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

- anonymous

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

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- anonymous

What have you tried?

- anonymous

factor out the 9, youre going to get arctan something

- 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

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## More answers

- anonymous

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

- anonymous

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

- anonymous

whoa, where did 4/2 come from?

- 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

- anonymous

dont use 4/2, thats weird

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Integral tables are for babies.

- anonymous

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

- anonymous

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

- anonymous

i know that formula :P

- anonymous

whats your table say

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

the 1/3 is squared to 1/9

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- anonymous

i think newton thats simpler,

- anonymous

paul seems to say you want a square root though

- anonymous

oh that makes so much more sense :)

- 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)

- anonymous

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

- anonymous

you can also do partial fractions i bet

- anonymous

i mean what you wrote.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

why not?

- anonymous

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

- anonymous

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

- anonymous

so those are your factors irreducible quadratic

- anonymous

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

- anonymous

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

- 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

- anonymous

avnis yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

not if you use complex root factoring

- anonymous

what

- anonymous

avnis just talking to newton

- 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?

- anonymous

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

- anonymous

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

- anonymous

how do i solve the rest of it

- anonymous

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

- anonymous

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

- anonymous

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

- 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.

- 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.

- anonymous

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

- anonymous

great job

- anonymous

you have an extra 1/2 there

- anonymous

oh yes

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

- anonymous

ill solve it

- anonymous

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

- anonymous

you make a triangle

- anonymous

what would the side be would one side be 3x

- anonymous

...

##### 1 Attachment

- anonymous

that is what i got

- anonymous

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

- anonymous

What did you get as sin(2x) ?

- anonymous

2sin(x)cos(x)

- 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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

:D

- anonymous

got it :)

- anonymous

Woo

- anonymous

i have learnt alot today :)

- anonymous

Good to hear. Goodbye.

- anonymous

thanks :)

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