Pawanyadav
  • Pawanyadav
Integrate √(4+x^2)/(x^6) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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hartnn
  • hartnn
Have you tried integration by parts? \(u = \sqrt{4+x^2 }, v = x^{-6}\)
Pawanyadav
  • Pawanyadav
Yes, it' don't helping.
hartnn
  • hartnn
hmm... x = 2 tan y ?

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Pawanyadav
  • Pawanyadav
OK checking it ....
Pawanyadav
  • Pawanyadav
Then how we will solve for Denominator.....
hartnn
  • hartnn
with that even i am stuck at cos^3 y/ sin^6 y with some constants outside
Pawanyadav
  • Pawanyadav
Same here..
ganeshie8
  • ganeshie8
\[\dfrac{\cos^3y}{\sin^6y} = \cot^3y\csc^3y = \cot y (\csc^2y-1)\csc^3y\] let \(u =\csc y\)
hartnn
  • hartnn
nice! would it be a challenge to bring x back?
ganeshie8
  • ganeshie8
sure it would be messy haha
Pawanyadav
  • Pawanyadav
I can't reach my answer with it.
ganeshie8
  • ganeshie8
how ?
Pawanyadav
  • Pawanyadav
OK I'm trying.....
Pawanyadav
  • Pawanyadav
The cot y is confusing me....
triciaal
  • triciaal
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mathmale
  • mathmale
Have you considered using trigonometric substitution?
Pawanyadav
  • Pawanyadav
I got another good approach.... First Divide numerator and denominator by x ,we get.. [4/x^2 +1]^1/2=numerator x^5=denominator Now let numerator=t So. 4/x^2 +1= t^2 -8/x^3 dx=2t dt So dx/x^3= -t/4 dt And 1/x^2=(t^2-1)/4 Substitute all these values we get. Integral of t(-2t/8){(t^2-1)/4}dt.., And we can solve it easily....
hartnn
  • hartnn
good one, same question though....would it be easier in this method to get back to x?
Pawanyadav
  • Pawanyadav
Yes,
triciaal
  • triciaal
|dw:1450028258369:dw|
Pawanyadav
  • Pawanyadav
I got my answer with this...
triciaal
  • triciaal
|dw:1450028603288:dw|
triciaal
  • triciaal
what was the correct original?
hartnn
  • hartnn
the first one
triciaal
  • triciaal
@hartnn thanks
Pawanyadav
  • Pawanyadav
Yes ,the first
hartnn
  • hartnn
that approach is not trivial....like you'll have to remember/memorize that for this particular problem we need to do all that...
Pawanyadav
  • Pawanyadav
Yes,.....is their any other way which applied to every question...

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