anonymous
  • anonymous
integrate 1/((x^5)*sqrt(9*x^2-1))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I've gotten pretty far I think, set x= 1/3 sec(U)
anonymous
  • anonymous
then I got to the point where I had \[81 \int\limits_{}^{} (\cos(x))^4 dx\]
anonymous
  • anonymous
the I apply the reduction formula twice and sub back in values for x to get (81/4)*(1/(3*x))^(3)*((sqrt(9 x^2-1))/(3*x))+243/(8)*(1/(3*x))*((sqrt(9 x^2-1))/(3*x))+243

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anonymous
  • anonymous
|dw:1327038086064:dw|
anonymous
  • anonymous
maple and wolfram tell me that there is supposed to be an arctan in there somehow. But I can't figure out where it would get added in.
anonymous
  • anonymous
I can show more steps in between, but what I have there matches maple except for the last term 243/8 is multiplied by -arctan(1/(sqrt(9 x^2-1))) for some reason.
Mr.Math
  • Mr.Math
It seems like if you have made a mistake when substituting back for x.
anonymous
  • anonymous
I did this |dw:1327038548902:dw|
anonymous
  • anonymous
I know I've done it correctly right up until the substitution
anonymous
  • anonymous
|dw:1327038727194:dw| then all I did was substitute the values for sin and cos I found from the triangle
anonymous
  • anonymous
oops that last term should be over 8
Mr.Math
  • Mr.Math
There's more than a way to substitute back, seems to me like wolfarm did this \[\tan{u}=\sqrt{9x^2-1} \implies u=\tan^{-1}{\sqrt{9x^2-1}}.\]
Mr.Math
  • Mr.Math
There should be a term in the form \(ku\) in your integral, k is a constant.
anonymous
  • anonymous
Back in the step where you had \[81\int\limits_{}^{}\cos ^{4}(u) du\] did you consider substituting \[\cos^{2}u = (1 + \cos2u) / 2\] and solving from there?
anonymous
  • anonymous
is the integral |dw:1327039058333:dw|
anonymous
  • anonymous
I think that's where my missing term might be
Mr.Math
  • Mr.Math
Yeah.
anonymous
  • anonymous
ok because I had it set to 1. Then I will do the tan substitution you mentioned before and it should work out
Mr.Math
  • Mr.Math
And this is the term where the arctan came from, because \(u=\tan^{-1}{\sqrt{9x^2-1}}\).
Mr.Math
  • Mr.Math
Exactly.
anonymous
  • anonymous
Yeah, thanks a lot for your help!
Mr.Math
  • Mr.Math
You're welcome.
anonymous
  • anonymous
If you guys are still online I ran into problems with how this was working out.
anonymous
  • anonymous
I somehow need u= -arctan(1/sqrt(9x^2−1))
Mr.Math
  • Mr.Math
Why would you need that?
anonymous
  • anonymous
|dw:1327040345712:dw|
anonymous
  • anonymous
that is what it's telling me I'm missing from the end
anonymous
  • anonymous
|dw:1327040463342:dw|
anonymous
  • anonymous
any ideas where this would come from. It's confusing since my first substitutions look right, but the u one isn't consistent with the same triangle

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