revise: posting a wrong question

http://img3.freeimagehosting.net/uploads/7808ccf785.gif

http://img3.freeimagehosting.net/uploads/286f2c4a07.jpg

Printable View

- Oct 27th 2007, 12:10 AMafeasfaerw23231233integration no.4
revise: posting a wrong question

http://img3.freeimagehosting.net/uploads/7808ccf785.gif

http://img3.freeimagehosting.net/uploads/286f2c4a07.jpg - Oct 27th 2007, 04:19 AMgalactus
Just use trig sub.

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

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

Make the subs and it simplifies down to:

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

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

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

You're finished. - Oct 27th 2007, 04:35 AMqyzren
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, 08:32 AMafeasfaerw23231233
revise: posting a wrong question

- Oct 27th 2007, 08:55 AMKrizalid
Does this make sense?