How can I evaluate the integral using the substitution u=1/x?

[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 I am 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.

 

[Dick]

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

 

[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.

 

[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

 

[Dick]

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

 

[danago]

oh yea ofcourse. We just went through trig subs in class, so that's all I've been thinking of trying now lol

Thanks very much for the help