# Using the substitution u=1/x

1. Jul 17, 2007

### danago

Using the substitution u=1/x, evaluate:

$$\int {\frac{{dx}}{{x^2 \sqrt {1 - x^2 } }}}$$

I was able to do it making the substitution $$x=cos\theta$$, but im supposed to show a worked solution using the given substitution.

$$\int {\frac{{dx}}{{x^2 \sqrt {1 - x^2 } }}} = \int {\frac{{ - x^2 du}}{{x^2 \sqrt {1 - x^2 } }}} = \int {\frac{{ - du}}{{\sqrt {1 - x^2 } }}}$$

Thats about as far as i was able to get.

Any help?

Thanks,
Dan.

2. Jul 17, 2007

### Dick

Don't stop there. Replace x with 1/u.

3. Jul 17, 2007

### TheoMcCloskey

Hint: if $$u=1/x$$, what is $$du$$ ?

OPPS: I see you already got that.

Now, complete the substitution process in you last integral ( $$x = 1/u$$ ) and simplify. You'll then find that another simple substitution will yield further progress.

Last edited: Jul 17, 2007
4. Jul 17, 2007

### danago

Ahh i think i see what to do. Doing what Dick said, i ended up with:

$$\int {\frac{u}{{\sqrt {u^2 - 1} }}} du$$

From that, it looks like i can make the sub $$u=sec \theta$$, which ill go and try now

5. Jul 17, 2007

### Dick

Easier yet. Try v=u^2-1.

6. Jul 17, 2007

### danago

oh yea ofcourse. We just went through trig subs in class, so thats all ive been thinking of trying now lol :yuck:

Thanks very much for the help