Question

find the intergral of 1/((1+9(x^2))^2)

Answer 1

What have you tried?

Answer 2

factor out the 9, youre going to get arctan something

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

whoa, where did 4/2 come from?

Answer 7

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

Answer 8

dont use 4/2, thats weird

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

Integral tables are for babies.

Answer 13

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

Answer 14

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

Answer 15

i know that formula :P

Answer 16

whats your table say

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

the 1/3 is squared to 1/9

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

i think newton thats simpler,

Answer 30

paul seems to say you want a square root though

Answer 31

oh that makes so much more sense :)

Answer 32

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)

Answer 33

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

Answer 34

you can also do partial fractions i bet

Answer 35

i mean what you wrote.

Answer 36

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

Answer 37

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

Answer 38

here integrals involving quadratics
http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithQuadratics.aspx

Answer 39

why not?

Answer 40

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

Answer 41

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

Answer 42

so those are your factors irreducible quadratic

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

avnis yes

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

not if you use complex root factoring

Answer 51

what

Answer 52

avnis just talking to newton

Answer 53

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?

Answer 54

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

Answer 55

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

Answer 56

how do i solve the rest of it

Answer 57

@anvis

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

Answer 58

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

Answer 59

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

Answer 60

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.

Answer 61

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.

Answer 62

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

Answer 63

great job

Answer 64

you have an extra 1/2 there

Answer 65

oh yes

Answer 66

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
\[\int_0^1\frac{1}{x^4+4} = \frac{1}{16}\ln 5 + \frac{1}{8}\tan^{-1}2\]

Answer 67

ill solve it

Answer 68

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

Answer 69

you make a triangle

Answer 70

what would the side be would one side be 3x

Answer 71

...

Answer 72

that is what i got

Answer 73

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

Answer 74

What did you get as sin(2x) ?

Answer 75

2sin(x)cos(x)

Answer 76

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.

Answer 77

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

Answer 78

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

Answer 79

Yeah the typing on my diagram should be a, too.

Indeed. And cos(a)? And then 2sin(2a) ?

Answer 80

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

Answer 81

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

Answer 82

:D

Answer 83

got it :)

Answer 84

Woo

Answer 85

i have learnt alot today :)

Answer 86

Good to hear. Goodbye.

Answer 87

thanks :)