## anonymous 5 years ago find the intergral of 1/((1+9(x^2))^2)

1. anonymous

What have you tried?

2. anonymous

factor out the 9, youre going to get arctan something

3. anonymous

alright i know what i need to do i need to use trig subsitution and i wasn't sure 1/((1+9(x^2))^4/2) or should i just use tan i mean what is going to be the co-effiecinet but that seems incorrect

4. anonymous

oh nevermind, i didnt see the square . you can do trig substitution

5. anonymous

ohk i can factor out the 9 that seems a good think thanks i just wasn't doing that :| i guess

6. anonymous

whoa, where did 4/2 come from?

7. anonymous

4/2 is the same thing as 2 so when i see all the examples in book they have roots under the section of trig subsitution

8. anonymous

dont use 4/2, thats weird

9. anonymous

It is weird. Very weird. Use a tan sub, as above.

10. anonymous

if youre using an integral table look for 1 / ( x^2 + a^2)^2 dx

11. anonymous

yup i am using a tan sub just sure what to put as a coefficient for tan

12. anonymous

Integral tables are for babies.

13. anonymous

yes that i what i am following cantroset but in that the a will 1/3

14. anonymous

errr you can have 1 / ( ax^2 + b^2) ^2 dx,

15. anonymous

i know that formula :P

16. anonymous

17. anonymous

:) so 1/3 should be my coefficient right :)

18. anonymous

lets use trig sub just to be thorough, i use pauls online math course, here one sec i will post link

19. anonymous

$\text{Let } x = \frac{1}{3}\cdot \tan \alpha$

20. anonymous

Paul is the man!! http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx scroll down

21. anonymous

yes why is that a 1/3 that is what i am lost about shouldn't it be 1/9

22. anonymous

ok one sec, i hate this thing it keeps freezing. yeesh why doesnt google help with the appearance of this

23. anonymous

the 1/3 is squared to 1/9

24. anonymous

ohk so it is (1+ 1/3 tan (theta))^2

25. anonymous

(1+ 1/3 tan (theta)^2)^2

26. anonymous

oh i think i see what youre doing, youre forcing it to be a square root ok sqrt a^2 + b^2x^2 , let x = a/b tan t

27. anonymous

$1+9x^2 = 1+ 9\left(\frac{1}{3}\tan \alpha \right)^2 = 1+\tan^2 \alpha$

28. anonymous

so you want ( 1 + 9x^2)^2 = [ (1+9x^2)^4] ^1/2 ?

29. anonymous

i think newton thats simpler,

30. anonymous

paul seems to say you want a square root though

31. anonymous

oh that makes so much more sense :)

32. anonymous

you don't really need to have it i guess you can force it but you don't need it you need it for inital i guess it won't really change anything ( i am referring to the square root)

33. anonymous

xD how on earth does that make more sense. Crazy Americans.

34. anonymous

you can also do partial fractions i bet

35. anonymous

i mean what you wrote.

36. anonymous

yeah i can do partial fraction, but i am required to do trig :)

37. anonymous

Oh, good. And you can't split it in to partial fractions (I don't think).

38. anonymous

39. anonymous

why not?

40. anonymous

Because it's the same term squared on the bottom and no others.

41. anonymous

But assuming you could put it into partial fractions, it would not simplify it at all.

42. anonymous

43. anonymous

(9x^2+1)^2 = (tan(a)^2+1) = sec(a)^4 is this correct

44. anonymous

but probably not the easiest approach. youre tan sub is fine , one sec

45. anonymous

No, because it would be $\frac{ax + b}{1+9x^2} + \frac{cx + d}{\left(1+9x^2\right)^2}$ But they all equal 0 except fro d = 1, because ... etc

46. anonymous

avnis yes

47. anonymous

Now include the change of what you are integrating with respect to and go from there.

48. anonymous

ohk so next 1/sec(a)^4 = cos(a)^4

49. anonymous

Yes, but$\frac{\mathbb{d}\alpha}{\mathbb{d}x}$ don't forget to include

50. anonymous

not if you use complex root factoring

51. anonymous

what

52. anonymous

avnis just talking to newton

53. anonymous

Oh, OK, you can;t split it up into partial fractions in a way which would make it a retardedly different problem - is that better?

54. anonymous

newton, it would be like partial decomp'ing 1/(x^2 + 1) = A/ (x+i) + B/ ( x -i)

55. anonymous

i prefer approaches , one size fits all. hehe, the one size hammer that is

56. anonymous

how do i solve the rest of it

57. anonymous

@anvis If you are now integrating with respect something other than x, you need to account for that. Note: $x = \frac{1}{3}\tan\alpha \implies \frac{\mathbb{d}x}{\mathbb{d}\alpha} = \frac{1}{3}\sec^2 \alpha$

58. anonymous

i see which will make a equal to tan^-1(3x)

59. anonymous

so i wll be left with 1/3 integral of cos^2(a)

60. anonymous

It will, but you don't need to include that yet. $\int\limits \frac{1}{(1+9x^2)^2} \mathbb{d}x$ becomes $\int\limits \frac{1}{(1+\tan^2\alpha)^2} \cdot \frac{1}{3} \sec^2 \alpha \cdot \mathbb{d}a = \int \frac{1}{3} \cos^2\alpha\ \cdot \mathbb{d}a$ yes.

61. anonymous

Now do whatever you feel like with cos^2 to make it easier to integrate, sub back in x with the tan^(-1) thing you had above and you're done.

62. anonymous

1/3inetgarl (1/2 (cos 2(a)+1/2) du

63. anonymous

great job

64. anonymous

you have an extra 1/2 there

65. anonymous

oh yes

66. anonymous

Sorry gang, got to go. But here's a nice one for you to try if you feel up to it: By finding constants a, b, c and d such that: $\frac{ax+b}{x^2+2x+2} + \frac{cx+d}{x^2-2x+2} \equiv \frac{1}{x^4+4}$ Show $\int_0^1\frac{1}{x^4+4} = \frac{1}{16}\ln 5 + \frac{1}{8}\tan^{-1}2$

67. anonymous

ill solve it

68. anonymous

hmm so i got a/6 + (sin 2a)+12 and i know that a= tan^-1(3x) how i do i solve that

69. anonymous

you make a triangle

70. anonymous

what would the side be would one side be 3x

71. anonymous

...

72. anonymous

that is what i got

73. anonymous

but there is this online answer from wolfaram and it doesn't match it so i am so lost

74. anonymous

What did you get as sin(2x) ?

75. anonymous

2sin(x)cos(x)

76. anonymous

OK, what what did you get as sin(x) and cos(x)? Hint, use the trinagle I uploaded, with the facts opp/hyp = sin(x) and adj/hpy = cos(x) NOTE these should all be as, NOT xs, sorry for the confusion.

77. anonymous

ya that is why i was thinking i don't have x:|

78. anonymous

sin(a) =3x/ sqrt (9x^2+1)

79. anonymous

Yeah the typing on my diagram should be a, too. Indeed. And cos(a)? And then 2sin(2a) ?

80. anonymous

sorry, only sin(2a), not 2 sin(2a)

81. anonymous

cos (a)= 1/ sqrt(9x^2+1)

82. anonymous

:D

83. anonymous

got it :)

84. anonymous

Woo

85. anonymous

i have learnt alot today :)

86. anonymous

Good to hear. Goodbye.

87. anonymous

thanks :)