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Mathematics 47 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

What have you tried?

OpenStudy (anonymous):

factor out the 9, youre going to get arctan something

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

whoa, where did 4/2 come from?

OpenStudy (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

OpenStudy (anonymous):

dont use 4/2, thats weird

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Integral tables are for babies.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i know that formula :P

OpenStudy (anonymous):

whats your table say

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

the 1/3 is squared to 1/9

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i think newton thats simpler,

OpenStudy (anonymous):

paul seems to say you want a square root though

OpenStudy (anonymous):

oh that makes so much more sense :)

OpenStudy (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)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you can also do partial fractions i bet

OpenStudy (anonymous):

i mean what you wrote.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

why not?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so those are your factors irreducible quadratic

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

avnis yes

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

not if you use complex root factoring

OpenStudy (anonymous):

what

OpenStudy (anonymous):

avnis just talking to newton

OpenStudy (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?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

how do i solve the rest of it

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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.

OpenStudy (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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

great job

OpenStudy (anonymous):

you have an extra 1/2 there

OpenStudy (anonymous):

oh yes

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

OpenStudy (anonymous):

ill solve it

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you make a triangle

OpenStudy (anonymous):

what would the side be would one side be 3x

OpenStudy (anonymous):

...

OpenStudy (anonymous):

that is what i got

OpenStudy (anonymous):

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

OpenStudy (anonymous):

What did you get as sin(2x) ?

OpenStudy (anonymous):

2sin(x)cos(x)

OpenStudy (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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

:D

OpenStudy (anonymous):

got it :)

OpenStudy (anonymous):

Woo

OpenStudy (anonymous):

i have learnt alot today :)

OpenStudy (anonymous):

Good to hear. Goodbye.

OpenStudy (anonymous):

thanks :)

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