How do I solve an indefinite integral with a trigonometric substitution?

[mat331760298]

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

 

[Mark44]

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

 

[mat331760298]

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

 

[Mark44]

Yes.

 

[mat331760298]

then you integrate to get 27(sint-(1/3)(sint)^3) and use opp side over hypotenuse to replace with sint?

 

[Mark44]

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