Question

Can anyone assist with the remaining steps in this trig substitution integral: Integral of 2/x^3(sqrtx^2-1) dx
Ill post where I'm up to. thanx

Answer 1

\[\int\limits_{}\frac{ 2dx }{ x^3\sqrt{x^2-1} }\]

I've solved to this point: =∫1+cos(2Ɵ)dƟ

Answer 2

\[=∫1+\cos(2Ɵ)dƟ\]
=∫1dƟ+∫cos(2Ɵ)dƟ

Answer 3

I am not sure if I would obtain the same result, which substitution did you choose?.

I would say:
\[\Large \tan^2\alpha+1=\sec^2\alpha \]
so this holds true:
\[\Large \sec^2\alpha-1=\tan^2\alpha \]

which looks like a perfect substitution for our integral problem:

let:
\[ \Large x=\sec\alpha\]
such that:
\[\Large dx=\sec\alpha\tan\alpha d\alpha \]

Answer 4

I didn't mean to disagree by the way, just had to carry out the steps myself first.

Answer 5

So it seems like I endup with :\[ \Large 2 \int \cos^2\alpha d\alpha \] if I didn't make any careless mistakes

Answer 6

and your integration will again lead to:
\[\Large \alpha +\sin\alpha\cos\alpha +C\]

Answer 7

Not at all, I did the above and I got to that point. then i thought i had to use the sub 1+cos2Ɵ

Answer 8

that's correct, the two cancels out at this step, so it seems, and you should obtain the same result as I have above, or maybe you have obtained this (double angular expression):

\[\Large \alpha + \frac{1}{2}\sin(2\alpha)+C\]

Answer 9

The directly above is what I got. Again, I forget how to do the triangle part, to back sub for Ɵ :(

Answer 10

alright, again we use the relation I posted above:

\[\Large \alpha +\sin\alpha\cos\alpha+C\]

Because in here it will be easier to back substitute, in the other expression, our result will turn out unpractical and optically displeasing.

Answer 11

We have performed that substitution:

\[\Large x=\sec\alpha=\frac{1}{\cos\alpha} \longrightarrow \cos\alpha=\frac{1}{x}=\frac{ank}{hyp}\]

Answer 12

im not following you how 1/cosƟ goes to just cosƟ...?