5 years ago find dy/dx for y=integral(arctan)tdt from (1/x,4x)

1. amistre64

$y = \int\limits_{} \tan^{-1}(t) dt$ is this the equation?

2. anonymous

yes but it is a definite integral 1/x to 4x

3. amistre64

we might have to convert this to a trig of "u"... maybe. I dont know any derivatives that become an arctan...

4. amistre64

if we let u = tan^-1(t) then; t = tan(u) ; dt = sec^2(u) du which gives us... [S] u sec^2(u) du ; to solve

5. amistre64

does that make sense so far?

6. amistre64

i wonder if we can decompose this......

7. amistre64

I dont see any easy ways of doing this :)

8. anonymous

i think i'm just making it more difficult than it needs to be in some way

9. anonymous

do you understand piecewise functions?

10. amistre64

is there an answer that you can chaeck against?

11. amistre64

i know if piecewise functions; but havenet come across a way to integrate them

12. anonymous

mmk and there isn't an answer to check :/ thanks for trying though!

13. anonymous

.-. I don't think that's right amistre though I didn't try it myself lol >_<

14. anonymous

wh are you letting tan^-1(t) = u? ._.

15. amistre64

this is prolly an integration by parts method; but I dont know much about when it goes into multiple integration by parts :)

16. amistre64

becasue there is no function that derives down to a tan^-1

17. amistre64

it is equivalent to: [S] u sec^2(u) du tho.... and thats doable somehow

18. anonymous

integration by parts is something like this:$\int\limits f'gdx = fg - \int\limits fg'dx$ where f' = sec^2(u) and g = u then find f and g' and plug it in the equation ^_^

19. anonymous

lol I got you amistre :)

20. amistre64

int by parts is still a mystery to me for some functions :)

21. anonymous

it's alright ^_^ all you have to do is find f and g' and then plug it in the equation, with that, you have solved the problem. g' = 1 and f = tan x then you'll get : $\int\limits f'gdx = (\sec^2(u))(u) - \int\limits \tan(u) du$ then just find the integral lol

22. amistre64

i usually do a table like this: | v up ----------- u dn | sec^2 ----------- +u | tan ----------- -1 | ?? ---------- +0 | N/A ---------- And so I get for the integral: u tan(u) - 1(????)

23. anonymous

$\int\limits f'gdx = (\sec^2(u))(u) + \ln|\cos(u)| + c$ I hope I'm right, I just followed your given lol

24. anonymous

O_O what kind of table is THAT LOL

25. amistre64

its an integration by parts table; but it lines everything up as tho it was multiple integration by parts :)

26. anonymous

.-. why so complicated? LOL

27. anonymous

you can do it my way :) much easier ^_^

28. amistre64

its easier when doing multiple integration by parts; which is all the table amounts to. You shift between + and - your u derives down and your v suits up.... its the same process :) just a different format to keep everything in line :)

29. anonymous

>_< lol, I've learned something new today! thank you :)

30. amistre64

:) all I need is an integral for tan(u) and I got it solved lol

31. amistre64

the table of integration says: tan(x) ints to ln(sec(x)) cool :)

32. amistre64

u tan(u) - ln(sec(u))

33. anonymous

hmm: $\tan x = \frac{sinx}{cosx}$ let u = cos x and du = -sinx and solve lol

34. amistre64

we re-stitute for u and get: tan^-1(t) tan(tan^-1(t)) - tan(tan^-1(t))

35. anonymous

.-.

36. amistre64

well...... ln(sec(tan^-1(t))) lol got excited ;)

37. anonymous

not convinced with the sec u amistre :(

38. amistre64

can always try to derive it back ;)

39. anonymous

tan x = sinx/cosx integrating that will give you ln|cosx|

40. anonymous

>_<lol

41. anonymous

$\tan ^{-1}t=u, tanu = t$ $dt=\sec ^{2}u du=(1 + \tan ^{2}u)d u$

42. anonymous

._.

43. anonymous

students have faced such a question last semester, thankfully I wasn't there lol , anyhow, thanks for your time amistre and uzma ^_^ but I gotta study communications , take care both :)

44. anonymous

:) you too

45. anonymous

:)

46. anonymous

$\int\limits_{?}^{?}\tan ^{-1}u=\int\limits_{?}^{?}1.\tan ^{-1}u$

47. anonymous

integrating by parts $\tan ^{-1}u.u-\int\limits_{?}^{?}u.1/(1+u ^{2})d u$

48. anonymous

$utan ^{-1}u -2 \int\limits_{?}^{?}2u/(1+u ^{2})$

49. amistre64

if I did it right; mine derives back to tan^-1(t)....

50. anonymous

$utan ^{-1}u - 2 \ln (1+u ^{2})$

51. anonymous

i used integration by parts method, not substitution

52. amistre64

$D[\tan^{-1}(t).\tan(\tan^{-1}(t)) - \ln(\sec(\tan^{-1}(t)))]$ $\frac{\tan^{-1}(t)\sec^2(\tan^{-1}(t))}{t^2 +1}+\frac{\tan(\tan^{-1}(t))}{t^2+1}- \frac{\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}$

53. amistre64

$\frac{\tan(\tan^{-1}(t)).\sec^2 (\tan^{-1}(t)).\sec(\tan^{-1}(t))+\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))-\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}$

54. amistre64

all that work and it cuts it short lol...; that last part is zeros the middle part to get:$\frac{\tan^{-1}(t).\sec^2(\tan^{-1}(t)).\sec(\tan^{-1}())}{\sec(\tan^{-1}(t)).(t^2+1)}$ had to fix a typo in that second round... it aint a tan(tan^-1(t)))

55. anonymous

so the result?

56. amistre64

$\frac{\tan^{-1}(t).\sec(^2(\tan^{-1}(t))}{t^2+1}$

57. anonymous

n this is what u wanted?

58. anonymous

i guess i should terminate at Tan^(-1)t?

59. amistre64

since t^2+2 = sec^2(a) = sec^2(tan^-1(t)); those cancel and we are left with.. tan^-1(t)

60. amistre64

...since t^2 + 1.....typoed it :)

61. anonymous

hmmmm,right :)

62. anonymous

did u check mine?is it wrong?

63. amistre64

i had enough trouble proofing mine lol...my computer is bogging with all thisequation editor stuff....lol

64. anonymous

yup...hats off to ur patience :P

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