anonymous
  • anonymous
find the intergral of 1/((1+9(x^2))^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
What have you tried?
anonymous
  • anonymous
factor out the 9, youre going to get arctan something
anonymous
  • 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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anonymous
  • anonymous
oh nevermind, i didnt see the square . you can do trig substitution
anonymous
  • anonymous
ohk i can factor out the 9 that seems a good think thanks i just wasn't doing that :| i guess
anonymous
  • anonymous
whoa, where did 4/2 come from?
anonymous
  • 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
  • anonymous
dont use 4/2, thats weird
anonymous
  • anonymous
It is weird. Very weird. Use a tan sub, as above.
anonymous
  • anonymous
if youre using an integral table look for 1 / ( x^2 + a^2)^2 dx
anonymous
  • anonymous
yup i am using a tan sub just sure what to put as a coefficient for tan
anonymous
  • anonymous
Integral tables are for babies.
anonymous
  • anonymous
yes that i what i am following cantroset but in that the a will 1/3
anonymous
  • anonymous
errr you can have 1 / ( ax^2 + b^2) ^2 dx,
anonymous
  • anonymous
i know that formula :P
anonymous
  • anonymous
whats your table say
anonymous
  • anonymous
:) so 1/3 should be my coefficient right :)
anonymous
  • anonymous
lets use trig sub just to be thorough, i use pauls online math course, here one sec i will post link
anonymous
  • anonymous
\[\text{Let } x = \frac{1}{3}\cdot \tan \alpha\]
anonymous
  • anonymous
Paul is the man!! http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx scroll down
anonymous
  • anonymous
yes why is that a 1/3 that is what i am lost about shouldn't it be 1/9
anonymous
  • anonymous
ok one sec, i hate this thing it keeps freezing. yeesh why doesnt google help with the appearance of this
anonymous
  • anonymous
the 1/3 is squared to 1/9
anonymous
  • anonymous
ohk so it is (1+ 1/3 tan (theta))^2
anonymous
  • anonymous
(1+ 1/3 tan (theta)^2)^2
anonymous
  • 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
  • anonymous
\[1+9x^2 = 1+ 9\left(\frac{1}{3}\tan \alpha \right)^2 = 1+\tan^2 \alpha\]
anonymous
  • anonymous
so you want ( 1 + 9x^2)^2 = [ (1+9x^2)^4] ^1/2 ?
anonymous
  • anonymous
i think newton thats simpler,
anonymous
  • anonymous
paul seems to say you want a square root though
anonymous
  • anonymous
oh that makes so much more sense :)
anonymous
  • 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
  • anonymous
xD how on earth does that make more sense. Crazy Americans.
anonymous
  • anonymous
you can also do partial fractions i bet
anonymous
  • anonymous
i mean what you wrote.
anonymous
  • anonymous
yeah i can do partial fraction, but i am required to do trig :)
anonymous
  • anonymous
Oh, good. And you can't split it in to partial fractions (I don't think).
anonymous
  • anonymous
here integrals involving quadratics http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithQuadratics.aspx
anonymous
  • anonymous
why not?
anonymous
  • anonymous
Because it's the same term squared on the bottom and no others.
anonymous
  • anonymous
But assuming you could put it into partial fractions, it would not simplify it at all.
anonymous
  • anonymous
so those are your factors irreducible quadratic
anonymous
  • anonymous
(9x^2+1)^2 = (tan(a)^2+1) = sec(a)^4 is this correct
anonymous
  • anonymous
but probably not the easiest approach. youre tan sub is fine , one sec
anonymous
  • 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
  • anonymous
avnis yes
anonymous
  • anonymous
Now include the change of what you are integrating with respect to and go from there.
anonymous
  • anonymous
ohk so next 1/sec(a)^4 = cos(a)^4
anonymous
  • anonymous
Yes, but\[\frac{\mathbb{d}\alpha}{\mathbb{d}x}\] don't forget to include
anonymous
  • anonymous
not if you use complex root factoring
anonymous
  • anonymous
what
anonymous
  • anonymous
avnis just talking to newton
anonymous
  • 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
  • anonymous
newton, it would be like partial decomp'ing 1/(x^2 + 1) = A/ (x+i) + B/ ( x -i)
anonymous
  • anonymous
i prefer approaches , one size fits all. hehe, the one size hammer that is
anonymous
  • anonymous
how do i solve the rest of it
anonymous
  • 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
  • anonymous
i see which will make a equal to tan^-1(3x)
anonymous
  • anonymous
so i wll be left with 1/3 integral of cos^2(a)
anonymous
  • 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
  • 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
  • anonymous
1/3inetgarl (1/2 (cos 2(a)+1/2) du
anonymous
  • anonymous
great job
anonymous
  • anonymous
you have an extra 1/2 there
anonymous
  • anonymous
oh yes
anonymous
  • 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
  • anonymous
ill solve it
anonymous
  • 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
  • anonymous
you make a triangle
anonymous
  • anonymous
what would the side be would one side be 3x
anonymous
  • anonymous
...
1 Attachment
anonymous
  • anonymous
that is what i got
anonymous
  • anonymous
but there is this online answer from wolfaram and it doesn't match it so i am so lost
anonymous
  • anonymous
What did you get as sin(2x) ?
anonymous
  • anonymous
2sin(x)cos(x)
anonymous
  • 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
  • anonymous
ya that is why i was thinking i don't have x:|
anonymous
  • anonymous
sin(a) =3x/ sqrt (9x^2+1)
anonymous
  • anonymous
Yeah the typing on my diagram should be a, too. Indeed. And cos(a)? And then 2sin(2a) ?
anonymous
  • anonymous
sorry, only sin(2a), not 2 sin(2a)
anonymous
  • anonymous
cos (a)= 1/ sqrt(9x^2+1)
anonymous
  • anonymous
:D
anonymous
  • anonymous
got it :)
anonymous
  • anonymous
Woo
anonymous
  • anonymous
i have learnt alot today :)
anonymous
  • anonymous
Good to hear. Goodbye.
anonymous
  • anonymous
thanks :)

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