Integral step by step help

  • #1
1. indefinite integral of 1/[(x^4)(sqrt(9x^2 -1))]

I'm looking for step by step help here. I use x=3secy and dx=3secytanydy to convert the top function to integral of: [3secytanydy]/[(tany)(3secy)^4]

I am not sure if this is right/how to get further
 

Answers and Replies

  • #2
34,687
6,394


1. indefinite integral of 1/[(x^4)(sqrt(9x^2 -1))]

I'm looking for step by step help here. I use x=3secy and dx=3secytanydy to convert the top function to integral of: [3secytanydy]/[(tany)(3secy)^4]

I am not sure if this is right/how to get further

I labelled by triangle with hypotenuse = 3x, adjacent side = 1, opp. side = sqrt(9x^2 - 1), with angle t.

So sec t = 3x ==> x = 1/3 sec t
dx = 1/3 sec t * tan t * dt
sqrt(9x^2 -1) = tan t

Then,
[tex]\int \frac{dx}{x^4 \sqrt{9x^2 - 1}} = \int \frac{1/3 sec(t)tan(t)dt}{(1/3)^4 sec^4(t)~tan(t)}[/tex]
Can you continue from there?
 
  • #3


do you get integral of:27*(cost)^3 dt ?
 
  • #5


then you integrate to get 27(sint-(1/3)(sint)^3) and use opp side over hypotenuse to replace with sint?
 
  • #6
34,687
6,394


Yes. After you get the antiderivative, undo your substitution, and you'll be done.
 

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