# integration no.4

• Oct 27th 2007, 01:10 AM
afeasfaerw23231233
integration no.4
• Oct 27th 2007, 05:19 AM
galactus
Just use trig sub.

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

Let $x=sin{\theta}, \;\ dx=cos{\theta}d{\theta}$

Make the subs and it simplifies down to:

$\int{cot^{2}{\theta}}d{\theta}$

$\int{cot^{n}u}du=\frac{-1}{n-1}cot^{n-1}u-\int{cot^{n-2}u}du$

After you integrate cot, then make the sub ${\theta}=sin^{-1}(x)$ to get it back in terms of x.

You're finished.
• Oct 27th 2007, 05:35 AM
qyzren
simple answer: coz you did it wrong.
split it up first: √[1/x^4 + 1/x²] dx
then do the substitution, u = 1/x, dx = -1/u² du

-∫√[u^4 + u²] * 1/u² du

factor out u from inside the square root and cancel one from denominator, you get
-∫√[u²-1]/u du
you can use a trig substitution to solve this.
• Oct 27th 2007, 09:32 AM
afeasfaerw23231233
revise: posting a wrong question
• Oct 27th 2007, 09:55 AM
Krizalid
Does this make sense?